The Calculus Course

1. Distance is the accumulation of velocity over time.
2. On a graph of velocity versus time, the total distance is represented by the area between the curve and the time axis.
3. Therefore, this is the area problem.

Answer: The area problem.

1. The slope between two points is a secant line.
2. As the points merge into one, the line no longer crosses the curve but just touches it.
3. A line that touches a curve at a single point is called a tangent line.

Answer: The tangent line.

1. The formula $\text{length} \times \text{width}$ only works for rectangles.
2. A rectangle has a constant height.
3. The curve $f(x) = x^2$ has a height that changes at every single point along the x axis.
4. Because the height is not constant, there is no single 'width' or 'length' that describes the whole shape.

Answer: Because the height of the shape is constantly changing, so it is not a rectangle.

1. Substitute $-4$ for every $x$ in the expression: $f(-4) = (-4)^{2} + 3(-4) - 5$.
2. Calculate the square: $(-4)^{2} = 16$.
3. Calculate the multiplication: $3(-4) = -12$.
4. Combine the terms: $16 - 12 - 5 = 4 - 5 = -1$.

Answer: -1

1. Identify the operation that could be undefined: division by zero.
2. Set the denominator equal to zero to find the restricted value: $x - 3 = 0$.
3. Solve for $x$: $x = 3$.
4. The function is defined for all real numbers except $x = 3$.

Answer: x \neq 3

1. Find the domain by looking at the x-coordinates of the endpoints: the leftmost is $-1$ and the rightmost is $5$.
2. The domain is $[-1, 5]$.
3. Find the range by looking at the y-coordinates of the endpoints: the lowest is $2$ and the highest is $10$.
4. The range is $[2, 10]$.

Answer: Domain: [-1, 5], Range: [2, 10]

1. Calculate the slope $m = \frac{10 - 4}{2 - (-1)} = \frac{6}{3} = 2$.
2. Substitute $m = 2$ and the point $(2, 10)$ into $y = mx + c$.
3. $10 = 2(2) + c \implies 10 = 4 + c \implies c = 6$.
4. Write the final equation.

Answer: $y = 2x + 6$

1. Find $f(0) = -\frac{1}{2}(0) + 3 = 3$.
2. Find $f(4) = -\frac{1}{2}(4) + 3 = -2 + 3 = 1$.
3. Calculate the average rate of change: $\frac{f(4) - f(0)}{4 - 0} = \frac{1 - 3}{4} = \frac{-2}{4} = -\frac{1}{2}$.

Answer: $-\frac{1}{2}$

1. Start with the equation $y = mx + c$.
2. Substitute $m = 3$, $x = 4$, and $y = -2$.
3. $-2 = 3(4) + c$.
4. $-2 = 12 + c$.
5. Subtract $12$ from both sides to find $c = -14$.

Answer: $-14$

1. Set the function to zero: $x^2 - 5x + 6 = 0$.
2. Factorise the quadratic: $(x - 2)(x - 3) = 0$.
3. Solve for $x$: $x = 2$ and $x = 3$.
4. Identify the degree as 2 (even) and the leading coefficient as 1 (positive), so both ends point upwards.

Answer: Roots: $x=2, 3$; End behavior: both ends point up.

1. Factor out the common term $x$: $f(x) = x(x^2 - 5x + 6)$.
2. Factorise the quadratic part: $f(x) = x(x - 2)(x - 3)$.
3. Set each factor to zero: $x = 0$, $x = 2$, and $x = 3$.
4. Identify the degree as 3 (odd) and the leading coefficient as 1 (positive), so the graph goes down to the left and up to the right.

Answer: Roots: $x=0, 2, 3$; End behavior: down to the left, up to the right.

1. Factor out the common term $x$: $f(x) = x(x^2 + 1)$.
2. Set the factors to zero: $x = 0$ or $x^2 + 1 = 0$.
3. Analyze $x^2 + 1 = 0$: this implies $x^2 = -1$.
4. Since no real number squared is negative, $x^2 + 1$ has no real roots.

Answer: No, the only real root is $x = 0$.

1. The starting amount is $20$ and the growth factor is $2$, so the function is $f(x) = 20 \times 2^x$.
2. Substitute $x = 5$ into the function: $f(5) = 20 \times 2^5$.
3. Calculate $2^5 = 32$.
4. Multiply $20 \times 32 = 640$.

Answer: 640

1. Continuous growth at $100\%$ is modeled by the base $e$.
2. The function is $f(t) = 1 \times e^t$.
3. Substitute $t = 3$: $f(3) = e^3$.
4. Calculate $2.718 \times 2.718 \times 2.718 \approx 20.079$.

Answer: 20.079

1. Evaluate $g(4) = 4^3 = 4 \times 4 \times 4 = 64$.
2. Evaluate $h(4) = 3^4 = 3 \times 3 \times 3 \times 3 = 81$.
3. Compare the two: $81 > 64$.
4. Find the difference: $81 - 64 = 17$.

Answer: h(x) is larger by 17

1. Take the natural logarithm of both sides: $\ln(3^x) = \ln(20)$
2. Apply the power law to bring $x$ to the front: $x \ln(3) = \ln(20)$
3. Divide both sides by $\ln(3)$ to isolate $x$: $x = \frac{\ln(20)}{\ln(3)}$

Answer: $x = \frac{\ln(20)}{\ln(3)}$

1. Use the product law: $\log_5(125) + \log_5(5^2)$
2. Evaluate $\log_5(125)$ since $5^3 = 125$, so $\log_5(125) = 3$
3. Use the power law for the second term: $2 \log_5(5) = 2 \cdot 1 = 2$
4. Add the results: $3 + 2 = 5$

Answer: 5

1. Take the natural logarithm of both sides: $\ln(e^{2x}) = \ln(7)$
2. Since $\ln$ and $e$ are inverses, $\ln(e^{2x}) = 2x$
3. The equation becomes $2x = \ln(7)$
4. Divide by 2: $x = \frac{\ln(7)}{2}$

Answer: $x = \frac{\ln(7)}{2}$

1. The angle $\frac{5\pi}{6}$ is in the second quadrant because it is between $\frac{\pi}{2}$ and $\pi$.
2. The reference angle is $\pi - \frac{5\pi}{6} = \frac{\pi}{6}$ (or $30^\circ$).
3. For $\frac{\pi}{6}$, the coordinates are $(\frac{\sqrt{3}}{2}, \frac{1}{2})$.
4. In the second quadrant, $x$ is negative and $y$ is positive, so the point is $(-\frac{\sqrt{3}}{2}, \frac{1}{2})$.
5. Therefore, $\sin(\frac{5\pi}{6}) = \frac{1}{2}$, $\cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2}$, and $\tan(\frac{5\pi}{6}) = \frac{1/2}{-\sqrt{3}/2} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$.

Answer: \sin(\frac{5\pi}{6}) = \frac{1}{2}, \cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2}, \tan(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{3}

1. To convert to radians, multiply by $\frac{\pi}{180}$: $225 \times \frac{\pi}{180} = \frac{225\pi}{180} = \frac{5\pi}{4}$.
2. The angle $\frac{5\pi}{4}$ is in the third quadrant.
3. The reference angle is $\frac{5\pi}{4} - \pi = \frac{\pi}{4}$.
4. The cosine of $\frac{\pi}{4}$ is $\frac{\sqrt{2}}{2}$.
5. In the third quadrant, cosine is negative, so $\cos(225^\circ) = -\frac{\sqrt{2}}{2}$.

Answer: \frac{5\pi}{4} \text{ radians}, \cos(225^\circ) = -\frac{\sqrt{2}}{2}

1. Use the identity $\sin^2(\theta) + \cos^2(\theta) = 1$.
2. Substitute the known value: $(\frac{3}{5})^2 + \cos^2(\theta) = 1 \implies \frac{9}{25} + \cos^2(\theta) = 1$.
3. Solve for $\cos^2(\theta)$: $\cos^2(\theta) = 1 - \frac{9}{25} = \frac{16}{25}$.
4. Take the square root: $\cos(\theta) = \pm \frac{4}{5}$.
5. Since $\theta$ is in the second quadrant, $\cos(\theta)$ must be negative, so $\cos(\theta) = -\frac{4}{5}$.
6. Find tangent: $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{3/5}{-4/5} = -\frac{3}{4}$.

Answer: \cos(\theta) = -\frac{4}{5}, \tan(\theta) = -\frac{3}{4}

1. We look for an angle $\theta$ such that $\cos(\theta) = -\frac{\sqrt{2}}{2}$ within the range $[0, \pi]$.
2. The reference angle for $\frac{\sqrt{2}}{2}$ is $\frac{\pi}{4}$.
3. Since the cosine is negative, the angle must be in the second quadrant.
4. The angle is $\pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

Answer: \frac{3\pi}{4}

1. We look for an angle $\theta$ such that $\tan(\theta) = -1$ within the range $(-\frac{\pi}{2}, \frac{\pi}{2})$.
2. We know $\tan(\frac{\pi}{4}) = 1$.
3. Since tangent is an odd function, $\tan(-\frac{\pi}{4}) = -1$.
4. The angle $-\frac{\pi}{4}$ is within the required range.

Answer: -\frac{\pi}{4}

1. First, evaluate the inner part: $\sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2}$.
2. Now we need to find $\arcsin(\frac{\sqrt{3}}{2})$.
3. We look for an angle $\theta$ in the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$ such that $\sin(\theta) = \frac{\sqrt{3}}{2}$.
4. The angle $\frac{\pi}{3}$ satisfies this and is within the range.

Answer: \frac{\pi}{3}

1. Identify the inner function $g(x) = 3x - 2$.
2. Substitute $g(x)$ into $f(x)$: $f(g(x)) = (3x - 2)^2$.
3. Expand the square: $(3x)^2 - 2(3x)(2) + 2^2$.
4. Simplify the terms: $9x^2 - 12x + 4$.

Answer: $f(g(x)) = 9x^2 - 12x + 4$

1. The multiplier $3$ outside the function indicates a vertical stretch by a factor of $3$.
2. The term $-\frac{\pi}{4}$ inside the function indicates a horizontal shift to the right by $\frac{\pi}{4}$ units.
3. The constant $+1$ at the end indicates a vertical shift upwards by $1$ unit.

Answer: Vertical stretch by 3, horizontal shift right by $\frac{\pi}{4}$, and vertical shift up by 1.

1. The composite function is $f(g(x)) = \frac{1}{\sin(x)}$.
2. The inner function $g(x) = \sin(x)$ is defined for all real numbers.
3. The outer function $f(u) = \frac{1}{u}$ is undefined where $u = 0$. Thus, we need $\sin(x) \neq 0$.
4. In the interval $[0, \pi]$, $\sin(x) = 0$ at $x = 0$ and $x = \pi$.

Answer: $x \neq 0, \pi$

1. Add $7$ to both sides to get $3x \ge 21$.
2. Divide both sides by $3$ to get $x \ge 7$.
3. Since $x$ is greater than or equal to $7$, the interval starts at $7$ and goes to infinity.
4. Use a square bracket for $7$ because it is included.

Answer: [7, \infty)

1. Set up two cases: $2x + 5 = 11$ and $2x + 5 = -11$.
2. For the first case, subtract $5$ to get $2x = 6$, then divide by $2$ to get $x = 3$.
3. For the second case, subtract $5$ to get $2x = -16$, then divide by $2$ to get $x = -8$.
4. Both values satisfy the original equation.

Answer: x = 3, x = -8

1. The distance from $x$ to $3$ must be greater than $2$.
2. This means $x - 3 > 2$ or $x - 3 < -2$.
3. Solving the first part: $x > 5$, which is $(5, \infty)$.
4. Solving the second part: $x < 1$, which is $(-\infty, 1)$.
5. Combine these two disjoint sets using the union symbol.

Answer: (-\infty, 1) \cup (5, \infty)

1. Step 1: Plug in values close to $1$ from the left, such as $0.9$ and $0.99$. $f(0.9) = 0.81 + 2 = 2.81$ and $f(0.99) = 0.9801 + 2 = 2.9801$.
2. Step 2: Plug in values close to $1$ from the right, such as $1.1$ and $1.01$. $f(1.1) = 1.21 + 2 = 3.21$ and $f(1.01) = 1.0201 + 2 = 3.0201$.
3. Step 3: Observe that both sides are approaching the value $3$.

Answer: 3

1. Step 1: Notice that plugging in $x = 4$ results in $0/0$, which is undefined.
2. Step 2: Factor the numerator as a difference of squares: $x^2 - 16 = (x - 4)(x + 4)$.
3. Step 3: Simplify the function: $f(x) = \frac{(x - 4)(x + 4)}{x - 4} = x + 4$ for $x \neq 4$.
4. Step 4: As $x$ approaches $4$, the expression $x + 4$ approaches $4 + 4 = 8$.

Answer: 8

1. Step 1: Approach $x = 2$ from the left. For all $x < 2$, $f(x) = 5$, so the left-hand limit is $5$.
2. Step 2: Approach $x = 2$ from the right. For all $x \ge 2$, $f(x) = 10$, so the right-hand limit is $10$.
3. Step 3: Compare the two sides. Since $5 \neq 10$, the function does not approach a single value.

Answer: No, the limit does not exist because the left-hand and right-hand limits are different.

1. Factorise the numerator: $x^2 + 5x + 6 = (x + 2)(x + 3)$
2. Rewrite the limit: $\lim_{x \to -2} \frac{(x + 2)(x + 3)}{x + 2}$
3. Cancel the common factor $(x + 2)$ for $x \neq -2$: $\lim_{x \to -2} (x + 3)$
4. Substitute $x = -2$ into the simplified expression: $-2 + 3 = 1$

Answer: 1

1. Split the absolute value into two cases: $3x + 1 > 7$ or $3x + 1 < -7$
2. Solve the first case: $3x > 6 \implies x > 2$
3. Solve the second case: $3x < -8 \implies x < -8/3$
4. Combine the results into interval notation: $(-\infty, -8/3) \cup (2, \infty)$

Answer: (-\infty, -8/3) \cup (2, \infty)

1. Substitute $g(x)$ into $f(x)$: $(f \circ g)(x) = f(2x - 5) = e^{2x - 5}$
2. Substitute $x = 3$ into the composite function: $e^{2(3) - 5}$
3. Simplify the exponent: $e^{6 - 5} = e^1$
4. The final value is $e$

Answer: e

1. Pick values approaching 3 from the left: $x = 2.9, 2.99, 2.999$.
2. Calculate $f(2.9) = \frac{8.41 - 9}{2.9 - 3} = \frac{-0.59}{-0.1} = 5.9$.
3. Calculate $f(2.99) = \frac{8.9401 - 9}{2.99 - 3} = \frac{-0.0599}{-0.01} = 5.99$.
4. Pick values approaching 3 from the right: $x = 3.1, 3.01, 3.001$.
5. Calculate $f(3.1) = \frac{9.61 - 9}{3.1 - 3} = \frac{0.61}{0.1} = 6.1$.
6. Calculate $f(3.01) = \frac{9.0601 - 9}{3.01 - 3} = \frac{0.0601}{0.01} = 6.01$.
7. Observe that both sides approach 6.

Answer: 6

1. Find the left-hand limit: $\lim_{x \to 1^{-}} (x + 5) = 1 + 5 = 6$.
2. Find the right-hand limit: $\lim_{x \to 1^{+}} (2x + 4) = 2(1) + 4 = 6$.
3. Since the left-hand limit equals the right-hand limit, the general limit exists.

Answer: 6

1. Approach 0 from the right: as $x \to 0^{+}$, $1/x$ becomes very large, so $\lim_{x \to 0^{+}} \frac{1}{x} = \infty$.
2. Approach 0 from the left: as $x \to 0^{-}$, $1/x$ becomes very large and negative, so $\lim_{x \to 0^{-}} \frac{1}{x} = -\infty$.
3. Since the left-hand and right-hand limits are not equal (and are not finite), the limit does not exist.

Answer: Does Not Exist

1. Compute the left-hand limit: $\lim_{x \to -1^{-}} x^2 = (-1)^2 = 1$.
2. Compute the right-hand limit: $\lim_{x \to -1^{+}} (-x) = -(-1) = 1$.
3. Since the left-hand limit equals the right-hand limit, the overall limit exists.

Answer: 1

1. As $x$ approaches $2$ from the right, $(x-2)^2$ is a very small positive number, so $1/(x-2)^2$ goes to $\infty$.
2. As $x$ approaches $2$ from the left, $(x-2)^2$ is also a very small positive number, so $1/(x-2)^2$ goes to $\infty$.
3. Because the function is unbounded, it does not approach a finite real number.

Answer: DNE (Unbounded)

1. For $x > -3$, $|x+3| = x+3$, so $f(x) = (x+3)/(x+3) = 1$. The right-hand limit is $1$.
2. For $x < -3$, $|x+3| = -(x+3)$, so $f(x) = -(x+3)/(x+3) = -1$. The left-hand limit is $-1$.
3. Since $1 \neq -1$, the limit does not exist.

Answer: DNE (Jump)

1. Substitute $x = 5$ to check for indeterminate form: $\frac{5^2 - 25}{5 - 5} = \frac{0}{0}$.
2. Factor the numerator as a difference of squares: $x^2 - 25 = (x - 5)(x + 5)$.
3. Cancel the common factor $(x - 5)$ from the top and bottom: $\lim_{x \to 5} (x + 5)$.
4. Use direct substitution: $5 + 5 = 10$.

Answer: 10

1. Recognize that the function is a polynomial, so we can use direct substitution.
2. Substitute $x = 2$ into the expression: $4(2)^3 - 2(2) + 7$.
3. Calculate the powers: $4(8) - 4 + 7$.
4. Perform the multiplication and addition: $32 - 4 + 7 = 35$.

Answer: 35

1. Substitute $x = 1$ to check: $\frac{1 + 2 - 3}{1 - 1} = \frac{0}{0}$.
2. Factor the numerator: $x^2 + 2x - 3 = (x - 1)(x + 3)$.
3. Factor the denominator: $x^2 - 1 = (x - 1)(x + 1)$.
4. Cancel the common factor $(x - 1)$: $\lim_{x \to 1} \frac{x + 3}{x + 1}$.
5. Substitute $x = 1$: $\frac{1 + 3}{1 + 1} = \frac{4}{2} = 2$.

Answer: 2

1. Start with the known bound for sine: $-1 \le \sin\left(\frac{5}{x}\right) \le 1$.
2. Multiply the entire inequality by $x^2$. Since $x^2 \ge 0$ for all $x$, the inequality signs remain the same: $-x^2 \le x^2 \sin\left(\frac{5}{x}\right) \le x^2$.
3. Evaluate the limit of the lower bound: $\lim_{x \to 0} -x^2 = 0$.
4. Evaluate the limit of the upper bound: $\lim_{x \to 0} x^2 = 0$.
5. Since both bounds approach $0$, by the Squeeze Theorem, the middle limit must also be $0$.

Answer: 0

1. Identify the lower bound function $g(x) = 2x - 1$ and the upper bound function $h(x) = x^2 + 1$.
2. Calculate the limit of the lower bound as $x$ approaches $1$: $\lim_{x \to 1} (2x - 1) = 2(1) - 1 = 1$.
3. Calculate the limit of the upper bound as $x$ approaches $1$: $\lim_{x \to 1} (x^2 + 1) = 1^2 + 1 = 2$.
4. Observe that the limits are $1$ and $2$. Since $1 \neq 2$, the Squeeze Theorem cannot be used to find a specific value for the limit.

Answer: The limit cannot be determined using the Squeeze Theorem because the bounds do not approach the same value.

1. We want the expression inside the sine to match the denominator. We have $3x$ inside the sine, but only $x$ in the denominator.
2. Multiply the expression by $\frac{3}{3}$ to manipulate the denominator: $\lim_{x \to 0} 3 \cdot \frac{\sin(3x)}{3x}$.
3. Use the constant multiple law to move the $3$ outside the limit: $3 \cdot \lim_{x \to 0} \frac{\sin(3x)}{3x}$.
4. As $x$ approaches $0$, the term $3x$ also approaches $0$. Therefore, $\lim_{x \to 0} \frac{\sin(3x)}{3x} = 1$ based on the fundamental limit proven today.
5. The final result is $3 \cdot 1 = 3$.

Answer: 3

1. Identify the highest power of $x$ in the denominator, which is $x^2$.
2. Divide every term in the numerator and denominator by $x^2$: $\frac{6 - 3/x + 1/x^2}{2 + 5/x^2}$.
3. Apply the limit as $x \to \infty$, noting that $3/x$, $1/x^2$, and $5/x^2$ all approach $0$.
4. The expression simplifies to $6/2 = 3$.

Answer: 3

1. Compare the degrees of the numerator and denominator. The numerator degree is $1$ and the denominator degree is $2$.
2. Since the degree of the denominator is higher, the limit as $x \to \infty$ is $0$.
3. Similarly, the limit as $x \to {-\infty}$ is $0$.
4. The horizontal asymptote is the line $y = 0$.

Answer: y = 0

1. Divide all terms by the highest power in the denominator, $x^3$.
2. The expression becomes $\frac{2 - 4/x^2}{1 + 7/x + 1/x^3}$.
3. As $x \to \infty$, the terms $4/x^2$, $7/x$, and $1/x^3$ all approach $0$.
4. The limit is $2/1 = 2$.

Answer: 2

1. Set the denominator to zero: $x + 5 = 0$, which gives $x = -5$.
2. Since the numerator is a non-zero constant, $x = -5$ is a vertical asymptote.
3. Test the right-hand limit: for $x = -4.9$, $f(-4.9) = \frac{4}{0.1} = 40$ (positive), so $\lim_{x \to -5^+} f(x) = \infty$.
4. Test the left-hand limit: for $x = -5.1$, $f(-5.1) = \frac{4}{-0.1} = -40$ (negative), so $\lim_{x \to -5^-} f(x) = -\infty$.

Answer: Vertical asymptote at $x = -5$; $\lim_{x \to -5^+} f(x) = \infty$ and $\lim_{x \to -5^-} f(x) = -\infty$.

1. The denominator is zero when $(x-2)^2 = 0$, which means $x = 2$.
2. The numerator is $1$, so $x = 2$ is a vertical asymptote.
3. Test the right-hand limit: for $x = 2.1$, $(2.1-2)^2 = 0.01$, and $1/0.01 = 100$ (positive), so $\lim_{x \to 2^+} f(x) = \infty$.
4. Test the left-hand limit: for $x = 1.9$, $(1.9-2)^2 = (-0.1)^2 = 0.01$, and $1/0.01 = 100$ (positive), so $\lim_{x \to 2^-} f(x) = \infty$.

Answer: Vertical asymptote at $x = 2$; $\lim_{x \to 2^+} f(x) = \infty$ and $\lim_{x \to 2^-} f(x) = \infty$.

1. Factor the denominator: $x^2 - 5x + 6 = (x-2)(x-3)$.
2. The function is $f(x) = \frac{x-3}{(x-2)(x-3)}$.
3. The value $x = 3$ makes both numerator and denominator zero, so it is a hole.
4. The value $x = 2$ makes only the denominator zero, so it is a vertical asymptote.
5. Simplify the function to $f(x) = \frac{1}{x-2}$ for $x \neq 3$ to confirm the behavior at $x=2$.

Answer: The only vertical asymptote is $x = 2$.

1. Find the function value: $f(2) = 2^2 = 4$.
2. Find the left-hand limit: $\lim_{x \to 2^-} x^2 = 4$.
3. Find the right-hand limit: $\lim_{x \to 2^+} 2x = 2(2) = 4$.
4. Since $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) = 4$, the function is continuous.

Answer: Continuous

1. Check the function value: $f(3) = \frac{0}{0}$, which is undefined.
2. Compute the limit: $\lim_{x \to 3} \frac{x-3}{(x-3)(x+3)} = \lim_{x \to 3} \frac{1}{x+3} = \frac{1}{6}$.
3. Since the limit exists as a finite number but the function is undefined, it is a removable discontinuity.

Answer: Removable Discontinuity

1. Check the function value: $f(5) = \frac{1}{0}$, which is undefined.
2. Check the right-hand limit: $\lim_{x \to 5^+} \frac{1}{x-5} = \infty$.
3. Check the left-hand limit: $\lim_{x \to 5^-} \frac{1}{x-5} = -\infty$.
4. Since the limits are infinite, this is an infinite discontinuity.

Answer: Infinite Discontinuity

1. Step 1: Note that $f(x)$ is a polynomial, so it is continuous on the closed interval $[1, 2]$.
2. Step 2: Evaluate the function at the left endpoint: $f(1) = 1^3 + 2(1) - 5 = 1 + 2 - 5 = -2$.
3. Step 3: Evaluate the function at the right endpoint: $f(2) = 2^3 + 2(2) - 5 = 8 + 4 - 5 = 7$.
4. Step 4: Since $f(1) = -2$ and $f(2) = 7$, and $0$ is between $-2$ and $7$, the Intermediate Value Theorem guarantees there is at least one $c$ in $(1, 2)$ such that $f(c) = 0$.

Answer: A root exists because $f(1) = -2$ and $f(2) = 7$ and the function is continuous.

1. Step 1: Define a new function $g(x) = \tan(x) - 2x$. We want to find where $g(c) = 0$.
2. Step 2: Check continuity. $\tan(x)$ is continuous on $[1, 1.5]$ because the first vertical asymptote is at $\pi/2 \approx 1.57$, which is outside this interval.
3. Step 3: Evaluate $g(1) = \tan(1) - 2(1) \approx 1.557 - 2 = -0.443$.
4. Step 4: Evaluate $g(1.5) = \tan(1.5) - 2(1.5) \approx 14.101 - 3 = 11.101$.
5. Step 5: Since $g(1) < 0$ and $g(1.5) > 0$, the IVT guarantees a root in $(1, 1.5)$.

Answer: A root exists in $(1, 1.5)$ because $g(1) \approx -0.443$ and $g(1.5) \approx 11.101$ and the function is continuous.

1. Step 1: $f(x)$ is continuous on $[0, 2]$ as it is the difference of an exponential and a linear function.
2. Step 2: Evaluate at $x=0$: $f(0) = e^0 - 3(0) = 1 - 0 = 1$ (positive).
3. Step 3: Evaluate at $x=1$: $f(1) = e^1 - 3(1) \approx 2.718 - 3 = -0.282$ (negative).
4. Step 4: Evaluate at $x=2$: $f(2) = e^2 - 3(2) \approx 7.389 - 6 = 1.389$ (positive).
5. Step 5: There is a sign change between $x=0$ and $x=1$, so there is a root in $(0, 1)$.
6. Step 6: There is another sign change between $x=1$ and $x=2$, so there is a second root in $(1, 2)$.

Answer: Roots exist in $(0, 1)$ and $(1, 2)$ because the function values change signs twice.

1. Start with the output inequality: $|(4x - 5) - 7| < \epsilon$.
2. Simplify the expression: $|4x - 12| < \epsilon$.
3. Factor out the constant: $4|x - 3| < \epsilon$.
4. Isolate the input distance: $|x - 3| < \frac{\epsilon}{4}$.
5. Choose $\delta = \frac{\epsilon}{4}$. If $0 < |x - 3| < \delta$, then $4|x - 3| < \epsilon$, which implies $|(4x - 5) - 7| < \epsilon$.

Answer: \delta = \frac{\epsilon}{4}

1. Set up the inequality: $|(2x + 8) - 6| < \epsilon$.
2. Simplify: $|2x + 2| < \epsilon$.
3. Factor: $2|x + 1| < \epsilon$.
4. Solve for the distance from $x$ to $-1$: $|x - (-1)| < \frac{\epsilon}{2}$.
5. Choose $\delta = \frac{\epsilon}{2}$.

Answer: \delta = \frac{\epsilon}{2}

1. Start with $|(-5x + 2) - L| < \epsilon$.
2. Since $L = -5a + 2$, substitute this in: $|-5x + 2 - (-5a + 2)| < \epsilon$.
3. Simplify: $|-5x + 5a| < \epsilon$.
4. Factor out $-5$: $|-5(x - a)| < \epsilon$.
5. Use absolute value properties: $5|x - a| < \epsilon$.
6. Divide by 5: $|x - a| < \frac{\epsilon}{5}$.

Answer: \delta = \frac{\epsilon}{5}

1. Start with $|(3x + 2) - 14| < \epsilon$.
2. Simplify to $|3x - 12| < \epsilon$, which is $3|x - 4| < \epsilon$.
3. This suggests $|x - 4| < \epsilon/3$.
4. Choose $\delta = \epsilon/3$.
5. Verify: if $|x - 4| < \delta$, then $3|x - 4| < 3(\epsilon/3) = \epsilon$, so $|(3x + 2) - 14| < \epsilon$.

Answer: \delta = \frac{\epsilon}{3}

1. Start with $|x^2 - 9| = |x - 3||x + 3| < \epsilon$.
2. Assume $|x - 3| < 1$, so $2 < x < 4$.
3. Then $|x + 3| < 4 + 3 = 7$.
4. We need $|x - 3| \cdot 7 < \epsilon$, so $|x - 3| < \epsilon/7$.
5. Choose $\delta = \min(1, \epsilon/7)$.

Answer: \delta = \min(1, \frac{\epsilon}{7})

1. Start with $|(x^2 + 2x) - 3| = |(x - 1)(x + 3)| < \epsilon$.
2. Assume $|x - 1| < 1$, so $0 < x < 2$.
3. Then $|x + 3| < 2 + 3 = 5$.
4. We need $|x - 1| \cdot 5 < \epsilon$, so $|x - 1| < \epsilon/5$.
5. Choose $\delta = \min(1, \epsilon/5)$.

Answer: \delta = \min(1, \frac{\epsilon}{5})

1. Let $m = \frac{n}{3}$, which implies $n = 3m$.
2. As $n \to \infty$, $m \to \infty$.
3. Rewrite the limit as $\lim_{m \to \infty} (1 + \frac{1}{m})^{3m}$.
4. Apply the power rule for limits: $[\lim_{m \to \infty} (1 + \frac{1}{m})^m]^3$.
5. Substitute the definition of $e$: $e^3$.

Answer: e^3

1. Let $m = 2n$.
2. As $n \to \infty$, $m \to \infty$.
3. Substitute $m$ into the expression: $\lim_{m \to \infty} (1 + \frac{1}{m})^m$.
4. This is the exact definition of $e$.

Answer: e

1. Rewrite the expression as $(1 + \frac{-2}{n})^n$.
2. Let $m = \frac{n}{-2}$, so $n = -2m$.
3. As $n \to \infty$, $m \to -\infty$.
4. The limit becomes $\lim_{m \to -\infty} (1 + \frac{1}{m})^{-2m}$.
5. This is $[\lim_{m \to -\infty} (1 + \frac{1}{m})^m]^{-2}$.
6. Substitute $e$: $e^{-2}$ or $\frac{1}{e^2}$.

Answer: e^{-2}

1. Identify that the argument of sine is $7x$ but the denominator is $2x$.
2. Multiply the numerator and denominator by $7$ to match the argument: $\frac{7}{2} \cdot \frac{\sin(7x)}{7x}$.
3. Apply the limit $\lim_{u \to 0} \frac{\sin(u)}{u} = 1$ where $u = 7x$.
4. The result is $\frac{7}{2} \cdot 1 = 3.5$.

Answer: 3.5

1. Identify that the argument of cosine is $4x$.
2. To use the formula $\lim_{u \to 0} \frac{1 - \cos(u)}{u} = 0$, we need $4x$ in the denominator.
3. Multiply the top and bottom by $4$: $4 \cdot \frac{1 - \cos(4x)}{4x}$.
4. The limit of the fraction part is $0$, so $4 \cdot 0 = 0$.

Answer: 0

1. Split the limit into two separate fractions: $\lim_{x \to 0} \frac{\sin(3x)}{x} + \lim_{x \to 0} \frac{1 - \cos(x)}{x}$.
2. For the first part, multiply by $3/3$ to get $3 \cdot \lim_{x \to 0} \frac{\sin(3x)}{3x} = 3 \cdot 1 = 3$.
3. For the second part, use the known limit $\lim_{x \to 0} \frac{1 - \cos(x)}{x} = 0$.
4. Add the results: $3 + 0 = 3$.

Answer: 3

1. Plugging in $x = -2$ gives $\frac{(-2)^2 + 5(-2) + 6}{-2 + 2} = \frac{4 - 10 + 6}{0} = \frac{0}{0}$, which is an indeterminate form.
2. Factor the numerator: $x^2 + 5x + 6 = (x + 2)(x + 3)$.
3. Rewrite the limit: $\lim_{x \to -2} \frac{(x + 2)(x + 3)}{x + 2}$.
4. Cancel the common factor $(x + 2)$ from the top and bottom: $\lim_{x \to -2} (x + 3)$.
5. Substitute $x = -2$ into the simplified expression: $-2 + 3 = 1$.

Answer: 1

1. Identify the highest power of $x$ in the denominator, which is $x^2$.
2. Divide every term by $x^2$: $\lim_{x \to \infty} \frac{3 - \frac{2}{x} + \frac{1}{x^2}}{5 + \frac{7}{x^2}}$.
3. Apply the limit laws: $\frac{2}{x}$, $\frac{1}{x^2}$, and $\frac{7}{x^2}$ all approach $0$ as $x$ approaches infinity.
4. The expression becomes $\frac{3 - 0 + 0}{5 + 0} = \frac{3}{5}$.

Answer: \frac{3}{5}

1. Check the limit from the right ($x > 0$): $\lim_{x \to 0^+} \frac{x}{x(x + 1)} = \lim_{x \to 0^+} \frac{1}{x + 1} = 1$.
2. Check the limit from the left ($x < 0$): $\lim_{x \to 0^-} \frac{-x}{x(x + 1)} = \lim_{x \to 0^-} \frac{-1}{x + 1} = -1$.
3. Since the left-hand limit ($-1$) and right-hand limit ($1$) are not equal, the two-sided limit does not exist.

Answer: \text{Does Not Exist}

1. Substitute $x = 4$ to get $\frac{16-16}{4-4} = \frac{0}{0}$, an indeterminate form.
2. Factor the numerator as a difference of squares: $x^2 - 16 = (x - 4)(x + 4)$.
3. Cancel the common factor $(x - 4)$ from the top and bottom.
4. Evaluate the remaining limit: $\lim_{x \to 4} (x + 4) = 4 + 4 = 8$.

Answer: 8

1. Identify the form $\frac{\sin(ax)}{bx}$.
2. Rewrite the expression to isolate the $\frac{\sin(7x)}{7x}$ form: $\frac{7}{2} \cdot \frac{\sin(7x)}{7x}$.
3. Apply the special limit $\lim_{u \to 0} \frac{\sin u}{u} = 1$ where $u = 7x$.
4. The result is $\frac{7}{2} \cdot 1 = 3.5$.

Answer: 7/2

1. Identify the highest power of $x$ in the denominator, which is $x^2$.
2. Divide every term by $x^2$: $\frac{3 - 5/x^2}{2 + 1/x + 1/x^2}$.
3. Take the limit as $x \to \infty$, noting that $5/x^2$, $1/x$, and $1/x^2$ all approach $0$.
4. The limit is $\frac{3 - 0}{2 + 0 + 0} = \frac{3}{2}$.

Answer: y = 3/2

1. Set up the limit: $\lim_{h \to 0} \frac{(1+h)^2 + 2(1+h) - (1^2 + 2(1))}{h}$
2. Expand the numerator: $\lim_{h \to 0} \frac{1 + 2h + h^2 + 2 + 2h - 3}{h}$
3. Simplify the numerator: $\lim_{h \to 0} \frac{4h + h^2}{h}$
4. Factor and cancel $h$: $\lim_{h \to 0} (4 + h)$
5. Evaluate the limit as $h \to 0$: $4 + 0 = 4$

Answer: 4

1. Set up the limit: $\lim_{h \to 0} \frac{((2+h)^2 - 3(2+h)) - (2^2 - 3(2))}{h}$
2. Expand the terms: $\lim_{h \to 0} \frac{4 + 4h + h^2 - 6 - 3h - (4 - 6)}{h}$
3. Simplify the numerator: $\lim_{h \to 0} \frac{h^2 + h - 2 - (-2)}{h} = \lim_{h \to 0} \frac{h^2 + h}{h}$
4. Factor and cancel $h$: $\lim_{h \to 0} (h + 1)$
5. Evaluate the limit: $0 + 1 = 1$

Answer: 1

1. Set up the limit: $\lim_{h \to 0} \frac{3(-1+h)^2 - 5 - (3(-1)^2 - 5)}{h}$
2. Expand the squared term: $\lim_{h \to 0} \frac{3(1 - 2h + h^2) - 5 - (3 - 5)}{h}$
3. Simplify the numerator: $\lim_{h \to 0} \frac{3 - 6h + 3h^2 - 5 - (-2)}{h}$
4. Combine constants: $\lim_{h \to 0} \frac{-6h + 3h^2}{h}$
5. Factor and cancel $h$: $\lim_{h \to 0} (-6 + 3h)$
6. Evaluate the limit: $-6$

Answer: -6

1. Set up the limit: $v(1) = \lim_{h \to 0} \frac{s(1+h) - s(1)}{h}$
2. Calculate $s(1) = 3(1)^2 - 2(1) = 1$
3. Calculate $s(1+h) = 3(1+h)^2 - 2(1+h) = 3(1 + 2h + h^2) - 2 - 2h = 3 + 6h + 3h^2 - 2 - 2h = 1 + 4h + 3h^2$
4. Substitute into the limit: $\lim_{h \to 0} \frac{(1 + 4h + 3h^2) - 1}{h} = \lim_{h \to 0} \frac{4h + 3h^2}{h}$
5. Simplify: $\lim_{h \to 0} (4 + 3h) = 4$

Answer: 4 \text{ m/s}

1. Set up the limit for the derivative at $x=10$: $C'(10) = \lim_{h \to 0} \frac{C(10+h) - C(10)}{h}$
2. Calculate $C(10) = 0.5(100) + 10(10) + 100 = 50 + 100 + 100 = 250$
3. Calculate $C(10+h) = 0.5(10+h)^2 + 10(10+h) + 100 = 0.5(100 + 20h + h^2) + 100 + 10h + 100 = 50 + 10h + 0.5h^2 + 200 + 10h = 250 + 20h + 0.5h^2$
4. Substitute into the limit: $\lim_{h \to 0} \frac{(250 + 20h + 0.5h^2) - 250}{h} = \lim_{h \to 0} \frac{20h + 0.5h^2}{h}$
5. Simplify: $\lim_{h \to 0} (20 + 0.5h) = 20$

Answer: 20 \text{ dollars per unit}

1. Find the general derivative $s'(t)$: $\lim_{h \to 0} \frac{2(t+h)^2 - 2t^2}{h}$
2. Expand $2(t+h)^2 = 2(t^2 + 2th + h^2) = 2t^2 + 4th + 2h^2$
3. Simplify the limit: $\lim_{h \to 0} \frac{4th + 2h^2}{h} = \lim_{h \to 0} (4t + 2h) = 4t$
4. Set the velocity equal to 12: $4t = 12$
5. Solve for $t$: $t = 3$

Answer: t = 3 \text{ seconds}

1. Check continuity: $f(0) = 0^2 + 1 = 1$. The limit from the right is $0 + 1 = 1$. It is continuous.
2. Find the left-hand derivative: $\lim_{h \to 0^-} \frac{(0+h)^2+1 - (0^2+1)}{h} = \lim_{h \to 0^-} \frac{h^2}{h} = 0$.
3. Find the right-hand derivative: $\lim_{h \to 0^+} \frac{(0+h)+1 - (0+1)}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1$.
4. Compare the slopes: $0 \neq 1$.
5. Since the one-sided derivatives are different, the function is not differentiable at $x = 0$.

Answer: Not differentiable at $x = 0$ (it has a corner).

1. Check continuity: $f(3) = |2(3) - 6| = 0$. The function is continuous.
2. For $x > 3$, $f(x) = 2x - 6$. The difference quotient is $\frac{2(3+h)-6 - 0}{h} = \frac{2h}{h} = 2$.
3. For $x < 3$, $f(x) = -(2x - 6) = -2x + 6$. The difference quotient is $\frac{-2(3+h)+6 - 0}{h} = \frac{-2h}{h} = -2$.
4. The left-hand slope $-2$ does not equal the right-hand slope $2$.

Answer: Not differentiable at $x = 3$.

1. Check continuity: $f(0) = 0^{1/5} = 0$. It is continuous.
2. Set up the difference quotient: $\lim_{h \to 0} \frac{h^{1/5} - 0}{h}$.
3. Simplify the expression: $\frac{h^{1/5}}{h} = h^{1/5 - 1} = h^{-4/5} = \frac{1}{h^{4/5}}$.
4. Evaluate the limit: as $h \to 0$, $1/h^{4/5}$ approaches infinity.
5. Since the slope becomes infinite, there is a vertical tangent.

Answer: It has a vertical tangent at $x = 0$.

1. Identify the exponent $n = 7$.
2. Apply the power rule: bring the $7$ to the front and subtract $1$ from the exponent.
3. The result is $7x^{7-1}$.

Answer: f'(x) = 7x^6

1. Rewrite the function using a negative exponent: $f(x) = x^{-4}$.
2. Apply the power rule: bring $-4$ to the front and subtract $1$ from the exponent.
3. Compute $-4 - 1 = -5$.
4. The derivative is $-4x^{-5}$, which can be rewritten as $-\frac{4}{x^5}$.

Answer: f'(x) = -\frac{4}{x^5}

1. Rewrite the radical as a fractional exponent: $f(x) = (x^2)^{1/5} = x^{2/5}$.
2. Apply the power rule: bring $2/5$ to the front and subtract $1$ from the exponent.
3. Compute $2/5 - 1 = 2/5 - 5/5 = -3/5$.
4. The derivative is $\frac{2}{5}x^{-3/5}$.

Answer: f'(x) = \frac{2}{5}x^{-3/5}

1. Apply the power rule to $7x^4$ to get $7(4x^3) = 28x^3$.
2. Apply the power rule to $-2x^3$ to get $-2(3x^2) = -6x^2$.
3. Apply the power rule to $5x$ to get $5(1) = 5$.
4. The derivative of the constant $-12$ is $0$.

Answer: f'(x) = 28x^3 - 6x^2 + 5

1. Differentiate $\frac{1}{4}x^4$ as $\frac{1}{4}(4x^3) = x^3$.
2. Differentiate $\frac{2}{3}x^3$ as $\frac{2}{3}(3x^2) = 2x^2$.
3. The derivative of $-8$ is $0$.

Answer: g'(x) = x^3 + 2x^2

1. First, expand the expression: $h(x) = 2x^2 - 10x + x - 5 = 2x^2 - 9x - 5$.
2. Differentiate $2x^2$ to get $4x$.
3. Differentiate $-9x$ to get $-9$.
4. The derivative of $-5$ is $0$.

Answer: h'(x) = 4x - 9

1. Identify $u = x^3$ and $v = \ln x$.
2. Compute derivatives: $u' = 3x^2$ and $v' = \frac{1}{x}$.
3. Apply product rule: $f'(x) = (3x^2)(\ln x) + (x^3)(\frac{1}{x})$.
4. Simplify the second term: $x^3 \cdot \frac{1}{x} = x^2$.
5. Combine: $f'(x) = 3x^2 \ln x + x^2$.

Answer: f'(x) = x^2(3 \ln x + 1)

1. Identify $u = x^2$ and $v = \sin x$.
2. Compute derivatives: $u' = 2x$ and $v' = \cos x$.
3. Apply product rule: $g'(x) = (2x)(\sin x) + (x^2)(\cos x)$.
4. Factor out $x$: $g'(x) = x(2 \sin x + x \cos x)$.

Answer: g'(x) = 2x \sin x + x^2 \cos x

1. Identify $u = x^2 + 1$ and $v = e^x$.
2. Compute derivatives: $u' = 2x$ and $v' = e^x$.
3. Apply product rule: $h'(x) = (2x)(e^x) + (x^2 + 1)(e^x)$.
4. Factor out $e^x$: $h'(x) = e^x(2x + x^2 + 1)$.
5. Recognize the quadratic as a perfect square: $x^2 + 2x + 1 = (x+1)^2$.

Answer: h'(x) = (x+1)^2 e^x

1. Identify $f(x) = x+2$ and $g(x) = x-3$.
2. Compute derivatives: $f'(x) = 1$ and $g'(x) = 1$.
3. Apply quotient rule: $\frac{(1)(x-3) - (x+2)(1)}{(x-3)^2}$.
4. Simplify numerator: $x - 3 - x - 2 = -5$.

Answer: f'(x) = \frac{-5}{(x-3)^2}

1. Identify high as $x^2$ (derivative $2x$) and low as $2x+1$ (derivative $2$).
2. Apply rule: $\frac{(2x)(2x+1) - (x^2)(2)}{(2x+1)^2}$.
3. Expand numerator: $4x^2 + 2x - 2x^2$.
4. Simplify: $2x^2 + 2x$.

Answer: y' = \frac{2x^2 + 2x}{(2x+1)^2}

1. Rewriting as $g(x) = e^x \cdot x^{-2}$ allows the product rule.
2. Using product rule: $g'(x) = (e^x)(x^{-2}) + (e^x)(-2x^{-3})$.
3. Factor out $e^x$ and $x^{-3}$: $e^x x^{-3}(x - 2)$.
4. Convert back to fraction: $\frac{e^x(x-2)}{x^3}$.

Answer: g'(x) = \frac{e^x(x-2)}{x^3}

1. Apply the linearity of the derivative to separate the terms: $f'(x) = 3\frac{d}{dx}\sin(x) - 4\frac{d}{dx}\cos(x)$.
2. Use the known derivatives: $\frac{d}{dx}\sin(x) = \cos(x)$ and $\frac{d}{dx}\cos(x) = -\sin(x)$.
3. Substitute these into the expression: $f'(x) = 3\cos(x) - 4(-\sin(x))$.
4. Simplify the signs: $f'(x) = 3\cos(x) + 4\sin(x)$.

Answer: f'(x) = 3\cos(x) + 4\sin(x)

1. Identify this as a product of $u(x) = x^2$ and $v(x) = \cos(x)$.
2. Apply the product rule: $g'(x) = u'(x)v(x) + u(x)v'(x)$.
3. Compute the individual derivatives: $u'(x) = 2x$ and $v'(x) = -\sin(x)$.
4. Substitute back: $g'(x) = (2x)\cos(x) + (x^2)(-\sin(x))$.
5. Simplify the expression: $g'(x) = 2x\cos(x) - x^2\sin(x)$.

Answer: g'(x) = 2x\cos(x) - x^2\sin(x)

1. Identify this as a quotient of $u(x) = \cos(x)$ and $v(x) = x+1$.
2. Apply the quotient rule: $h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$.
3. Compute the individual derivatives: $u'(x) = -\sin(x)$ and $v'(x) = 1$.
4. Substitute into the formula: $h'(x) = \frac{(-\sin(x))(x+1) - (\cos(x))(1)}{(x+1)^2}$.
5. Simplify the numerator: $h'(x) = \frac{-(x+1)\sin(x) - \cos(x)}{(x+1)^2}$.

Answer: h'(x) = \frac{-(x+1)\sin(x) - \cos(x)}{(x+1)^2}

1. Identify the constant multiple as $5$.
2. Recall that the derivative of $\sec(x)$ is $\sec(x)\tan(x)$.
3. Apply the linearity rule: $f'(x) = 5 \cdot \frac{d}{dx}\sec(x)$.

Answer: f'(x) = 5\sec(x)\tan(x)

1. Differentiate $\tan(x)$ to get $\sec^2(x)$.
2. Differentiate $\cos(x)$ to get $-\sin(x)$.
3. Combine the results using the sum rule.

Answer: g'(x) = \sec^2(x) - \sin(x)

1. Use the product rule with $u = x^2$ and $v = \cot(x)$.
2. Compute $u' = 2x$ and $v' = -\csc^2(x)$.
3. Apply the formula $h'(x) = u'v + uv'$.
4. Substitute the values: $h'(x) = (2x)\cot(x) + (x^2)(-\csc^2(x))$.

Answer: h'(x) = 2x\cot(x) - x^2\csc^2(x)

1. Identify the outer function $f(u) = \ln(u)$ and the inner function $u = x^2 + 3$.
2. Find the derivative of the outer function: $f'(u) = \frac{1}{u}$.
3. Find the derivative of the inner function: $u' = 2x$.
4. Apply the chain rule: $\frac{dy}{dx} = \frac{1}{x^2 + 3} \cdot 2x$.

Answer: \frac{2x}{x^2 + 3}

1. Identify the outer function as cosine and the inner function as $4x^3$.
2. The derivative of the outer function is $-\sin(u)$, evaluated at $u = 4x^3$.
3. The derivative of the inner function $4x^3$ is $12x^2$.
4. Multiply them together: $-\sin(4x^3) \cdot 12x^2$.

Answer: -12x^2\sin(4x^3)

1. Identify the outer function as $u^7$ and the inner function as $u = 5x^2 - 2$.
2. Differentiate the outer layer: $7u^6$, which is $7(5x^2 - 2)^6$.
3. Differentiate the inner layer: $\frac{d}{dx}(5x^2 - 2) = 10x$.
4. Multiply the results: $7(5x^2 - 2)^6 \cdot 10x$.

Answer: 70x(5x^2 - 2)^6

1. Identify this as a product of $u = x$ and $v = \ln(x^2 + 1)$.
2. The derivative of $u$ is $u' = 1$.
3. The derivative of $v$ requires the chain rule: $v' = \frac{1}{x^2 + 1} \cdot 2x = \frac{2x}{x^2 + 1}$.
4. Apply the product rule: $y' = u'v + uv' = 1 \cdot \ln(x^2 + 1) + x \cdot \frac{2x}{x^2 + 1}$.
5. Simplify the expression to $\ln(x^2 + 1) + \frac{2x^2}{x^2 + 1}$.

Answer: \frac{dy}{dx} = \ln(x^2 + 1) + \frac{2x^2}{x^2 + 1}

1. Rewrite the function as $y = (\sin(3x))^4$.
2. The outermost layer is the power of 4. Its derivative is $4(\sin(3x))^3$.
3. The next layer is the sine function. Its derivative is $\cos(3x)$.
4. The innermost layer is $3x$. Its derivative is $3$.
5. Multiply all layers together: $y' = 4\sin^3(3x) \cdot \cos(3x) \cdot 3$.
6. Simplify to $12\sin^3(3x)\cos(3x)$.

Answer: \frac{dy}{dx} = 12\sin^3(3x)\cos(3x)

1. Identify this as a quotient with $u = e^{2x}$ and $v = \cos(x)$.
2. Find $u'$ using the chain rule: $u' = e^{2x} \cdot 2 = 2e^{2x}$.
3. Find $v'$: $v' = -\sin(x)$.
4. Apply the quotient rule: $y' = \frac{u'v - uv'}{v^2} = \frac{(2e^{2x})(\cos(x)) - (e^{2x})(-\sin(x))}{(\cos(x))^2}$.
5. Factor out $e^{2x}$ in the numerator: $y' = \frac{e^{2x}(2\cos(x) + \sin(x))}{\cos^2(x)}$.

Answer: \frac{dy}{dx} = \frac{e^{2x}(2\cos(x) + \sin(x))}{\cos^2(x)}

1. Identify the base $a = 7$.
2. Apply the formula $\frac{d}{dx} a^x = a^x \ln a$.
3. Substitute $7$ for $a$ to get $7^x \ln 7$.

Answer: f'(x) = 7^x \ln 7

1. Identify the outer function as $e^u$ and the inner function as $u = 5x^3$.
2. The derivative of the outer function is $e^{5x^3}$.
3. The derivative of the inner function is $15x^2$.
4. Multiply them together using the chain rule: $15x^2 e^{5x^3}$.

Answer: f'(x) = 15x^2 e^{5x^3}

1. Use the product rule with $u = x$ and $v = e^{2x}$.
2. Compute $u' = 1$ and $v' = 2e^{2x}$ using the chain rule.
3. Apply the product rule: $f'(x) = u'v + uv' = (1)(e^{2x}) + (x)(2e^{2x})$.
4. Factor out $e^{2x}$ to simplify: $e^{2x}(1 + 2x)$.

Answer: f'(x) = e^{2x}(1 + 2x)

1. Identify the inner function $u = x^3 + 7x$.
2. Compute the derivative of the inner function: $u' = 3x^2 + 7$.
3. Apply the formula $\frac{d}{dx} \ln(u) = \frac{u'}{u}$.
4. Substitute the expressions: $f'(x) = \frac{3x^2 + 7}{x^3 + 7x}$.

Answer: $f'(x) = \frac{3x^2 + 7}{x^3 + 7x}$

1. Take the natural log of both sides: $\ln y = \ln(x^5 \sin x)$.
2. Use log laws to expand: $\ln y = 5 \ln x + \ln(\sin x)$.
3. Differentiate implicitly: $\frac{1}{y} \frac{dy}{dx} = \frac{5}{x} + \frac{\cos x}{\sin x}$.
4. Simplify the trig term: $\frac{1}{y} \frac{dy}{dx} = \frac{5}{x} + \cot x$.
5. Multiply by $y$: $\frac{dy}{dx} = (x^5 \sin x) (\frac{5}{x} + \cot x)$.

Answer: $\frac{dy}{dx} = (x^5 \sin x) (\frac{5}{x} + \cot x)$

1. Set $y = x^{\cos x}$ and take the natural log: $\ln y = \ln(x^{\cos x})$.
2. Bring the exponent down: $\ln y = (\cos x) \ln x$.
3. Differentiate using the product rule: $\frac{1}{y} \frac{dy}{dx} = (-\sin x)\ln x + (\cos x)\frac{1}{x}$.
4. Isolate $\frac{dy}{dx}$ by multiplying by $y$: $\frac{dy}{dx} = x^{\cos x} (\frac{\cos x}{x} - \sin x \ln x)$.

Answer: $f'(x) = x^{\cos x} (\frac{\cos x}{x} - \sin x \ln x)$

1. Identify the outer function as $\arcsin(u)$ and the inner function as $u = \sqrt{x}$.
2. The derivative of $\arcsin(u)$ is $\frac{1}{\sqrt{1-u^2}}$.
3. The derivative of $\sqrt{x}$ is $\frac{1}{2\sqrt{x}}$.
4. Apply the chain rule: $f'(x) = \frac{1}{\sqrt{1-(\sqrt{x})^2}} \cdot \frac{1}{2\sqrt{x}}$.
5. Simplify the expression: $f'(x) = \frac{1}{2\sqrt{x}\sqrt{1-x}}$.

Answer: f'(x) = \frac{1}{2\sqrt{x-x^2}}

1. Identify the outer function as $\arctan(u)$ and the inner function as $u = e^{2x}$.
2. The derivative of $\arctan(u)$ is $\frac{1}{1+u^2}$.
3. The derivative of $e^{2x}$ is $2e^{2x}$ using the chain rule.
4. Apply the chain rule: $g'(x) = \frac{1}{1+(e^{2x})^2} \cdot 2e^{2x}$.
5. Simplify the exponent: $(e^{2x})^2 = e^{4x}$.

Answer: g'(x) = \frac{2e^{2x}}{1+e^{4x}}

1. Identify the outer function as $\ln(u)$ and the inner function as $u = \arctan(x)$.
2. The derivative of $\ln(u)$ is $\frac{1}{u}$.
3. The derivative of $\arctan(x)$ is $\frac{1}{1+x^2}$.
4. Apply the chain rule: $h'(x) = \frac{1}{\arctan(x)} \cdot \frac{1}{1+x^2}$.

Answer: h'(x) = \frac{1}{(1+x^2)\arctan(x)}

1. Differentiate both sides with respect to $x$: $2x + 6y \frac{dy}{dx} = 0$.
2. Subtract $2x$ from both sides: $6y \frac{dy}{dx} = -2x$.
3. Divide by $6y$: $\frac{dy}{dx} = \frac{-2x}{6y}$.
4. Simplify the fraction: $\frac{dy}{dx} = -\frac{x}{3y}$.

Answer: \frac{dy}{dx} = -\frac{x}{3y}

1. Apply the product rule to $x^2 y$: $(2xy + x^2 \frac{dy}{dx})$.
2. Apply the product rule to $y^2 x$: $(2yx \frac{dy}{dx} + y^2)$.
3. Combine: $2xy + x^2 \frac{dy}{dx} + 2xy \frac{dy}{dx} + y^2 = 0$.
4. Group $\frac{dy}{dx}$ terms: $\frac{dy}{dx}(x^2 + 2xy) = -2xy - y^2$.
5. Isolate $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{-2xy - y^2}{x^2 + 2xy}$.

Answer: \frac{dy}{dx} = -\frac{2xy + y^2}{x^2 + 2xy}

1. Differentiate both sides: $e^y \frac{dy}{dx} + 1 = \frac{dy}{dx}$.
2. Move $\frac{dy}{dx}$ terms to one side: $e^y \frac{dy}{dx} - \frac{dy}{dx} = -1$.
3. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(e^y - 1) = -1$.
4. Isolate $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{-1}{e^y - 1}$.
5. Alternatively, multiply top and bottom by $-1$: $\frac{dy}{dx} = \frac{1}{1 - e^y}$.

Answer: \frac{dy}{dx} = \frac{1}{1 - e^y}

1. First, find the first derivative using the power rule: $f'(x) = 9x^2 - 14x$.
2. Next, differentiate the first derivative: $f''(x) = 18x - 14$.

Answer: f''(x) = 18x - 14

1. The first derivative of $\cos(x)$ is $-\sin(x)$.
2. The derivative of $-\sin(x)$ is $-\cos(x)$.

Answer: g''(x) = -\cos(x)

1. The first derivative is $h'(x) = 5e^{5x}$ using the chain rule.
2. The second derivative is $h''(x) = 5 \cdot 5e^{5x} = 25e^{5x}$.
3. The third derivative is $h'''(x) = 25 \cdot 5e^{5x} = 125e^{5x}$.

Answer: h'''(x) = 125e^{5x}

1. Identify the outer rule as the product rule with $u = x^3$ and $v = \cos(x^2)$.
2. Compute $u' = 3x^2$.
3. Compute $v'$ using the chain rule: $v' = -\sin(x^2) \cdot 2x = -2x \sin(x^2)$.
4. Apply the product rule: $f'(x) = (3x^2)(\cos(x^2)) + (x^3)(-2x \sin(x^2))$.
5. Simplify the expression: $f'(x) = 3x^2 \cos(x^2) - 2x^4 \sin(x^2)$.

Answer: f'(x) = 3x^2 \cos(x^2) - 2x^4 \sin(x^2)

1. Identify the outer rule as the quotient rule with $u = \ln(x)$ and $v = x^2$.
2. Compute $u' = 1/x$ and $v' = 2x$.
3. Apply the quotient rule: $g'(x) = \frac{(1/x)(x^2) - (\ln x)(2x)}{(x^2)^2}$.
4. Simplify the numerator: $(1/x)(x^2) = x$, so we have $\frac{x - 2x \ln x}{x^4}$.
5. Factor out $x$ from the numerator and cancel with the denominator: $g'(x) = \frac{x(1 - 2 \ln x)}{x^4} = \frac{1 - 2 \ln x}{x^3}$.

Answer: g'(x) = \frac{1 - 2 \ln x}{x^3}

1. Identify the outer rule as the chain rule for the exponential function $e^u$ where $u = x \sin(x)$.
2. The derivative is $h'(x) = e^{x \sin(x)} \cdot \frac{d}{dx}(x \sin(x))$.
3. To find the derivative of $x \sin(x)$, use the product rule: $(1)(\sin x) + (x)(\cos x)$.
4. Combine the results: $h'(x) = e^{x \sin(x)} (\sin x + x \cos x)$.

Answer: h'(x) = e^{x \sin(x)} (\sin x + x \cos x)

1. The relation is $A = s^2$.
2. Differentiate with respect to $t$: $\frac{dA}{dt} = 2s \frac{ds}{dt}$.
3. Substitute $s = 8$ and $\frac{ds}{dt} = 5$.
4. $\frac{dA}{dt} = 2(8)(5) = 80$.

Answer: 80 \text{ cm}^2/\text{s}

1. The relation is $V = \frac{4}{3}\pi r^3$.
2. Differentiate with respect to $t$: $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.
3. Substitute $r = 3$ and $\frac{dr}{dt} = 2$.
4. $\frac{dV}{dt} = 4\pi(3^2)(2) = 4\pi(9)(2) = 72\pi$.

Answer: 72\pi \text{ cm}^3/\text{s}

1. The relation is $x^2 + y^2 = z^2$.
2. Differentiate with respect to $t$: $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 2z \frac{dz}{dt}$.
3. Find $z$ when $x=3, y=4$: $z = \sqrt{3^2 + 4^2} = 5$.
4. Substitute $\frac{dx}{dt} = 3$ and $\frac{dy}{dt} = -4$ (since it moves toward the origin).
5. $2(3)(3) + 2(4)(-4) = 2(5) \frac{dz}{dt} \implies 18 - 32 = 10 \frac{dz}{dt} \implies -14 = 10 \frac{dz}{dt}$.
6. $\frac{dz}{dt} = -1.4$.

Answer: -1.4 \text{ units/s}

1. The area of a circle is $A = \pi r^2$.
2. Differentiate with respect to time $t$: $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.
3. Substitute the given values $r = 5$ and $\frac{dr}{dt} = 2$.
4. $\frac{dA}{dt} = 2\pi(5)(2) = 20\pi$.

Answer: $20\pi\text{ cm}^2\text{/s}$

1. Equation: $x^2 + y^2 = 13^2$.
2. Differentiate: $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$.
3. Find $x$ when $y=5$: $x^2 + 5^2 = 13^2 \implies x^2 = 144 \implies x = 12$.
4. Substitute $x=12, y=5, \frac{dy}{dt}=-1$: $2(12)\frac{dx}{dt} + 2(5)(-1) = 0$.
5. $24\frac{dx}{dt} - 10 = 0 \implies \frac{dx}{dt} = \frac{10}{24} = \frac{5}{12}$.

Answer: $\frac{5}{12}\text{ ft/s}$

1. Volume $V = \frac{1}{3}\pi r^2 h$. Ratio $\frac{r}{h} = \frac{4}{10} = \frac{2}{5}$, so $r = \frac{2}{5}h$.
2. Substitute $r$: $V = \frac{1}{3}\pi (\frac{2}{5}h)^2 h = \frac{4}{75}\pi h^3$.
3. Differentiate: $\frac{dV}{dt} = \frac{4}{25}\pi h^2 \frac{dh}{dt}$.
4. Substitute $\frac{dV}{dt} = 2$ and $h = 5$: $2 = \frac{4}{25}\pi (5)^2 \frac{dh}{dt}$.
5. $2 = 4\pi \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{1}{2\pi}$.

Answer: $\frac{1}{2\pi}\text{ m/min}$

1. Let $f(x) = x^{1/3}$. We choose $a = 27$ because it is a perfect cube near $28$.
2. Calculate $f(27) = 27^{1/3} = 3$.
3. Find the derivative: $f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3(\sqrt[3]{x})^2}$.
4. Evaluate the derivative at $a = 27$: $f'(27) = \frac{1}{3(3^2)} = \frac{1}{27}$.
5. Apply the formula: $L(28) = f(27) + f'(27)(28 - 27) = 3 + \frac{1}{27}(1)$.
6. Simplify to get $3 + 0.037 = 3.037$.

Answer: 3.037

1. The volume of a sphere is $V = \frac{4}{3}\pi r^3$.
2. The differential is $dV = V'(r) dr = 4\pi r^2 dr$.
3. Substitute the known values: $r = 5$ and $dr = 0.1$.
4. Calculate $dV = 4\pi(5^2)(0.1) = 4\pi(25)(0.1)$.
5. Simplify to $100\pi(0.1) = 10\pi$.

Answer: 10\pi \text{ cm}^3 \approx 31.4 \text{ cm}^3

1. The area of a square is $A = x^2$.
2. The differential is $dA = 2x dx$.
3. Substitute the values: $x = 10$ and $dx = 0.05$.
4. Calculate $dA = 2(10)(0.05)$.
5. Simplify to $20(0.05) = 1$.

Answer: 1 \text{ cm}^2

1. Check the form: as $x \to 0$, $x^2 \to 0$ and $\sin(x) \to 0$. This is a $0/0$ form.
2. Apply L'Hopital's rule: differentiate the numerator to get $2x$ and the denominator to get $\cos(x)$.
3. Evaluate the new limit: $\lim_{x \to 0} \frac{2x}{\cos(x)}$.
4. Substitute $x = 0$: $\frac{2(0)}{\cos(0)} = \frac{0}{1} = 0$.

Answer: 0

1. Check the form: $e^0 - 1 - 0 = 0$ and $0^2 = 0$. This is $0/0$.
2. Apply L'Hopital's rule: the derivative of $e^x - 1 - x$ is $e^x - 1$, and the derivative of $x^2$ is $2x$.
3. Check the new form: $e^0 - 1 = 0$ and $2(0) = 0$. This is still $0/0$.
4. Apply L'Hopital's rule again: the derivative of $e^x - 1$ is $e^x$, and the derivative of $2x$ is $2$.
5. Evaluate the limit: $\lim_{x \to 0} \frac{e^x}{2} = \frac{e^0}{2} = \frac{1}{2}$.

Answer: 1/2

1. Check the form: $\tan(0) - 0 = 0$ and $0^3 = 0$. This is $0/0$.
2. Apply L'Hopital's rule: the derivative of $\tan(x) - x$ is $\sec^2(x) - 1$, and the derivative of $x^3$ is $3x^2$.
3. Note that $\sec^2(x) - 1 = \tan^2(x)$. The limit is $\lim_{x \to 0} \frac{\tan^2(x)}{3x^2}$.
4. This is still $0/0$. Apply L'Hopital's rule again: the derivative of $\tan^2(x)$ is $2\tan(x)\sec^2(x)$, and the derivative of $3x^2$ is $6x$.
5. The limit is $\lim_{x \to 0} \frac{2\tan(x)\sec^2(x)}{6x} = \lim_{x \to 0} \frac{\tan(x)}{3x} \cdot \sec^2(x)$.
6. Apply L'Hopital's rule one last time to $\frac{\tan(x)}{3x}$: the derivative of $\tan(x)$ is $\sec^2(x)$ and the derivative of $3x$ is $3$.
7. The limit becomes $\frac{\sec^2(0)}{3} \cdot \sec^2(0) = \frac{1}{3} \cdot 1 = \frac{1}{3}$.

Answer: 1/3

1. Identify the form: as $x \to 0^+$, $x^2 \to 0$ and $\ln x \to -\infty$. This is a $0 \cdot \infty$ form.
2. Rewrite as a quotient: $\lim_{x \to 0^+} \frac{\ln x}{1/x^2}$. This is now $-\infty / \infty$.
3. Apply L'Hopital's rule: the derivative of $\ln x$ is $1/x$ and the derivative of $1/x^2$ is $-2/x^3$.
4. Simplify the expression: $\frac{1/x}{-2/x^3} = \frac{1}{x} \cdot \frac{x^3}{-2} = -\frac{x^2}{2}$.
5. Evaluate the limit: as $x \to 0^+$, $-x^2/2 \to 0$.

Answer: 0

1. Identify the form: as $x \to \infty$, both terms go to infinity, so this is $\infty - \infty$.
2. Rationalize the expression by multiplying by the conjugate: $\frac{(\sqrt{x^2+x}-x)(\sqrt{x^2+x}+x)}{\sqrt{x^2+x}+x}$.
3. Simplify the numerator: $(\sqrt{x^2+x})^2 - x^2 = x^2 + x - x^2 = x$.
4. The limit is now $\lim_{x \to \infty} \frac{x}{\sqrt{x^2+x}+x}$, which is $\infty / \infty$.
5. Apply L'Hopital's rule: the derivative of the top is $1$. The derivative of the bottom is $\frac{2x+1}{2\sqrt{x^2+x}} + 1$.
6. As $x \to \infty$, the fraction $\frac{2x+1}{2\sqrt{x^2+x}}$ approaches $1$.
7. The limit becomes $1 / (1 + 1) = 1/2$.

Answer: 1/2

1. Identify the form: as $x \to 0^+$, the base $(1+x) \to 1$ and the exponent $1/x \to \infty$. This is a $1^\infty$ form.
2. Set $y = (1+x)^{1/x}$ and take the natural log: $\ln y = \ln((1+x)^{1/x}) = \frac{1}{x} \ln(1+x)$.
3. Rewrite as a quotient: $\lim_{x \to 0^+} \frac{\ln(1+x)}{x}$. This is a $0/0$ form.
4. Apply L'Hopital's rule: the derivative of $\ln(1+x)$ is $1/(1+x)$ and the derivative of $x$ is $1$.
5. Evaluate the limit: $\lim_{x \to 0^+} \frac{1/(1+x)}{1} = 1$.
6. Since $\ln y \to 1$, then $y \to e^1 = e$.

Answer: e

1. Calculate the average slope: $f(0) = 0$ and $f(2) = 2^3 - 2 = 6$. The average slope is $(6 - 0) / (2 - 0) = 3$.
2. Find the derivative: $f'(x) = 3x^2 - 1$.
3. Set the derivative equal to the average slope: $3c^2 - 1 = 3$.
4. Solve for $c$: $3c^2 = 4$, so $c^2 = 4/3$, which means $c = \pm 2/\sqrt{3}$.
5. Check which value is in the interval $[0, 2]$. Only $c = 2/\sqrt{3}$ is valid.

Answer: $c = \frac{2}{\sqrt{3}}$

1. Check continuity: $f(x)$ is continuous on $[-1, 1]$.
2. Check endpoints: $f(-1) = (-1)^{2/3} = 1$ and $f(1) = 1^{2/3} = 1$. Endpoints are equal.
3. Check differentiability: $f'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}}$.
4. The derivative is undefined at $x = 0$, which is inside the open interval $(-1, 1)$.
5. Since the function is not differentiable on the open interval, Rolle's theorem does not apply.

Answer: No, because $f(x)$ is not differentiable at $x = 0$.

1. Calculate the average slope: $f(1) = \ln(1) = 0$ and $f(e) = \ln(e) = 1$. The average slope is $(1 - 0) / (e - 1) = 1 / (e - 1)$.
2. Find the derivative: $f'(x) = 1/x$.
3. Set the derivative equal to the average slope: $1/c = 1 / (e - 1)$.
4. Solve for $c$: $c = e - 1$.
5. Check if $c$ is in the interval $[1, e]$. Since $e \approx 2.718$, $c \approx 1.718$, which is between $1$ and $2.718$.

Answer: $c = e - 1$

1. Find the derivative: $f'(x) = 6x^2 - 6x - 12$.
2. Factor the derivative: $f'(x) = 6(x^2 - x - 2) = 6(x - 2)(x + 1)$.
3. Identify critical points where $f'(x) = 0$: $x = 2$ and $x = -1$.
4. Test intervals: For $x < -1$, $f'(-2) = 6(-4)(-1) = 24 > 0$ (Increasing). For $-1 < x < 2$, $f'(0) = -12 < 0$ (Decreasing). For $x > 2$, $f'(3) = 6(1)(4) = 24 > 0$ (Increasing).

Answer: Increasing on $(-\infty, -1) \cup (2, \infty)$, decreasing on $(-1, 2)$

1. Find the derivative: $f'(x) = e^x + 1$.
2. Analyze the range of the exponential function: $e^x$ is always greater than $0$ for all $x$.
3. Therefore, $f'(x) = e^x + 1$ must always be greater than $1$, which means $f'(x) > 0$ for all $x$.
4. By the Mean Value Theorem, since the derivative is always positive, the function is strictly increasing.

Answer: Since $f'(x) = e^x + 1 > 0$ for all $x$, the function is strictly increasing.

1. Set the derivative to be less than zero: $\frac{x-3}{x^2} < 0$.
2. Since $x^2$ is always positive for $x > 0$, the sign of the fraction depends only on the numerator.
3. Solve $x - 3 < 0$, which gives $x < 3$.
4. Combine this with the domain $x > 0$ to get the interval $(0, 3)$.

Answer: The function is decreasing on the interval $(0, 3)$

1. Find the derivative: $f'(x) = 3x^2 - 6x - 9$.
2. Set $f'(x) = 0$: $3(x^2 - 2x - 3) = 0$, which factors to $3(x-3)(x+1) = 0$. Critical points are $x = -1$ and $x = 3$.
3. Test intervals: For $x < -1$, $f'(-2) = 15 > 0$ (increasing). For $-1 < x < 3$, $f'(0) = -9 < 0$ (decreasing). For $x > 3$, $f'(4) = 15 > 0$ (increasing).
4. Classify: At $x = -1$, sign changes $(+) \to (-)$, so it is a local max. At $x = 3$, sign changes $(-) \to (+)$, so it is a local min.
5. Calculate values: $f(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = 10$. $f(3) = (3)^3 - 3(3)^2 - 9(3) + 5 = -22$.

Answer: Local maximum of $10$ at $x = -1$; local minimum of $-22$ at $x = 3$.

1. Find the derivative: $f'(x) = 4x^3 - 8x$.
2. Set $f'(x) = 0$: $4x(x^2 - 2) = 0$. Critical points are $x = 0, x = \sqrt{2}, x = -\sqrt{2}$.
3. Test intervals: $x < -\sqrt{2}$ (e.g., $-2$): $f'(-2) = -32 + 16 = -16 < 0$ (decreasing).
4. Interval $-\sqrt{2} < x < 0$ (e.g., $-1$): $f'(-1) = -4 + 8 = 4 > 0$ (increasing).
5. Interval $0 < x < \sqrt{2}$ (e.g., $1$): $f'(1) = 4 - 8 = -4 < 0$ (decreasing).
6. Interval $x > \sqrt{2}$ (e.g., $2$): $f'(2) = 32 - 16 = 16 > 0$ (increasing).

Answer: Increasing on $(-\sqrt{2}, 0) \cup (\sqrt{2}, \infty)$; decreasing on $(-\infty, -\sqrt{2}) \cup (0, \sqrt{2})$.

1. Find the derivative: $f'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}}$.
2. Identify critical points: $f'(x)$ is never zero, but it is undefined at $x = 0$. Since $x = 0$ is in the domain of $f$, it is a critical point.
3. Test intervals: For $x < 0$, $\sqrt[3]{x}$ is negative, so $f'(x) < 0$ (decreasing).
4. For $x > 0$, $\sqrt[3]{x}$ is positive, so $f'(x) > 0$ (increasing).
5. Classify: The sign changes from negative to positive at $x = 0$, so it is a local minimum.

Answer: Critical point at $x = 0$, which is a local minimum.

1. Find the first derivative: $f'(x) = 4x^3 - 12x$.
2. Find the second derivative: $f''(x) = 12x^2 - 12$.
3. Set $f''(x) = 0$ to find potential inflection points: $12(x^2 - 1) = 0$, so $x = 1$ and $x = -1$.
4. Test intervals: for $x < -1$, $f''(-2) = 36 > 0$ (Up); for $-1 < x < 1$, $f''(0) = -12 < 0$ (Down); for $x > 1$, $f''(2) = 36 > 0$ (Up).
5. Since the sign changes at both $x = -1$ and $x = 1$, both are inflection points.

Answer: Concave up on $(-\infty, -1) \cup (1, \infty)$, concave down on $(-1, 1)$, inflection points at $x = -1, 1$.

1. Find the first derivative using the product rule: $f'(x) = 1 \cdot e^x + x \cdot e^x = e^x(1 + x)$.
2. Set $f'(x) = 0$ to find critical points: $e^x(1 + x) = 0$ implies $x = -1$.
3. Find the second derivative: $f''(x) = e^x(1 + x) + e^x(1) = e^x(x + 2)$.
4. Evaluate at the critical point: $f''(-1) = e^{-1}(-1 + 2) = e^{-1} \approx 0.368$.
5. Since $f''(-1) > 0$, the function is concave up at $x = -1$.

Answer: Local minimum at $x = -1$.

1. Find the first derivative: $f'(x) = 5x^4$. Setting $f'(0) = 0$ confirms $x = 0$ is a critical point.
2. Find the second derivative: $f''(x) = 20x^3$.
3. Evaluate at $x = 0$: $f''(0) = 0$. The second derivative test is inconclusive.
4. Use the first derivative test: for $x < 0$, $f'(x) = 5x^4 > 0$; for $x > 0$, $f'(x) = 5x^4 > 0$.
5. Since the first derivative does not change sign, it is not a local extremum. Check concavity: for $x < 0$, $f''(x) = 20x^3 < 0$ (Down); for $x > 0$, $f''(x) = 20x^3 > 0$ (Up).
6. Since concavity changes, it is an inflection point.

Answer: Inflection point at $x = 0$.

1. Domain is $x \neq 2$. Vertical asymptote at $x = 2$.
2. Limit as $x \to \infty$ is $1$, so horizontal asymptote is $y = 1$.
3. Using quotient rule, $f'(x) = \frac{1(x-2) - 1(x+1)}{(x-2)^2} = \frac{-3}{(x-2)^2}$. Since $f'(x) < 0$ everywhere, the function is always decreasing.
4. Using quotient rule on $f'(x)$, $f''(x) = \frac{6}{(x-2)^3}$. Concave down for $x < 2$ and concave up for $x > 2$.
5. Intercepts: $y$-intercept at $f(0) = -1/2$, $x$-intercept at $f(x) = 0 \implies x = -1$.

Answer: A hyperbola with asymptotes $x=2, y=1$, always decreasing, concave down on $(-\infty, 2)$ and concave up on $(2, \infty)$.

1. Domain is all real numbers. No vertical asymptotes. Horizontal asymptote at $y = 0$.
2. First derivative $f'(x) = \frac{-2x}{(x^2+1)^2}$. Critical point at $x = 0$. $f'(x) > 0$ for $x < 0$ and $f'(x) < 0$ for $x > 0$. Local max at $(0, 1)$.
3. Second derivative $f''(x) = \frac{2(3x^2-1)}{(x^2+1)^3}$. Inflection points where $3x^2 = 1$, so $x = \pm 1/\sqrt{3}$.
4. Concave up for $|x| > 1/\sqrt{3}$ and concave down for $|x| < 1/\sqrt{3}$.

Answer: A bell-shaped curve with a maximum at $(0, 1)$, horizontal asymptote $y=0$, and inflection points at $x = \pm 1/\sqrt{3}$.

1. Domain is $x \neq \pm 2$. Vertical asymptotes at $x = -2$ and $x = 2$.
2. Limit as $x \to \infty$ is $1$, so horizontal asymptote is $y = 1$.
3. First derivative $f'(x) = \frac{2x(x^2-4) - 2x(x^2-1)}{(x^2-4)^2} = \frac{-6x}{(x^2-4)^2}$.
4. Critical point at $x = 0$. $f'(x) > 0$ for $x < 0$ and $f'(x) < 0$ for $x > 0$. Local max at $(0, 1/4)$.
5. Intercepts: $y$-intercept at $(0, 1/4)$, $x$-intercepts at $x = \pm 1$.

Answer: A graph with vertical asymptotes $x=\pm 2$, horizontal asymptote $y=1$, and a local maximum at $(0, 1/4)$.

1. Find $f'(x) = 2x e^{-x} - x^2 e^{-x} = x e^{-x}(2-x)$.
2. Set $f'(x) = 0$ to find critical points at $x=0$ and $x=2$.
3. Find $f''(x) = (2-2x)e^{-x} - (2x-x^2)e^{-x} = e^{-x}(x^2-4x+2)$.
4. Solve $x^2-4x+2=0$ using the quadratic formula to find inflection points at $x = 2 \pm \sqrt{2}$.
5. Analyze signs: $f$ increases on $(0,2)$ and decreases on $(2, \infty)$.
6. Plot the local maximum at $(2, 4e^{-2})$ and the inflection points.

Answer: Local max at $x=2$, inflection points at $x=2-\sqrt{2}$ and $x=2+\sqrt{2}$, horizontal asymptote $y=0$ as $x \to \infty$.

1. Find $f'(x) = e^x \cos(x) - e^x \sin(x) = e^x(\cos(x) - \sin(x))$.
2. Set $f'(x) = 0 \implies \tan(x) = 1$, so $x = \pi/4$ and $x = 5\pi/4$.
3. Find $f''(x) = e^x(\cos(x) - \sin(x)) + e^x(-\sin(x) - \cos(x)) = -2e^x \sin(x)$.
4. Set $f''(x) = 0 \implies \sin(x) = 0$, so $x = 0, \pi, 2\pi$.
5. Determine concavity: concave down on $(0, \pi)$ and concave up on $(\pi, 2\pi)$.

Answer: Local max at $x=\pi/4$, local min at $x=5\pi/4$, inflection point at $x=\pi$.

1. Find $f'(x) = \cos(x) - x \sin(x)$.
2. Set $f'(x) = 0 \implies \cot(x) = x$. This occurs at $x \approx 0.86$ radians.
3. Find $f''(x) = -\sin(x) - (\sin(x) + x \cos(x)) = -2\sin(x) - x \cos(x)$.
4. Set $f''(x) = 0 \implies \tan(x) = -x/2$. This occurs at $x \approx 2.28$ radians.
5. Check endpoints: $f(0)=0$ and $f(\pi)=-\pi$.

Answer: Local max at $x \approx 0.86$, inflection point at $x \approx 2.28$, decreasing from the max to the endpoint.

1. Let $L$ be length and $W$ be width. The constraint is $2L + 2W = 120$, so $L = 60 - W$.
2. The objective function is $A = L \times W = (60 - W)W = 60W - W^2$.
3. Find the derivative: $A'(W) = 60 - 2W$.
4. Set to zero: $60 - 2W = 0 \implies W = 30$.
5. Then $L = 60 - 30 = 30$.
6. Check endpoints $W=0$ and $W=60$, both give $A=0$. The maximum is at $W=30, L=30$.

Answer: L = 30 \text{ cm}, W = 30 \text{ cm}

1. Let $W$ be the two sides perpendicular to the wall and $L$ be the side parallel to the wall.
2. Constraint: $L + 2W = 100 \implies L = 100 - 2W$.
3. Objective: $A = W(100 - 2W) = 100W - 2W^2$.
4. Derivative: $A'(W) = 100 - 4W$.
5. Critical point: $100 - 4W = 0 \implies W = 25$.
6. Calculate area: $A(25) = 25(100 - 50) = 25 \times 50 = 1250$.

Answer: 1250 \text{ square meters}

1. The height is $x$ and the base sides are $10 - 2x$.
2. Objective: $V(x) = x(10 - 2x)^2 = x(100 - 40x + 4x^2) = 4x^3 - 40x^2 + 100x$.
3. Derivative: $V'(x) = 12x^2 - 80x + 100$.
4. Set to zero and divide by $4$: $3x^2 - 20x + 25 = 0$.
5. Factor: $(3x - 5)(x - 5) = 0$.
6. Critical points are $x = 5/3$ and $x = 5$.
7. Domain is $0 < x < 5$. At $x=5$, $V=0$. At $x=5/3$, $V = (5/3)(10 - 10/3)^2 = (5/3)(20/3)^2 = 2000/27$.

Answer: x = 5/3 \text{ inches}

1. Let $x$ be the side of the square base and $h$ be the height. The volume is $V = x^2 h = 10$, so $h = 10/x^2$.
2. The surface area is $A = x^2 + 4xh$. Substituting $h$, we get $A(x) = x^2 + 4x(10/x^2) = x^2 + 40/x$.
3. Find the derivative: $A'(x) = 2x - 40/x^2$.
4. Set $A'(x) = 0$: $2x = 40/x^2$, which means $x^3 = 20$, so $x = \sqrt[3]{20}$.
5. Find $h$: $h = 10/(\sqrt[3]{20})^2 = 10/20^{2/3} = (10 \cdot 20^{1/3})/20 = 20^{1/3}/2 = \sqrt[3]{20}/2$.

Answer: x = \sqrt[3]{20}, h = \frac{\sqrt[3]{20}}{2}

1. Let $x$ be the length of the wire used for the square. The remaining $20-x$ is used for the circle.
2. Side of square is $x/4$, area is $(x/4)^2 = x^2/16$.
3. Circumference of circle is $2\pi r = 20-x$, so $r = (20-x)/(2\pi)$, area is $\pi((20-x)/(2\pi))^2 = (20-x)^2/(4\pi)$.
4. Total area $A(x) = x^2/16 + (20-x)^2/(4\pi)$.
5. Derivative $A'(x) = x/8 - (20-x)/(2\pi)$.
6. Set $A'(x) = 0$: $x/8 = (20-x)/(2\pi) \implies \pi x = 4(20-x) \implies x(\pi+4) = 80 \implies x = 80/(\pi+4)$.

Answer: x = \frac{80}{\pi+4}

1. Use Snell's law: $\sin(\theta_1)/v_1 = \sin(\theta_2)/v_2$.
2. Rearrange for the ratio: $v_1/v_2 = \sin(\theta_1)/\sin(\theta_2)$.
3. Substitute the angles: $v_1/v_2 = \sin(30^\circ)/\sin(45^\circ)$.
4. Calculate values: $\sin(30^\circ) = 1/2$ and $\sin(45^\circ) = \sqrt{2}/2$.
5. Ratio: $(1/2) / (\sqrt{2}/2) = 1/\sqrt{2} = \sqrt{2}/2$.

Answer: \frac{\sqrt{2}}{2}

1. Calculate width: $\Delta x = \frac{2-0}{4} = 0.5$.
2. Identify right endpoints: $x_1=0.5, x_2=1.0, x_3=1.5, x_4=2.0$.
3. Evaluate function: $f(0.5)=3.5, f(1.0)=5, f(1.5)=6.5, f(2.0)=8$.
4. Sum and multiply: $R_4 = 0.5 \times (3.5 + 5 + 6.5 + 8) = 0.5 \times 23 = 11.5$.

Answer: 11.5

1. Calculate width: $\Delta x = \frac{3-1}{2} = 1$.
2. Identify left endpoints: $x_0=1, x_1=2$.
3. Evaluate function: $f(1)=1, f(2)=0.5$.
4. Sum and multiply: $L_2 = 1 \times (1 + 0.5) = 1.5$.

Answer: 1.5

1. Calculate width: $\Delta x = \frac{1-0}{2} = 0.5$.
2. Left endpoints: $0, 0.5$. $L_2 = 0.5 \times (0^3 + 0.5^3) = 0.5 \times 0.125 = 0.0625$.
3. Right endpoints: $0.5, 1$. $R_2 = 0.5 \times (0.5^3 + 1^3) = 0.5 \times 1.125 = 0.5625$.
4. Since $f(x)=x^3$ is increasing on $[0, 1]$, the right sum $R_2$ is the overestimate.

Answer: L_2 = 0.0625, R_2 = 0.5625; R_2 is the overestimate because the function is increasing.

1. The function is $f(x) = 2x - 4$.
2. The interval is from $a = 1$ to $b = 5$.
3. The definite integral is $\int_{1}^{5} (2x - 4) \, dx$.
4. The function crosses the x axis at $2x - 4 = 0$, which is $x = 2$.
5. From $x = 1$ to $x = 2$, the function is negative. From $x = 2$ to $x = 5$, the function is positive.
6. The positive region (from 2 to 5) is larger than the negative region (from 1 to 2), so the result will be positive.

Answer: \int_{1}^{5} (2x - 4) \, dx; \text{Positive}

1. An odd function has rotational symmetry about the origin.
2. This means for every point $(x, y)$ on the graph, there is a corresponding point $(-x, -y)$.
3. The area between the curve and the x axis from $0$ to $a$ is exactly the same size as the area from $-a$ to $0$.
4. However, because the function is odd, one of these areas is above the x axis and the other is below.
5. The signed area from $-a$ to $0$ is the negative of the signed area from $0$ to $a$.
6. Adding them together results in zero.

Answer: 0

1. The definite integral is $\int_{0}^{6} |x - 3| \, dx$.
2. The absolute value function $|x - 3|$ is always greater than or equal to zero for all $x$.
3. Since $f(x) \ge 0$ for the entire interval $[0, 6]$, there is no area below the x axis.
4. The signed area formula is $\text{Area}_{\text{above}} - \text{Area}_{\text{below}}$.
5. Since $\text{Area}_{\text{below}} = 0$, the signed area is simply the area above the x axis.
6. Therefore, the signed area equals the total geometric area.

Answer: \int_{0}^{6} |x - 3| \, dx; \text{The function is never negative.}

1. Apply the sum rule to split the integral into three parts: $\int_{1}^{3} 5x^3 \, dx - \int_{1}^{3} 2x \, dx + \int_{1}^{3} 7 \, dx$.
2. Apply the constant multiple rule to move the coefficients outside: $5 \int_{1}^{3} x^3 \, dx - 2 \int_{1}^{3} x \, dx + 7 \int_{1}^{3} 1 \, dx$.

Answer: 5 \int_{1}^{3} x^3 \, dx - 2 \int_{1}^{3} x \, dx + 7 \int_{1}^{3} 1 \, dx

1. Use the additivity property: $\int_{0}^{4} f(x) \, dx = \int_{0}^{2} f(x) \, dx + \int_{2}^{4} f(x) \, dx$.
2. Substitute the known values: $12 = 5 + \int_{2}^{4} f(x) \, dx$.
3. Subtract 5 from both sides to isolate the unknown integral: $\int_{2}^{4} f(x) \, dx = 12 - 5 = 7$.

Answer: 7

1. Use the comparison property: since $f(x) \ge 3$, then $\int_{0}^{5} f(x) \, dx \ge \int_{0}^{5} 3 \, dx$.
2. Calculate the integral of the constant function 3 over the interval $[0, 5]$.
3. The area of a rectangle with height 3 and width $5 - 0 = 5$ is $3 \times 5 = 15$.

Answer: 15

1. Apply linearity to split the integral into three parts: $5 \int x^4 \, dx - 6 \int x \, dx + 2 \int 1 \, dx$.
2. Use the power rule on $x^4$ to get $x^5/5$, so $5(x^5/5) = x^5$.
3. Use the power rule on $x$ to get $x^2/2$, so $-6(x^2/2) = -3x^2$.
4. The integral of the constant $2$ is $2x$.
5. Combine the terms and add the constant of integration $C$.

Answer: x^5 - 3x^2 + 2x + C

1. Rewrite the first term using a negative exponent: $\int (3x^{-2} + e^x) \, dx$.
2. Apply the power rule to $3x^{-2}$: the new exponent is $-1$, and we divide by $-1$, giving $3(x^{-1}/-1) = -3x^{-1}$.
3. The antiderivative of $e^x$ is $e^x$.
4. Combine the results and add $C$.

Answer: -3x^{-1} + e^x + C

1. Expand the squared binomial first: $(x+1)^2 = x^2 + 2x + 1$.
2. Set up the integral of the expanded form: $\int (x^2 + 2x + 1) \, dx$.
3. Integrate $x^2$ to get $x^3/3$.
4. Integrate $2x$ to get $x^2$.
5. Integrate $1$ to get $x$.
6. Combine all terms and add $C$.

Answer: \frac{1}{3}x^3 + x^2 + x + C

1. Identify the integrand as $f(t) = t^3 + 5t$.
2. The upper limit of integration is simply $x$.
3. Apply the Fundamental Theorem of Calculus part 1: $\frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x)$.
4. Substitute $x$ for $t$ in the integrand.

Answer: x^3 + 5x

1. Identify the integrand as $f(t) = \sin(t)$ and the upper limit as $g(x) = x^4$.
2. Apply the chain rule version of the theorem: $f(g(x)) \cdot g'(x)$.
3. The value of the integrand at the upper limit is $\sin(x^4)$.
4. The derivative of the upper limit $x^4$ is $4x^3$.
5. Multiply the two results together.

Answer: 4x^3 \sin(x^4)

1. Notice that the variable $x$ is in the lower limit of integration.
2. Use the property $\int_{x}^{a} f(t) \, dt = -\int_{a}^{x} f(t) \, dt$ to move $x$ to the upper limit.
3. Rewrite the function as $H(x) = -\int_{10}^{x} e^{t^2} \, dt$.
4. Apply the Fundamental Theorem of Calculus part 1 to the integral.
5. The derivative is the negative of the integrand evaluated at $x$.

Answer: -e^{x^2}

1. Find the antiderivative of $x^3 + 2x$ using the reverse power rule: $F(x) = \frac{1}{4}x^4 + x^2$.
2. Evaluate $F$ at the upper limit $x=2$: $F(2) = \frac{1}{4}(2)^4 + (2)^2 = \frac{16}{4} + 4 = 4 + 4 = 8$.
3. Evaluate $F$ at the lower limit $x=0$: $F(0) = \frac{1}{4}(0)^4 + (0)^2 = 0$.
4. Subtract the lower limit from the upper limit: $8 - 0 = 8$.

Answer: 8

1. Rewrite the function as $x^{-2}$ to use the power rule.
2. Find the antiderivative: $F(x) = \frac{x^{-1}}{-1} = -\frac{1}{x}$.
3. Evaluate at the upper limit $x=2$: $F(2) = -\frac{1}{2}$.
4. Evaluate at the lower limit $x=1$: $F(1) = -\frac{1}{1} = -1$.
5. Subtract: $-\frac{1}{2} - (-1) = -\frac{1}{2} + 1 = \frac{1}{2}$.

Answer: 1/2

1. Find the antiderivative of $\sin(x)$. Since the derivative of $\cos(x)$ is $-\sin(x)$, the antiderivative is $F(x) = -\cos(x)$.
2. Evaluate at the upper limit $x=\pi$: $F(\pi) = -\cos(\pi) = -(-1) = 1$.
3. Evaluate at the lower limit $x=0$: $F(0) = -\cos(0) = -(1) = -1$.
4. Subtract: $1 - (-1) = 2$.

Answer: 2

1. Apply linearity to split the integral into three parts: $6\int x^5 \, dx - 3\int \frac{1}{x} \, dx + 2\int e^x \, dx$.
2. Use the power rule for the first term: $6 \frac{x^6}{6} = x^6$.
3. Use the logarithm rule for the second term: $-3 \ln|x|$.
4. Use the exponential rule for the third term: $2e^x$.
5. Combine the results and add the constant of integration $C$.

Answer: x^6 - 3\ln|x| + 2e^x + C

1. Rewrite the first term as a power: $\int (x^{-3} + \cos x + 5^x) \, dx$.
2. Apply the power rule to $x^{-3}$: $\frac{x^{-2}}{-2} = -\frac{1}{2x^2}$.
3. Apply the trig rule to $\cos x$: $\sin x$.
4. Apply the general exponential rule to $5^x$: $\frac{5^x}{\ln 5}$.
5. Combine and add $C$.

Answer: -\frac{1}{2x^2} + \sin x + \frac{5^x}{\ln 5} + C

1. Rewrite the first term as $2x^{-1/2}$.
2. Integrate $2x^{-1/2}$ using the power rule: $2 \frac{x^{1/2}}{1/2} = 4\sqrt{x}$.
3. Integrate $-\sin x$: the antiderivative of $\sin x$ is $-\cos x$, so $-\sin x$ becomes $-(-\cos x) = \cos x$.
4. Integrate $\sec^2 x$: the antiderivative is $\tan x$.
5. Combine and add $C$.

Answer: 4\sqrt{x} + \cos x + \tan x + C

1. Let $u = x^3$. Then $\frac{du}{dx} = 3x^2$, which means $dx = \frac{du}{3x^2}$.
2. Substitute into the integral: $\int x^2 e^u \frac{du}{3x^2} = \frac{1}{3} \int e^u \, du$.
3. Using the known integral of $e^u$, we get $\frac{1}{3} e^u + C$.
4. Back-substitute $u = x^3$ to get $\frac{1}{3} e^{x^3} + C$.

Answer: \frac{1}{3} e^{x^3} + C

1. Let $u = 4x - 2$. Then $du = 4 \, dx$, so $dx = \frac{1}{4} du$.
2. Substitute into the integral: $\int \sin(u) \frac{1}{4} du = \frac{1}{4} \int \sin(u) \, du$.
3. The integral of $\sin(u)$ is $-\cos(u)$, so we have $-\frac{1}{4} \cos(u) + C$.
4. Back-substitute $u = 4x - 2$ to get $-\frac{1}{4} \cos(4x - 2) + C$.

Answer: -\frac{1}{4} \cos(4x - 2) + C

1. Let $u = 1 + e^x$. Then $du = e^x \, dx$.
2. The numerator $e^x \, dx$ is exactly $du$.
3. The integral becomes $\int \frac{1}{u} \, du = \ln|u| + C$.
4. Back-substitute $u = 1 + e^x$ to get $\ln(1 + e^x) + C$ (absolute value is not needed as $1+e^x$ is always positive).

Answer: \ln(1 + e^x) + C

1. Let $u = \cos x$. Then $du = -\sin x \, dx$, which means $\sin x \, dx = -du$.
2. Change the limits: when $x = 0$, $u = \cos(0) = 1$. When $x = \pi/2$, $u = \cos(\pi/2) = 0$.
3. The integral becomes $\int_{1}^{0} u^2 (-du) = \int_{0}^{1} u^2 \, du$.
4. Integrate $u^2$ to get $\frac{1}{3}u^3$ and evaluate from $0$ to $1$.
5. $\frac{1}{3}(1)^3 - \frac{1}{3}(0)^3 = \frac{1}{3}$.

Answer: 1/3

1. Let $u = 5x - 2$. Then $du = 5 \, dx$, which means $dx = \frac{1}{5} du$.
2. Substitute into the integral: $\int e^u \cdot \frac{1}{5} du$.
3. Pull out the constant: $\frac{1}{5} \int e^u \, du$.
4. Integrate $e^u$ to get $\frac{1}{5} e^u + C$.
5. Back-substitute $u = 5x - 2$ to get $\frac{1}{5} e^{5x-2} + C$.

Answer: \frac{1}{5} e^{5x-2} + C

1. Let $u = \sqrt{x+1}$. Then $u^2 = x + 1$, so $x = u^2 - 1$.
2. Differentiate $u^2 = x + 1$ to get $2u \, du = dx$.
3. Change limits: when $x = 0$, $u = \sqrt{1} = 1$. When $x = 1$, $u = \sqrt{2}$.
4. Substitute everything: $\int_{1}^{\sqrt{2}} (u^2 - 1) u (2u \, du) = \int_{1}^{\sqrt{2}} (2u^4 - 2u^2) \, du$.
5. Integrate: $[\frac{2}{5}u^5 - \frac{2}{3}u^3]_{1}^{\sqrt{2}}$.
6. Evaluate: $(\frac{2}{5}(\sqrt{2})^5 - \frac{2}{3}(\sqrt{2})^3) - (\frac{2}{5} - \frac{2}{3}) = (\frac{8\sqrt{2}}{5} - \frac{4\sqrt{2}}{3}) - (\frac{6-10}{15}) = \frac{24\sqrt{2}-20\sqrt{2}}{15} + \frac{4}{15} = \frac{4\sqrt{2}+4}{15}$.

Answer: \frac{4\sqrt{2}+4}{15}

1. Rewrite the integrand as $\sec^2 x (\sec x \tan x)$.
2. Let $u = \sec x$, then $du = \sec x \tan x \, dx$.
3. The integral becomes $\int u^2 \, du$.
4. Integrate to get $\frac{1}{3}u^3 + C$.
5. Substitute back $u = \sec x$.

Answer: \frac{1}{3}\sec^3 x + C

1. Peel off one sine: $\int (\sin^2 x)^2 \sin x \, dx$.
2. Use $\sin^2 x = 1 - \cos^2 x$ to get $\int (1 - \cos^2 x)^2 \sin x \, dx$.
3. Let $u = \cos x$, then $du = -\sin x \, dx$.
4. The integral becomes $-\int (1 - u^2)^2 \, du = -\int (1 - 2u^2 + u^4) \, du$.
5. Integrate: $-(u - \frac{2}{3}u^3 + \frac{1}{5}u^5) + C$.
6. Substitute back $u = \cos x$.

Answer: -\cos x + \frac{2}{3}\cos^3 x - \frac{1}{5}\cos^5 x + C

1. Both powers are odd, so we can choose either. Let's peel one cosine: $\int \cos^2 x \sin^3 x \cos x \, dx$.
2. Replace $\cos^2 x$ with $1 - \sin^2 x$: $\int (1 - \sin^2 x) \sin^3 x \cos x \, dx$.
3. Let $u = \sin x$, then $du = \cos x \, dx$.
4. The integral becomes $\int (1 - u^2) u^3 \, du = \int (u^3 - u^5) \, du$.
5. Integrate: $\frac{1}{4}u^4 - \frac{1}{6}u^6 + C$.
6. Substitute back $u = \sin x$.

Answer: \frac{1}{4}\sin^4 x - \frac{1}{6}\sin^6 x + C

1. Using LIATE, Logarithmic comes before Algebraic, so let $u = \ln x$ and $dv = x \, dx$.
2. Compute $du = \frac{1}{x} \, dx$ and $v = \frac{1}{2}x^2$.
3. Apply the formula: $\int u \, dv = \frac{1}{2}x^2 \ln x - \int \frac{1}{2}x^2 \cdot \frac{1}{x} \, dx$.
4. Simplify the integral: $\int \frac{1}{2}x \, dx = \frac{1}{4}x^2$.
5. Combine results: $\frac{1}{2}x^2 \ln x - \frac{1}{4}x^2 + C$.

Answer: \frac{1}{2}x^2 \ln x - \frac{1}{4}x^2 + C

1. Let $u = x$ and $dv = e^{2x} \, dx$.
2. Compute $du = dx$ and $v = \frac{1}{2}e^{2x}$.
3. Apply the formula: $\frac{1}{2}x e^{2x} - \int \frac{1}{2}e^{2x} \, dx$.
4. Integrate the remaining part: $\int \frac{1}{2}e^{2x} \, dx = \frac{1}{4}e^{2x}$.
5. Combine results: $\frac{1}{2}x e^{2x} - \frac{1}{4}e^{2x} + C$.

Answer: \frac{1}{2}x e^{2x} - \frac{1}{4}e^{2x} + C

1. Let $u = x^2$ and $dv = \cos x \, dx$, so $du = 2x \, dx$ and $v = \sin x$.
2. First pass: $x^2 \sin x - \int 2x \sin x \, dx$.
3. For the second integral, let $u = 2x$ and $dv = \sin x \, dx$, so $du = 2 \, dx$ and $v = -\cos x$.
4. Second pass: $\int 2x \sin x \, dx = -2x \cos x - \int -2 \cos x \, dx = -2x \cos x + 2 \sin x$.
5. Combine everything: $x^2 \sin x - (-2x \cos x + 2 \sin x) + C = x^2 \sin x + 2x \cos x - 2 \sin x + C$.

Answer: x^2 \sin x + 2x \cos x - 2 \sin x + C

1. Use integration by parts with $u = x^3$ and $dv = e^x dx$. This gives $x^3 e^x - \int 3x^2 e^x dx$.
2. Apply integration by parts again to $\int 3x^2 e^x dx$ with $u = 3x^2$ and $dv = e^x dx$. This gives $3x^2 e^x - \int 6x e^x dx$.
3. Apply integration by parts a third time to $\int 6x e^x dx$ with $u = 6x$ and $dv = e^x dx$. This gives $6x e^x - \int 6 e^x dx$.
4. The final integral is $6 e^x$. Combine all terms: $x^3 e^x - (3x^2 e^x - (6x e^x - 6 e^x))$.
5. Simplify the signs: $x^3 e^x - 3x^2 e^x + 6x e^x - 6 e^x + C$.

Answer: (x^3 - 3x^2 + 6x - 6)e^x + C

1. Let $I = \int e^{2x} \cos x dx$. Set $u = \cos x$ and $dv = e^{2x} dx$. Then $du = -\sin x dx$ and $v = \frac{1}{2} e^{2x}$.
2. First pass: $I = \frac{1}{2} e^{2x} \cos x + \int \frac{1}{2} e^{2x} \sin x dx$.
3. Second pass on $\int \frac{1}{2} e^{2x} \sin x dx$: set $u = \sin x$ and $dv = \frac{1}{2} e^{2x} dx$. Then $du = \cos x dx$ and $v = \frac{1}{4} e^{2x}$.
4. The second integral is $\frac{1}{4} e^{2x} \sin x - \int \frac{1}{4} e^{2x} \cos x dx$.
5. Substitute back: $I = \frac{1}{2} e^{2x} \cos x + \frac{1}{4} e^{2x} \sin x - \frac{1}{4} I$.
6. Add $\frac{1}{4} I$ to both sides: $\frac{5}{4} I = e^{2x} (\frac{1}{2} \cos x + \frac{1}{4} \sin x)$.
7. Multiply by $\frac{4}{5}$: $I = \frac{4}{5} e^{2x} (\frac{1}{2} \cos x + \frac{1}{4} \sin x) = \frac{e^{2x}}{5} (2 \cos x + \sin x) + C$.

Answer: \frac{e^{2x}}{5} (2 \cos x + \sin x) + C

1. Set $u = x^2$ and $dv = \sin x dx$. Then $du = 2x dx$ and $v = -\cos x$.
2. First pass: $-x^2 \cos x + \int 2x \cos x dx$.
3. Second pass on $\int 2x \cos x dx$: set $u = 2x$ and $dv = \cos x dx$. Then $du = 2 dx$ and $v = \sin x$.
4. The second integral is $2x \sin x - \int 2 \sin x dx$.
5. The final integral is $-2(-\cos x) = 2 \cos x$.
6. Combine: $-x^2 \cos x + 2x \sin x + 2 \cos x + C$.

Answer: (2 - x^2)\cos x + 2x \sin x + C

1. Since the power is odd, peel off one $\sin x$ factor: $\int (\sin^2 x)^2 \sin x \, dx$.
2. Use the identity $\sin^2 x = 1 - \cos^2 x$ to get $\int (1 - \cos^2 x)^2 \sin x \, dx$.
3. Let $u = \cos x$, then $du = -\sin x \, dx$. The integral becomes $-\int (1 - u^2)^2 \, du$.
4. Expand the binomial: $-\int (1 - 2u^2 + u^4) \, du$.
5. Integrate term by term: $-(u - \frac{2}{3}u^3 + \frac{1}{5}u^5) + C$.
6. Substitute back $u = \cos x$: $-\cos x + \frac{2}{3}\cos^3 x - \frac{1}{5}\cos^5 x + C$.

Answer: -\cos x + \frac{2}{3}\cos^3 x - \frac{1}{5}\cos^5 x + C

1. Both powers are odd, so we can choose either. Let's peel one $\cos x$ and let $u = \sin x$.
2. The integral becomes $\int \sin^3 x (\cos^2 x) \cos x \, dx = \int \sin^3 x (1 - \sin^2 x) \cos x \, dx$.
3. Substitute $u = \sin x$ and $du = \cos x \, dx$: $\int u^3(1 - u^2) \, du$.
4. Distribute: $\int (u^3 - u^5) \, du$.
5. Integrate: $\frac{1}{4}u^4 - \frac{1}{6}u^6 + C$.
6. Substitute back: $\frac{1}{4}\sin^4 x - \frac{1}{6}\sin^6 x + C$.

Answer: \frac{1}{4}\sin^4 x - \frac{1}{6}\sin^6 x + C

1. The power is even, so use the half-angle identity: $\int (\frac{1 + \cos(2x)}{2})^2 \, dx$.
2. Expand the square: $\int \frac{1 + 2\cos(2x) + \cos^2(2x)}{4} \, dx$.
3. Split the integral: $\frac{1}{4} \int 1 \, dx + \frac{1}{2} \int \cos(2x) \, dx + \frac{1}{4} \int \cos^2(2x) \, dx$.
4. Apply the half-angle identity again to $\cos^2(2x)$: $\int \frac{1 + \cos(4x)}{2} \, dx$.
5. Integrate all parts: $\frac{1}{4}x + \frac{1}{4}\sin(2x) + \frac{1}{4}(\frac{1}{2}x + \frac{1}{8}\sin(4x)) + C$.
6. Simplify: $\frac{1}{4}x + \frac{1}{4}\sin(2x) + \frac{1}{8}x + \frac{1}{32}\sin(4x) + C = \frac{3}{8}x + \frac{1}{4}\sin(2x) + \frac{1}{32}\sin(4x) + C$.

Answer: \frac{3}{8}x + \frac{1}{4}\sin(2x) + \frac{1}{32}\sin(4x) + C

1. Identify $a = 5$ and use the substitution $x = 5 \sin \theta$.
2. Compute the differential $dx = 5 \cos \theta \, d\theta$.
3. Substitute into the integral: $\int \frac{5 \cos \theta}{\sqrt{25 - 25 \sin^2 \theta}} \, d\theta = \int \frac{5 \cos \theta}{5 \cos \theta} \, d\theta$.
4. Integrate $1 \, d\theta$ to get $\theta + C$.
5. Back-substitute using $\sin \theta = x/5$, so $\theta = \arcsin(x/5)$.

Answer: \arcsin\left(\frac{x}{5}\right) + C

1. Use $x = 2 \sin \theta$, so $dx = 2 \cos \theta \, d\theta$.
2. The term $(4 - x^2)^{3/2}$ becomes $(4 \cos^2 \theta)^{3/2} = 8 \cos^3 \theta$.
3. The integral becomes $\int \frac{2 \cos \theta}{8 \cos^3 \theta} \, d\theta = \frac{1}{4} \int \sec^2 \theta \, d\theta$.
4. Integrate to get $\frac{1}{4} \tan \theta + C$.
5. Using the triangle, $\tan \theta = \frac{x}{\sqrt{4 - x^2}}$.

Answer: \frac{x}{4\sqrt{4 - x^2}} + C

1. Use $x = 3 \sin \theta$, so $dx = 3 \cos \theta \, d\theta$.
2. The integral becomes $\int (3 \cos \theta)(3 \cos \theta) \, d\theta = 9 \int \cos^2 \theta \, d\theta$.
3. Use the identity $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$ to get $\frac{9}{2} \int (1 + \cos(2\theta)) \, d\theta$.
4. Integrate to get $\frac{9}{2} \theta + \frac{9}{4} \sin(2\theta) + C$.
5. Convert back: $\theta = \arcsin(x/3)$ and $\sin(2\theta) = 2(x/3)(\sqrt{9-x^2}/3) = \frac{2x\sqrt{9-x^2}}{9}$.

Answer: \frac{9}{2}\arcsin\left(\frac{x}{3}\right) + \frac{x\sqrt{9 - x^2}}{2} + C

1. Identify the form $\sqrt{a^2 + x^2}$ with $a = 5$.
2. Substitute $x = 5 \tan \theta$, so $dx = 5 \sec^2 \theta \, d\theta$.
3. The denominator becomes $25 \tan^2 \theta + 25 = 25 \sec^2 \theta$.
4. The integral becomes $\int \frac{5 \sec^2 \theta}{25 \sec^2 \theta} \, d\theta = \int \frac{1}{5} \, d\theta = \frac{1}{5} \theta + C$.
5. Since $\tan \theta = \frac{x}{5}$, then $\theta = \arctan(\frac{x}{5})$.

Answer: \frac{1}{5} \arctan\left(\frac{x}{5}\right) + C

1. Identify the form $\sqrt{x^2 - a^2}$ with $a = 3$.
2. Substitute $x = 3 \sec \theta$, so $dx = 3 \sec \theta \tan \theta \, d\theta$.
3. The square root becomes $\sqrt{9 \sec^2 \theta - 9} = 3 \tan \theta$.
4. The integral becomes $\int \frac{3 \sec \theta \tan \theta}{3 \tan \theta} \, d\theta = \int \sec \theta \, d\theta$.
5. The antiderivative of $\sec \theta$ is $\ln|\sec \theta + \tan \theta| + C$.
6. Substitute back $\sec \theta = \frac{x}{3}$ and $\tan \theta = \frac{\sqrt{x^2 - 9}}{3}$, giving $\ln|\frac{x + \sqrt{x^2 - 9}}{3}| + C$.

Answer: \ln\left|\frac{x + \sqrt{x^2 - 9}}{3}\right| + C

1. Substitute $x = 2 \tan \theta$, so $dx = 2 \sec^2 \theta \, d\theta$.
2. The square root is $2 \sec \theta$ and $x^3 = 8 \tan^3 \theta$.
3. The integral becomes $\int \frac{8 \tan^3 \theta}{2 \sec \theta} (2 \sec^2 \theta) \, d\theta = 8 \int \tan^3 \theta \sec \theta \, d\theta$.
4. Rewrite as $8 \int \tan \theta (\tan^2 \theta \sec \theta) \, d\theta = 8 \int \tan \theta (\sec^2 \theta - 1) \sec \theta \, d\theta$.
5. Alternatively, use $u = \sec \theta$, then $du = \sec \theta \tan \theta \, d\theta$.
6. The integral is $8 \int (u^2 - 1) \, du = 8 (\frac{u^3}{3} - u) + C = 8 (\frac{\sec^3 \theta}{3} - \sec \theta) + C$.
7. Substitute back $\sec \theta = \frac{\sqrt{x^2 + 4}}{2}$.

Answer: \frac{1}{3}(x^2+4)^{3/2} - 4\sqrt{x^2+4} + C

1. Factor the denominator: $x^2-x-6 = (x-3)(x+2)$.
2. Set up decomposition: $\frac{3x+1}{(x-3)(x+2)} = \frac{A}{x-3} + \frac{B}{x+2}$.
3. Multiply by denominator: $3x+1 = A(x+2) + B(x-3)$.
4. Find $A$ by letting $x=3$: $3(3)+1 = A(3+2) \implies 10 = 5A \implies A=2$.
5. Find $B$ by letting $x=-2$: $3(-2)+1 = B(-2-3) \implies -5 = -5B \implies B=1$.
6. Integrate: $\int (\frac{2}{x-3} + \frac{1}{x+2}) \, dx = 2\ln|x-3| + \ln|x+2| + C$.

Answer: 2\ln|x-3| + \ln|x+2| + C

1. Factor the denominator: $x^2-4x = x(x-4)$.
2. Set up decomposition: $\frac{1}{x(x-4)} = \frac{A}{x} + \frac{B}{x-4}$.
3. Multiply by denominator: $1 = A(x-4) + Bx$.
4. Find $A$ by letting $x=0$: $1 = A(-4) \implies A = -1/4$.
5. Find $B$ by letting $x=4$: $1 = B(4) \implies B = 1/4$.
6. Integrate: $\int (\frac{-1/4}{x} + \frac{1/4}{x-4}) \, dx = \frac{1}{4}(\ln|x-4| - \ln|x|) + C$.

Answer: \frac{1}{4}\ln|\frac{x-4}{x}| + C

1. Factor the denominator: $x^3-x = x(x-1)(x+1)$.
2. Set up decomposition: $\frac{x^2+1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}$.
3. Multiply by denominator: $x^2+1 = A(x-1)(x+1) + Bx(x+1) + Cx(x-1)$.
4. Find $A$ by letting $x=0$: $1 = A(-1)(1) \implies A = -1$.
5. Find $B$ by letting $x=1$: $1^2+1 = B(1)(2) \implies 2 = 2B \implies B = 1$.
6. Find $C$ by letting $x=-1$: $(-1)^2+1 = C(-1)(-2) \implies 2 = 2C \implies C = 1$.
7. Integrate: $\int (\frac{-1}{x} + \frac{1}{x-1} + \frac{1}{x+1}) \, dx = -\ln|x| + \ln|x-1| + \ln|x+1| + C$.

Answer: \ln|\frac{x^2-1}{x}| + C

1. Factor the denominator: $x^2-4x+4 = (x-2)^2$.
2. Set up partial fractions: $\frac{1}{(x-2)^2} = \frac{A}{x-2} + \frac{B}{(x-2)^2}$.
3. Solve for constants: $1 = A(x-2) + B$. Setting $x=2$ gives $B=1$. Comparing coefficients of $x$ gives $A=0$.
4. Integrate: $\int \frac{1}{(x-2)^2} \, dx = -(x-2)^{-1} + C$.

Answer: -\frac{1}{x-2} + C

1. Factor the denominator: $x(x^2+1)$.
2. Set up partial fractions: $\frac{1}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1}$.
3. Solve for constants: $1 = A(x^2+1) + (Bx+C)x$.
4. Comparing coefficients: $x^2$ terms: $A+B=0$; $x$ terms: $C=0$; constants: $A=1$. Thus $B=-1$.
5. Integrate: $\int (\frac{1}{x} - \frac{x}{x^2+1}) \, dx = \ln|x| - \frac{1}{2}\ln(x^2+1) + C$.

Answer: \ln|x| - \frac{1}{2}\ln(x^2+1) + C

1. Check degree: Numerator degree is 2, denominator degree is 2. The fraction is improper.
2. Perform long division: $\frac{x^2}{x^2+1} = 1 - \frac{1}{x^2+1}$.
3. Integrate the resulting terms: $\int 1 \, dx - \int \frac{1}{x^2+1} \, dx$.
4. Apply basic integration rules: $x - \arctan(x) + C$.

Answer: x - \arctan(x) + C

1. Substitute $\cos x = \frac{1-t^2}{1+t^2}$ and $dx = \frac{2}{1+t^2} dt$.
2. The denominator becomes $1 - \frac{1-t^2}{1+t^2} = \frac{1+t^2 - (1-t^2)}{1+t^2} = \frac{2t^2}{1+t^2}$.
3. The integral is $\int \frac{1}{2t^2 / (1+t^2)} \cdot \frac{2}{1+t^2} \, dt = \int \frac{1}{t^2} \, dt$.
4. Integrating $t^{-2}$ gives $-t^{-1} + C = -\frac{1}{t} + C$.
5. Substitute back $t = \tan(x/2)$ to get $-\cot(x/2) + C$.

Answer: -\cot\left(\frac{x}{2}\right) + C

1. Substitute $\cos x = \frac{1-t^2}{1+t^2}$ and $dx = \frac{2}{1+t^2} dt$.
2. The denominator is $2 + \frac{1-t^2}{1+t^2} = \frac{2(1+t^2) + 1-t^2}{1+t^2} = \frac{3+t^2}{1+t^2}$.
3. The integral becomes $\int \frac{1+t^2}{3+t^2} \cdot \frac{2}{1+t^2} \, dt = \int \frac{2}{3+t^2} \, dt$.
4. This is an arctangent form: $2 \cdot \frac{1}{\sqrt{3}} \arctan\left(\frac{t}{\sqrt{3}}\right) + C$.
5. Substitute back $t = \tan(x/2)$.

Answer: \frac{2}{\sqrt{3}} \arctan\left(\frac{\tan(x/2)}{\sqrt{3}}\right) + C

1. Substitute $\sin x = \frac{2t}{1+t^2}$, $\cos x = \frac{1-t^2}{1+t^2}$, and $dx = \frac{2}{1+t^2} dt$.
2. The integral becomes $\int \frac{2t/(1+t^2)}{1 + (2t + 1 - t^2)/(1+t^2)} \cdot \frac{2}{1+t^2} \, dt$.
3. Simplify the denominator: $1 + \frac{2t+1-t^2}{1+t^2} = \frac{1+t^2+2t+1-t^2}{1+t^2} = \frac{2+2t}{1+t^2}$.
4. The integral is $\int \frac{2t}{1+t^2} \cdot \frac{1+t^2}{2+2t} \cdot \frac{2}{1+t^2} \, dt = \int \frac{2t}{(1+t)(1+t^2)} \, dt$.
5. Use partial fractions: $\frac{2t}{(1+t)(1+t^2)} = \frac{A}{1+t} + \frac{Bt+C}{1+t^2}$. Solving gives $A=-1, B=1, C=1$.
6. Integrate: $\int (\frac{-1}{1+t} + \frac{t}{1+t^2} + \frac{1}{1+t^2}) \, dt = -\ln|1+t| + \frac{1}{2}\ln(1+t^2) + \arctan(t) + C$.

Answer: -\ln|1 + \tan(x/2)| + \frac{1}{2}\ln(1 + \tan^2(x/2)) + \frac{x}{2} + C

1. Identify the integrand as a product of an algebraic function $x^2$ and a trigonometric function $\cos x$.
2. Since there is no clear u-substitution (the derivative of $x^2$ is $2x$, not $\cos x$), use integration by parts.
3. Set $u = x^2$ and $dv = \cos x \, dx$, which gives $du = 2x \, dx$ and $v = \sin x$.
4. Apply the formula: $\int x^2 \cos x \, dx = x^2 \sin x - \int 2x \sin x \, dx$.
5. Apply integration by parts again to $\int 2x \sin x \, dx$ with $u = 2x$ and $dv = \sin x \, dx$.
6. This gives $2x(-\cos x) - \int 2(-\cos x) \, dx = -2x \cos x + 2 \sin x$.

Answer: x^2 \sin x + 2x \cos x - 2 \sin x + C

1. Check for u-substitution: the derivative of $x^2+2x-3$ is $2x+2$, which is not a multiple of $x+5$.
2. Recognize this as a rational function. Factor the denominator: $x^2+2x-3 = (x+3)(x-1)$.
3. Use partial fraction decomposition: $\frac{x+5}{(x+3)(x-1)} = \frac{A}{x+3} + \frac{B}{x-1}$.
4. Solve for A: $A = \frac{-3+5}{-3-1} = \frac{2}{-4} = -\frac{1}{2}$.
5. Solve for B: $B = \frac{1+5}{1+3} = \frac{6}{4} = \frac{3}{2}$.
6. Integrate: $\int (\frac{-1/2}{x+3} + \frac{3/2}{x-1}) \, dx = -\frac{1}{2}\ln|x+3| + \frac{3}{2}\ln|x-1| + C$.

Answer: \frac{3}{2}\ln|x-1| - \frac{1}{2}\ln|x+3| + C

1. Identify the form $\sqrt{a^2-x^2}$ in the denominator, where $a=3$.
2. This is a cue for trigonometric substitution. Set $x = 3 \sin \theta$.
3. Then $dx = 3 \cos \theta \, d\theta$ and $\sqrt{9-x^2} = \sqrt{9(1-\sin^2 \theta)} = 3 \cos \theta$.
4. Substitute into the integral: $\int \frac{3 \cos \theta}{3 \cos \theta} \, d\theta = \int 1 \, d\theta = \theta + C$.
5. Convert back to x: since $x/3 = \sin \theta$, then $\theta = \arcsin(x/3)$.

Answer: \arcsin\left(\frac{x}{3}\right) + C

1. Identify that $x$ is a multiple of the derivative of the exponent $x^2$.
2. Let $u = x^2$, then $du = 2x \, dx$, which means $x \, dx = \frac{1}{2} du$.
3. Substitute into the integral: $\int \frac{1}{2} e^u \, du$.
4. Integrate to get $\frac{1}{2} e^u + C$.
5. Substitute back $u = x^2$.

Answer: \frac{1}{2} e^{x^2} + C

1. Factor the denominator as $(x-1)(x+1)$.
2. Use partial fraction decomposition: $\frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$.
3. Solve for constants: $1 = A(x+1) + B(x-1)$. Setting $x=1$ gives $A=1/2$; setting $x=-1$ gives $B=-1/2$.
4. Integrate $\int (\frac{1/2}{x-1} - \frac{1/2}{x+1}) \, dx$.
5. Result is $\frac{1}{2}\ln|x-1| - \frac{1}{2}\ln|x+1| + C$.

Answer: \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C

1. Use integration by parts with $u = x^3$ and $dv = \cos(x) \, dx$.
2. First pass: $u=x^3, v=\sin(x)$. Integral becomes $x^3\sin(x) - \int 3x^2\sin(x) \, dx$.
3. Second pass on $\int 3x^2\sin(x) \, dx$: $u=3x^2, v=-\cos(x)$. Integral becomes $-3x^2\cos(x) + \int 6x\cos(x) \, dx$.
4. Third pass on $\int 6x\cos(x) \, dx$: $u=6x, v=\sin(x)$. Integral becomes $6x\sin(x) - \int 6\sin(x) \, dx$.
5. Final integration: $- \int 6\sin(x) \, dx = 6\cos(x)$.
6. Combine all terms: $x^3\sin(x) + 3x^2\cos(x) - 6x\sin(x) - 6\cos(x) + C$.

Answer: (x^3 - 6x)\sin(x) + (3x^2 - 6)\cos(x) + C

1. Use integration by parts with $u = (\ln x)^2$ and $dv = x \, dx$.
2. Then $du = 2\ln x \cdot \frac{1}{x} \, dx$ and $v = \frac{1}{2}x^2$.
3. The formula gives $\frac{1}{2}x^2(\ln x)^2 - \int \frac{1}{2}x^2 \cdot \frac{2\ln x}{x} \, dx$, which simplifies to $\frac{1}{2}x^2(\ln x)^2 - \int x \ln x \, dx$.
4. Apply integration by parts again to $\int x \ln x \, dx$ with $u = \ln x$ and $dv = x \, dx$.
5. This gives $\frac{1}{2}x^2 \ln x - \int \frac{1}{2}x \, dx = \frac{1}{2}x^2 \ln x - \frac{1}{4}x^2$.
6. Combine all terms: $\frac{1}{2}x^2(\ln x)^2 - \frac{1}{2}x^2 \ln x + \frac{1}{4}x^2 + C$.

Answer: \frac{1}{2}x^2(\ln x)^2 - \frac{1}{2}x^2 \ln x + \frac{1}{4}x^2 + C

1. Use the Weierstrass substitution $t = \tan(x/2)$.
2. Substitute $\cos x = \frac{1-t^2}{1+t^2}$ and $dx = \frac{2}{1+t^2} \, dt$.
3. The integral becomes $\int \frac{1}{1 + \frac{1-t^2}{1+t^2}} \cdot \frac{2}{1+t^2} \, dt$.
4. Simplify the denominator: $1 + \frac{1-t^2}{1+t^2} = \frac{1+t^2+1-t^2}{1+t^2} = \frac{2}{1+t^2}$.
5. The integral simplifies to $\int \frac{1+t^2}{2} \cdot \frac{2}{1+t^2} \, dt = \int 1 \, dt = t + C$.
6. Substitute back $t = \tan(x/2)$.

Answer: \tan(x/2) + C

1. Use integration by parts with $u = \sin(2x)$ and $dv = e^x \, dx$.
2. Then $du = 2\cos(2x) \, dx$ and $v = e^x$.
3. The first pass gives $e^x \sin(2x) - \int 2e^x \cos(2x) \, dx$.
4. Apply integration by parts again to $\int 2e^x \cos(2x) \, dx$ with $u = 2\cos(2x)$ and $dv = e^x \, dx$.
5. Then $du = -4\sin(2x) \, dx$ and $v = e^x$.
6. This gives $2e^x \cos(2x) - \int -4e^x \sin(2x) \, dx = 2e^x \cos(2x) + 4\int e^x \sin(2x) \, dx$.
7. Let $I = \int e^x \sin(2x) \, dx$. We have $I = e^x \sin(2x) - (2e^x \cos(2x) + 4I)$.
8. Solve for $I$: $5I = e^x \sin(2x) - 2e^x \cos(2x) \implies I = \frac{e^x}{5}(\sin(2x) - 2\cos(2x)) + C$.

Answer: \frac{e^x}{5}(\sin(2x) - 2\cos(2x)) + C

1. Rewrite as a limit: $\lim_{R \to \infty} \int_{1}^{R} x^{-3} \, dx$.
2. Find the antiderivative: $\int x^{-3} \, dx = \frac{x^{-2}}{-2} = -\frac{1}{2x^2}$.
3. Evaluate from $1$ to $R$: $\lim_{R \to \infty} [-\frac{1}{2R^2} - (-\frac{1}{2(1)^2})]$.
4. Take the limit: $0 + \frac{1}{2} = \frac{1}{2}$.

Answer: Converges to $\frac{1}{2}$

1. Rewrite as a limit: $\lim_{R \to \infty} \int_{1}^{R} \frac{1}{x^2+1} \, dx$.
2. The antiderivative is $\arctan(x)$.
3. Evaluate from $1$ to $R$: $\lim_{R \to \infty} [\arctan(R) - \arctan(1)]$.
4. The limit of $\arctan(R)$ as $R \to \infty$ is $\pi/2$.
5. The value of $\arctan(1)$ is $\pi/4$.
6. Result: $\pi/2 - \pi/4 = \pi/4$.

Answer: Converges to $\frac{\pi}{4}$

1. Rewrite as $\lim_{R \to \infty} \int_{0}^{R} x e^{-x} \, dx$.
2. Use integration by parts: $u=x, dv=e^{-x}dx \implies du=dx, v=-e^{-x}$.
3. Integral is $[-xe^{-x}]_0^R - \int_0^R -e^{-x} dx = [-xe^{-x} - e^{-x}]_0^R$.
4. Evaluate: $\lim_{R \to \infty} (-Re^{-R} - e^{-R}) - (0 - e^0)$.
5. Since $Re^{-R} \to 0$ and $e^{-R} \to 0$ as $R \to \infty$, we are left with $0 - (-1) = 1$.

Answer: Converges to $1$

1. Identify the discontinuity at $x=0$.
2. Rewrite as a limit: $\lim_{t \to 0^{+}} \int_{t}^{1} x^{-3/2} \, dx$.
3. Find the antiderivative: $\int x^{-3/2} \, dx = -2x^{-1/2} = -\frac{2}{\sqrt{x}}$.
4. Evaluate the limit: $\lim_{t \to 0^{+}} [-\frac{2}{\sqrt{1}} - (-\frac{2}{\sqrt{t}})] = \lim_{t \to 0^{+}} (-2 + \frac{2}{\sqrt{t}})$.
5. As $t \to 0^{+}$, the term $\frac{2}{\sqrt{t}}$ approaches infinity.

Answer: Diverges

1. Observe that for $x \in (0, 1]$, $\sqrt{x} + x^2 > \sqrt{x}$.
2. Therefore, $\frac{1}{\sqrt{x} + x^2} < \frac{1}{\sqrt{x}}$.
3. The integral $\int_{0}^{1} \frac{1}{\sqrt{x}} \, dx$ is a p-integral with $p = 1/2$.
4. Since $p < 1$, the integral $\int_{0}^{1} \frac{1}{\sqrt{x}} \, dx$ converges.
5. By the comparison test, since the original function is smaller than a convergent function, it also converges.

Answer: Converges

1. Identify the discontinuity at $x=0$. Split the integral: $\int_{-1}^{0} x^{-1/3} \, dx + \int_{0}^{1} x^{-1/3} \, dx$.
2. Evaluate the right side: $\lim_{t \to 0^{+}} [\frac{3}{2}x^{2/3}]_{t}^{1} = \frac{3}{2}(1)^{2/3} - 0 = \frac{3}{2}$.
3. Evaluate the left side: $\lim_{t \to 0^{-}} [\frac{3}{2}x^{2/3}]_{-1}^{t} = 0 - \frac{3}{2}(-1)^{2/3} = -\frac{3}{2}$.
4. Sum the two parts: $\frac{3}{2} + (-\frac{3}{2}) = 0$.

Answer: 0

1. Calculate $h = (2-1)/4 = 0.25$.
2. Find x values: $1, 1.25, 1.5, 1.75, 2$.
3. Evaluate $f(x) = \ln(x)$: $\ln(1)=0, \ln(1.25) \approx 0.223, \ln(1.5) \approx 0.405, \ln(1.75) \approx 0.560, \ln(2) \approx 0.693$.
4. Apply formula: $\frac{0.25}{2} [0 + 2(0.223 + 0.405 + 0.560) + 0.693]$.
5. Compute: $0.125 [0 + 2(1.188) + 0.693] = 0.125 [2.376 + 0.693] = 0.125 [3.069] \approx 0.384$.

Answer: 0.384

1. Calculate $h = (1-0)/4 = 0.25$.
2. Find x values: $0, 0.25, 0.5, 0.75, 1$.
3. Evaluate $f(x) = 1/(1+x^2)$: $f(0)=1, f(0.25)=0.941, f(0.5)=0.8, f(0.75)=0.64, f(1)=0.5$.
4. Apply formula: $\frac{0.25}{3} [1 + 4(0.941) + 2(0.8) + 4(0.64) + 0.5]$.
5. Compute: $\frac{0.25}{3} [1 + 3.764 + 1.6 + 2.56 + 0.5] = \frac{0.25}{3} [9.424] \approx 0.785$.

Answer: 0.785

1. Simpson's rule is based on quadratic interpolation. The error term for Simpson's rule involves the fourth derivative of the function.
2. For $f(x) = x^3 + 2x$, the first derivative is $3x^2 + 2$, the second is $6x$, the third is $6$, and the fourth is $0$.
3. Since the fourth derivative is zero, the error is zero, making the approximation exact.
4. Using FTC: $\int_{0}^{2} (x^3 + 2x) \, dx = [\frac{1}{4}x^4 + x^2]_{0}^{2} = (\frac{16}{4} + 4) - (0) = 4 + 4 = 8$.

Answer: 8

1. Use the linearity of the integral to pull the constant 5 outside: $5 \int \sec x \, dx$.
2. Apply the standard formula $\int \sec x \, dx = \ln |\sec x + \tan x|$.
3. Combine the results and add the constant of integration.

Answer: $5 \ln |\sec x + \tan x| + C$

1. Find the antiderivative: $\ln |\sec x + \tan x|$.
2. Evaluate at the upper limit $\pi/6$: $\sec(\pi/6) = 2/\sqrt{3}$ and $\tan(\pi/6) = 1/\sqrt{3}$.
3. The sum is $3/\sqrt{3} = \sqrt{3}$, so the value is $\ln(\sqrt{3})$.
4. Evaluate at the lower limit $0$: $\sec(0) = 1$ and $\tan(0) = 0$, giving $\ln(1) = 0$.
5. Subtract the lower limit from the upper limit.

Answer: $\ln(\sqrt{3})$

1. Use u-substitution: let $u = 3x$, then $du = 3 \, dx$, which means $dx = \frac{1}{3} du$.
2. Rewrite the integral as $\frac{1}{3} \int \sec u \, du$.
3. Apply the formula: $\frac{1}{3} \ln |\sec u + \tan u| + C$.
4. Substitute $3x$ back in for $u$.

Answer: $\frac{1}{3} \ln |\sec(3x) + \tan(3x)| + C$

1. Identify the interval width: $b - a = 2 - 0 = 2$.
2. Set up the integral: $\int_{0}^{2} (4x^3 - 2x) \, dx$.
3. Find the antiderivative: $x^4 - x^2$.
4. Evaluate the definite integral: $(2^4 - 2^2) - (0^4 - 0^2) = 16 - 4 = 12$.
5. Divide by the width: $f_{\text{avg}} = \frac{12}{2} = 6$.

Answer: 6

1. Identify the interval width: $e - 1$.
2. Set up the integral: $\int_{1}^{e} \frac{1}{x} \, dx$.
3. Find the antiderivative: $\ln |x|$.
4. Evaluate the definite integral: $\ln(e) - \ln(1) = 1 - 0 = 1$.
5. Divide by the width: $f_{\text{avg}} = \frac{1}{e-1}$.

Answer: \frac{1}{e-1}

1. Interval width is $2 - 0 = 2$.
2. Integral is $\int_{0}^{2} 3x^2 \, dx = [x^3]_{0}^{2} = 8 - 0 = 8$.
3. Average value is $f_{\text{avg}} = \frac{8}{2} = 4$.
4. Set $f(c) = 4$, so $3c^2 = 4$.
5. Solve for $c$: $c^2 = \frac{4}{3}$, so $c = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$ (since $c$ must be in $[0, 2]$).

Answer: f_{\text{avg}} = 4, c = \frac{2}{\sqrt{3}}

1. Check the symmetry of $f(x) = x \sin(x)$.
2. Substitute $-x$: $f(-x) = (-x) \sin(-x)$.
3. Since $\sin(-x) = -\sin(x)$, we have $f(-x) = (-x)(-\sin(x)) = x \sin(x)$.
4. The function is even, so the integral is $2 \int_{0}^{\pi} x \sin(x) \, dx$.
5. Use integration by parts: $\int x \sin(x) \, dx = -x \cos(x) + \sin(x)$.
6. Evaluate from $0$ to $\pi$: $[-\pi \cos(\pi) + \sin(\pi)] - [0 + 0] = [-\pi(-1) + 0] = \pi$.
7. Multiply by $2$ to get $2\pi$.

Answer: 2\pi

1. Let $f(x) = (x^5 + x^3 + x) \cos(x)$.
2. Check symmetry: $f(-x) = ((-x)^5 + (-x)^3 + (-x)) \cos(-x)$.
3. This simplifies to $(-x^5 - x^3 - x) \cos(x) = -(x^5 + x^3 + x) \cos(x)$.
4. Since $f(-x) = -f(x)$, the function is odd.
5. The integral of an odd function over a symmetric interval $[-1, 1]$ is $0$.

Answer: 0

1. Let $I = \int_{0}^{\pi/4} \frac{1}{1 + \tan(x)} \, dx$.
2. Apply the King Property: replace $x$ with $\pi/4 - x$.
3. Use the identity $\tan(\pi/4 - x) = \frac{1 - \tan(x)}{1 + \tan(x)}$.
4. The integrand becomes $\frac{1}{1 + \frac{1 - \tan(x)}{1 + \tan(x)}} = \frac{1 + \tan(x)}{(1 + \tan(x)) + (1 - \tan(x))} = \frac{1 + \tan(x)}{2}$.
5. Add the two versions of $I$: $2I = \int_{0}^{\pi/4} (\frac{1}{1 + \tan(x)} + \frac{1 + \tan(x)}{2}) \, dx$.
6. Alternatively, rewrite the integrand as $\frac{\cos(x)}{\cos(x) + \sin(x)}$.
7. By King Property, $I = \int_{0}^{\pi/4} \frac{\sin(x)}{\sin(x) + \cos(x)} \, dx$.
8. Then $2I = \int_{0}^{\pi/4} 1 \, dx = \pi/4$.
9. Thus $I = \pi/8$.

Answer: \pi/8

1. Identify this as a product of a polynomial and a trig function, suggesting integration by parts.
2. Let $u = x^2$ and $dv = \cos x \, dx$. Then $du = 2x \, dx$ and $v = \sin x$.
3. Apply the formula: $\int x^2 \cos x \, dx = x^2 \sin x - \int 2x \sin x \, dx$.
4. Apply integration by parts again to $\int 2x \sin x \, dx$ with $u = 2x$ and $dv = \sin x \, dx$.
5. Then $du = 2 \, dx$ and $v = -\cos x$.
6. The second integral becomes $-2x \cos x - \int -2 \cos x \, dx = -2x \cos x + 2 \sin x$.
7. Combine everything: $x^2 \sin x - (-2x \cos x + 2 \sin x) + C = x^2 \sin x + 2x \cos x - 2 \sin x + C$.

Answer: x^2 \sin x + 2x \cos x - 2 \sin x + C

1. Recognize the integrand as the derivative of the arctangent function.
2. The antiderivative of $\frac{1}{1+x^2}$ is $\arctan(x)$.
3. Evaluate at the boundaries: $\arctan(1) - \arctan(0)$.
4. Since $\tan(\pi/4) = 1$ and $\tan(0) = 0$, the values are $\pi/4$ and $0$.

Answer: \frac{\pi}{4}

1. Factor the denominator: $x^2 - 5x + 6 = (x-2)(x-3)$.
2. Use partial fraction decomposition: $\frac{1}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$.
3. Solve for constants: $1 = A(x-3) + B(x-2)$.
4. Setting $x=3$ gives $B=1$. Setting $x=2$ gives $A=-1$.
5. The integral becomes $\int \frac{-1}{x-2} \, dx + \int \frac{1}{x-3} \, dx$.
6. Integrate to get $-\ln|x-2| + \ln|x-3| + C = \ln\left|\frac{x-3}{x-2}\right| + C$.

Answer: \ln\left|\frac{x-3}{x-2}\right| + C

1. Find intersections: $x^2 - 4 = 0 \implies x = \pm 2$.
2. Determine top/bottom: For $x=0$, $y = -4$, so the x-axis ($y=0$) is the top boundary.
3. Set up integral: $\int_{-2}^{2} (0 - (x^2 - 4)) \, dx = \int_{-2}^{2} (4 - x^2) \, dx$.
4. Integrate: $[4x - \frac{1}{3}x^3]_{-2}^{2}$.
5. Evaluate: $(8 - \frac{8}{3}) - (-8 + \frac{8}{3}) = \frac{16}{3} - (-\frac{16}{3}) = \frac{32}{3}$.

Answer: $\frac{32}{3}$

1. Find intersections: $\sqrt{x} = x^2 \implies x = x^4 \implies x(x^3 - 1) = 0 \implies x = 0, 1$.
2. Determine top/bottom: For $x=0.25$, $\sqrt{0.25} = 0.5$ and $0.25^2 = 0.0625$. So $\sqrt{x}$ is on top.
3. Set up integral: $\int_{0}^{1} (\sqrt{x} - x^2) \, dx$.
4. Integrate: $[\frac{2}{3}x^{3/2} - \frac{1}{3}x^3]_{0}^{1}$.
5. Evaluate: $(\frac{2}{3} - \frac{1}{3}) - 0 = \frac{1}{3}$.

Answer: $\frac{1}{3}$

1. Find intersections: $y^2 - 1 = 1 - y^2 \implies 2y^2 = 2 \implies y^2 = 1 \implies y = \pm 1$.
2. Determine right/left: For $y=0$, $x = -1$ for the first curve and $x = 1$ for the second. So $1 - y^2$ is on the right.
3. Set up integral: $\int_{-1}^{1} ((1 - y^2) - (y^2 - 1)) \, dy = \int_{-1}^{1} (2 - 2y^2) \, dy$.
4. Integrate: $[2y - \frac{2}{3}y^3]_{-1}^{1}$.
5. Evaluate: $(2 - \frac{2}{3}) - (-2 + \frac{2}{3}) = \frac{4}{3} - (-\frac{4}{3}) = \frac{8}{3}$.

Answer: $\frac{8}{3}$

1. Identify the radius: $R(x) = x^3$.
2. Set up the integral: $V = \int_{0}^{2} \pi (x^3)^2 \, dx = \pi \int_{0}^{2} x^6 \, dx$.
3. Integrate: $\pi [\frac{1}{7}x^7]_{0}^{2}$.
4. Evaluate: $\pi (\frac{128}{7} - 0) = \frac{128\pi}{7}$.

Answer: $\frac{128\pi}{7}$

1. Solve for $x$ in terms of $y$: $x = \sqrt{y}$.
2. Identify the radius: $R(y) = \sqrt{y}$.
3. Set up the integral from $y=0$ to $y=4$: $V = \int_{0}^{4} \pi (\sqrt{y})^2 \, dy = \pi \int_{0}^{4} y \, dy$.
4. Integrate and evaluate: $\pi [\frac{1}{2}y^2]_{0}^{4} = \pi (8 - 0) = 8\pi$.

Answer: $8\pi$

1. Find intersection points: $x = \sqrt{x} \implies x^2 = x \implies x(x-1)=0$, so $x=0$ and $x=1$.
2. Identify radii: Outer $R(x) = \sqrt{x}$, Inner $r(x) = x$.
3. Set up the washer integral: $V = \pi \int_{0}^{1} ((\sqrt{x})^2 - x^2) \, dx = \pi \int_{0}^{1} (x - x^2) \, dx$.
4. Integrate: $\pi [\frac{1}{2}x^2 - \frac{1}{3}x^3]_{0}^{1}$.
5. Evaluate: $\pi (\frac{1}{2} - \frac{1}{3}) = \pi (\frac{3-2}{6}) = \frac{\pi}{6}$.

Answer: $\frac{\pi}{6}$

1. Identify the radius as $x$ and the height as $f(x) = x^3$.
2. Set up the integral: $V = \int_{0}^{1} 2\pi x (x^3) \, dx$.
3. Simplify the integrand: $V = 2\pi \int_{0}^{1} x^4 \, dx$.
4. Integrate: $V = 2\pi [\frac{1}{5}x^5]_{0}^{1}$.
5. Evaluate: $V = 2\pi (\frac{1}{5} - 0) = \frac{2\pi}{5}$.

Answer: $\frac{2\pi}{5}$

1. Identify the radius as $x$ and the height as $f(x) = \sqrt{x}$.
2. Set up the integral: $V = \int_{0}^{4} 2\pi x (x^{1/2}) \, dx$.
3. Simplify the integrand: $V = 2\pi \int_{0}^{4} x^{3/2} \, dx$.
4. Integrate: $V = 2\pi [\frac{2}{5}x^{5/2}]_{0}^{4}$.
5. Evaluate: $V = 2\pi (\frac{2}{5} \cdot 32) = 2\pi (\frac{64}{5}) = \frac{128\pi}{5}$.

Answer: $\frac{128\pi}{5}$

1. Find intersection points: $y^2 = 2y \implies y(y-2) = 0$, so $y=0$ and $y=2$.
2. The intersection points in x are $x=0$ and $x=4$.
3. For rotation around the $y$-axis, shells use $dx$. The height is the difference between the upper curve $y = \sqrt{x}$ and the lower curve $y = x/2$.
4. Set up the integral: $V = \int_{0}^{4} 2\pi x (\sqrt{x} - \frac{x}{2}) \, dx$.
5. Simplify: $V = 2\pi \int_{0}^{4} (x^{3/2} - \frac{1}{2}x^2) \, dx$.
6. Integrate: $V = 2\pi [\frac{2}{5}x^{5/2} - \frac{1}{6}x^3]_{0}^{4}$.
7. Evaluate: $V = 2\pi (\frac{64}{5} - \frac{64}{6}) = 2\pi (\frac{384 - 320}{30}) = 2\pi (\frac{64}{30}) = \frac{64\pi}{15}$.

Answer: $\frac{64\pi}{15}$

1. Find intersection points: $x^3 = x \implies x(x^2-1) = 0$, so $x=0$ and $x=1$ in the first quadrant.
2. Using the washer method (integrate $dx$): Outer radius $R(x) = x$, inner radius $r(x) = x^3$.
3. Set up integral: $V = \int_{0}^{1} \pi (x^2 - (x^3)^2) \, dx = \pi \int_{0}^{1} (x^2 - x^6) \, dx$.
4. Integrate: $\pi [\frac{1}{3}x^3 - \frac{1}{7}x^7]_0^1 = \pi (\frac{1}{3} - \frac{1}{7}) = \pi (\frac{7-3}{21}) = \frac{4\pi}{21}$.

Answer: $\frac{4\pi}{21}$

1. Identify boundaries: $x$ goes from $0$ to $4$. $y$ goes from $0$ to $2$.
2. Using shell method (integrate $dx$): Radius is $x$, height is $\sqrt{x}$.
3. Set up integral: $V = \int_{0}^{4} 2\pi x \sqrt{x} \, dx = 2\pi \int_{0}^{4} x^{3/2} \, dx$.
4. Integrate: $2\pi [\frac{2}{5}x^{5/2}]_0^4 = 2\pi (\frac{2}{5} \cdot 32) = \frac{128\pi}{5}$.

Answer: $\frac{128\pi}{5}$

1. Intersection points: $x^2 = 2x \implies x=0, x=2$.
2. Shell method (integrate $dx$): Radius is $x - (-1) = x+1$. Height is $2x - x^2$.
3. Set up integral: $V = \int_{0}^{2} 2\pi (x+1)(2x - x^2) \, dx$.
4. Expand: $(x+1)(2x - x^2) = 2x^2 - x^3 + 2x - x^2 = -x^3 + x^2 + 2x$.
5. Integrate: $2\pi [-\frac{1}{4}x^4 + \frac{1}{3}x^3 + x^2]_0^2 = 2\pi (-4 + \frac{8}{3} + 4) = 2\pi (\frac{8}{3}) = \frac{16\pi}{3}$.

Answer: $\frac{16\pi}{3}$

1. Compute the derivative: $y' = \frac{3x^2}{6} - \frac{1}{2x^2} = \frac{x^2}{2} - \frac{1}{2x^2}$.
2. Compute $1 + (y')^2$: $1 + (\frac{x^4}{4} - \frac{1}{2} + \frac{1}{4x^4}) = \frac{x^4}{4} + \frac{1}{2} + \frac{1}{4x^4}$.
3. Recognize the perfect square: $1 + (y')^2 = (\frac{x^2}{2} + \frac{1}{2x^2})^2$.
4. Integrate: $L = \int_{1}^{2} (\frac{x^2}{2} + \frac{1}{2x^2}) \, dx = [\frac{x^3}{6} - \frac{1}{2x}]_{1}^{2}$.
5. Evaluate: $(\frac{8}{6} - \frac{1}{4}) - (\frac{1}{6} - \frac{1}{2}) = \frac{7}{6} + \frac{1}{4} = \frac{14+3}{12} = \frac{17}{12}$.

Answer: 17/12

1. Compute the derivative: $y' = \frac{2}{3} \cdot \frac{3}{2} x^{1/2} = x^{1/2}$.
2. Set up the integral: $L = \int_{0}^{3} \sqrt{1 + (x^{1/2})^2} \, dx = \int_{0}^{3} \sqrt{1 + x} \, dx$.
3. Use u-substitution where $u = 1+x$: $L = [\frac{2}{3}(1+x)^{3/2}]_{0}^{3}$.
4. Evaluate: $\frac{2}{3}(4)^{3/2} - \frac{2}{3}(1)^{3/2} = \frac{2}{3}(8) - \frac{2}{3}(1) = \frac{14}{3}$.

Answer: 14/3

1. Compute the derivative: $y' = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x$.
2. Set up the integral: $L = \int_{0}^{\pi/4} \sqrt{1 + (-\tan x)^2} \, dx$.
3. Use the identity $1 + \tan^2 x = \sec^2 x$: $L = \int_{0}^{\pi/4} \sqrt{\sec^2 x} \, dx = \int_{0}^{\pi/4} \sec x \, dx$.
4. Integrate: $L = [\ln|\sec x + \tan x|]_{0}^{\pi/4}$.
5. Evaluate: $\ln|\sec(\pi/4) + \tan(\pi/4)| - \ln|\sec(0) + \tan(0)| = \ln|\sqrt{2} + 1| - \ln|1 + 0| = \ln(1 + \sqrt{2})$.

Answer: \ln(1 + \sqrt{2})

1. Identify the radius for rotation around the $y$-axis: $r = x$.
2. Find the derivative: $y' = 2x$.
3. Set up the arc length element: $ds = \sqrt{1 + (2x)^2} dx = \sqrt{1 + 4x^2} dx$.
4. Set up the integral: $S = \int_{0}^{1} 2\pi x \sqrt{1 + 4x^2} \, dx$.
5. Use u-substitution: $u = 1 + 4x^2$, $du = 8x dx$, so $x dx = \frac{1}{8} du$.
6. Integrate: $S = \frac{2\pi}{8} \int_{1}^{5} u^{1/2} du = \frac{\pi}{4} [\frac{2}{3}u^{3/2}]_{1}^{5} = \frac{\pi}{6}(5\sqrt{5}-1)$.

Answer: \frac{\pi}{6}(5\sqrt{5}-1)

1. Find the derivative: $y' = 2$.
2. Calculate the arc length element: $\sqrt{1 + 2^2} = \sqrt{5}$.
3. Set up the integral: $S = \int_{0}^{3} 2\pi (2x) \sqrt{5} \, dx$.
4. Integrate: $S = 4\pi\sqrt{5} \int_{0}^{3} x \, dx = 4\pi\sqrt{5} [\frac{1}{2}x^2]_{0}^{3}$.
5. Evaluate: $4\pi\sqrt{5} \cdot \frac{9}{2} = 18\pi\sqrt{5}$.

Answer: 18\pi\sqrt{5}

1. The radius is $x = \sqrt{y}$.
2. Find the derivative of $x$ with respect to $y$: $x' = \frac{1}{2\sqrt{y}}$.
3. Set up the integral: $S = \int_{0}^{4} 2\pi \sqrt{y} \sqrt{1 + \frac{1}{4y}} \, dy$.
4. Simplify the integrand: $2\pi \sqrt{y} \frac{\sqrt{4y+1}}{2\sqrt{y}} = \pi \sqrt{4y+1}$.
5. Integrate: $S = \pi \int_{0}^{4} (4y+1)^{1/2} \, dy = \pi [ \frac{1}{4} \cdot \frac{2}{3} (4y+1)^{3/2} ]_{0}^{4}$.
6. Evaluate: $\frac{\pi}{6} (17^{3/2} - 1^{3/2}) = \frac{\pi}{6}(17\sqrt{17}-1)$.

Answer: \frac{\pi}{6}(17\sqrt{17}-1)

1. The force function is $F(x) = 500x$.
2. The work is the integral of $F(x)$ from $0.1$ to $0.3$: $W = \int_{0.1}^{0.3} 500x \, dx$.
3. The antiderivative is $250x^2$.
4. Evaluate at the boundaries: $250(0.3)^2 - 250(0.1)^2 = 250(0.09) - 250(0.01)$.
5. Calculate the final value: $22.5 - 2.5 = 20$.

Answer: 20 \text{ Joules}

1. The volume of a slice at height $x$ is $V = \pi(1)^2 dx = \pi dx$.
2. The weight of the slice is $F = 1000 \cdot 9.8 \cdot \pi dx = 9800\pi dx$.
3. The distance the slice is lifted is $4 - x$.
4. The work integral is $W = \int_{0}^{4} 9800\pi (4-x) \, dx$.
5. Integrate: $9800\pi [4x - \frac{1}{2}x^2]_{0}^{4} = 9800\pi (16 - 8) = 9800\pi (8)$.
6. Final result: $78400\pi$.

Answer: 78400\pi \text{ Joules}

1. Let $h$ be the depth from the surface. The plate extends from $h = 1$ to $h = 4$.
2. The pressure at depth $h$ is $P = 9800h$.
3. The area of a thin strip is $w \cdot dh = 2 \, dh$.
4. The force integral is $F = \int_{1}^{4} (9800h)(2) \, dh = 19600 \int_{1}^{4} h \, dh$.
5. Integrate: $19600 [\frac{1}{2}h^2]_{1}^{4} = 9800(4^2 - 1^2)$.
6. Calculate: $9800(16 - 1) = 9800(15) = 147000$.

Answer: 147000 \text{ Newtons}

1. Calculate area: $A = \int_{0}^{4} x^{1/2} \, dx = [\frac{2}{3}x^{3/2}]_{0}^{4} = \frac{2}{3}(8) = \frac{16}{3}$.
2. Calculate $M_y$: $M_y = \int_{0}^{4} x(x^{1/2}) \, dx = \int_{0}^{4} x^{3/2} \, dx = [\frac{2}{5}x^{5/2}]_{0}^{4} = \frac{2}{5}(32) = \frac{64}{5}$.
3. Calculate $M_x$: $M_x = \int_{0}^{4} \frac{1}{2}(\sqrt{x})^2 \, dx = \int_{0}^{4} \frac{1}{2}x \, dx = [\frac{1}{4}x^2]_{0}^{4} = 4$.
4. Find $\bar{x} = \frac{64/5}{16/3} = \frac{64}{5} \cdot \frac{3}{16} = \frac{12}{5} = 2.4$.
5. Find $\bar{y} = \frac{4}{16/3} = 4 \cdot \frac{3}{16} = \frac{3}{4} = 0.75$.

Answer: (\frac{12}{5}, \frac{3}{4})

1. Intersection points are $x=0$ and $x=1$.
2. Area $A = \int_{0}^{1} (x - x^2) \, dx = [\frac{1}{2}x^2 - \frac{1}{3}x^3]_{0}^{1} = \frac{1}{6}$.
3. Moment $M_y = \int_{0}^{1} x(x - x^2) \, dx = [\frac{1}{3}x^3 - \frac{1}{4}x^4]_{0}^{1} = \frac{1}{12}$.
4. Moment $M_x = \int_{0}^{1} \frac{1}{2}(x^2 - (x^2)^2) \, dx = \frac{1}{2}[\frac{1}{3}x^3 - \frac{1}{5}x^5]_{0}^{1} = \frac{1}{2}(\frac{2}{15}) = \frac{1}{15}$.
5. $\bar{x} = \frac{1/12}{1/6} = \frac{1}{2}$, $\bar{y} = \frac{1/15}{1/6} = \frac{2}{5}$.

Answer: (\frac{1}{2}, \frac{2}{5})

1. Area $A = \int_{0}^{1} e^x \, dx = [e^x]_{0}^{1} = e - 1$.
2. Moment $M_y = \int_{0}^{1} x e^x \, dx$. Using integration by parts: $[xe^x - e^x]_{0}^{1} = (e - e) - (0 - 1) = 1$.
3. Moment $M_x = \int_{0}^{1} \frac{1}{2}(e^x)^2 \, dx = \int_{0}^{1} \frac{1}{2}e^{2x} \, dx = [\frac{1}{4}e^{2x}]_{0}^{1} = \frac{1}{4}(e^2 - 1)$.
4. $\bar{x} = \frac{1}{e-1}$.
5. $\bar{y} = \frac{(e^2-1)/4}{e-1} = \frac{(e-1)(e+1)}{4(e-1)} = \frac{e+1}{4}$.

Answer: (\frac{1}{e-1}, \frac{e+1}{4})

1. Set up the volume integral: $V = \int_{1}^{\infty} \pi (\frac{1}{x^2})^2 \, dx = \int_{1}^{\infty} \frac{\pi}{x^4} \, dx$.
2. Evaluate as a limit: $\lim_{b \to \infty} \int_{1}^{b} \pi x^{-4} \, dx$.
3. Find the antiderivative: $\pi [\frac{x^{-3}}{-3}]_{1}^{b} = \pi [-\frac{1}{3b} - (-\frac{1}{3})]$.
4. Take the limit as $b \to \infty$: $\pi (0 + \frac{1}{3}) = \frac{\pi}{3}$.

Answer: V = \frac{\pi}{3}

1. The surface area formula is $S = \int_{1}^{\infty} 2\pi \frac{1}{x^2} \sqrt{1 + (-\frac{2}{x^3})^2} \, dx$.
2. Note that $\sqrt{1 + \frac{4}{x^6}} > 1$ for all $x \ge 1$.
3. Therefore, $S > \int_{1}^{\infty} \frac{2\pi}{x^2} \, dx$.
4. The integral $\int_{1}^{\infty} \frac{1}{x^2} \, dx$ is a p-integral with $p=2$, which converges.
5. Since the integrand $\frac{1}{x^2} \sqrt{1 + \frac{4}{x^6}}$ behaves like $\frac{1}{x^2}$ for large $x$, the surface area actually converges.

Answer: The surface area is finite.

1. Set up the volume integral: $V = \int_{1}^{\infty} \pi (\frac{1}{\sqrt{x}})^2 \, dx$.
2. Simplify the integrand: $V = \int_{1}^{\infty} \frac{\pi}{x} \, dx$.
3. Evaluate the improper integral: $\lim_{b \to \infty} [\pi \ln x]_{1}^{b}$.
4. The limit $\lim_{b \to \infty} \pi \ln b$ diverges to infinity.

Answer: V = \infty

1. Separate the variables by dividing by $y$ and multiplying by $dx$: $\frac{1}{y} \, dy = x \, dx$.
2. Integrate both sides: $\int \frac{1}{y} \, dy = \int x \, dx$.
3. Evaluate the integrals: $\ln|y| = \frac{1}{2}x^2 + C$.
4. Exponentiate both sides: $|y| = e^{\frac{1}{2}x^2 + C} = e^C e^{\frac{1}{2}x^2}$.
5. Replace $e^C$ with a constant $A$: $y = Ae^{\frac{1}{2}x^2}$.

Answer: $y = Ae^{\frac{1}{2}x^2}$

1. Separate the variables: $\frac{1}{y} \, dy = 3 \, dx$.
2. Integrate both sides: $\ln|y| = 3x + C$.
3. Solve for $y$: $y = Ae^{3x}$.
4. Apply the initial condition $y(0) = 5$: $5 = Ae^{3(0)} \implies 5 = A(1) \implies A = 5$.
5. Substitute $A$ back into the general solution: $y = 5e^{3x}$.

Answer: $y = 5e^{3x}$

1. Separate the variables: $y^2 \, dy = x^2 \, dx$.
2. Integrate both sides: $\int y^2 \, dy = \int x^2 \, dx$.
3. Evaluate the integrals: $\frac{1}{3}y^3 = \frac{1}{3}x^3 + C$.
4. Multiply by 3 to simplify: $y^3 = x^3 + 3C$.
5. Let $K = 3C$ and solve for $y$: $y = \sqrt[3]{x^3 + K}$.

Answer: $y = \sqrt[3]{x^3 + K}$

1. Find the decay constant $k$ using $k = \frac{-\ln 2}{1600}$.
2. Set up the model $y(t) = 50 e^{kt}$.
3. Substitute $t = 2000$ into the equation: $y(2000) = 50 e^{\frac{-\ln 2}{1600} \cdot 2000}$.
4. Simplify the exponent: $y(2000) = 50 e^{-1.25 \ln 2} = 50(2^{-1.25})$.
5. Calculate the final value: $y(2000) \approx 21.02$ grams.

Answer: 21.02 \text{ g}

1. Use the model $y(t) = 200 e^{kt}$ and the condition $y(2) = 350$.
2. Solve for $k$: $350 = 200 e^{2k} \implies 1.75 = e^{2k} \implies k = \frac{\ln(1.75)}{2} \approx 0.2798$.
3. Use the doubling time formula $t_{\text{double}} = \frac{\ln 2}{k}$.
4. Substitute $k$: $t_{\text{double}} = \frac{\ln 2}{0.2798} \approx 2.48$ years.

Answer: 2.48 \text{ years}

1. Use Newton's Law of Cooling: $T(t) = 20 + (150 - 20)e^{kt} = 20 + 130 e^{kt}$.
2. Find $k$ using $T(5) = 100$: $100 = 20 + 130 e^{5k} \implies 80 = 130 e^{5k} \implies e^{5k} = \frac{8}{13}$.
3. Solve for $k$: $k = \frac{\ln(8/13)}{5} \approx -0.0937$.
4. Set $T(t) = 40$ and solve for $t$: $40 = 20 + 130 e^{-0.0937t} \implies 20 = 130 e^{-0.0937t}$.
5. Divide by $130$: $\frac{2}{13} = e^{-0.0937t} \implies \ln(2/13) = -0.0937t$.
6. Calculate $t$: $t = \frac{\ln(2/13)}{-0.0937} \approx 19.4$ minutes.

Answer: 19.4 \text{ minutes}

1. Use the formula $y(t) = \frac{L}{1 + Ae^{-kt}}$.
2. Find $A$ using $y(0) = 500$: $500 = \frac{2000}{1 + A} \implies 1 + A = 4 \implies A = 3$.
3. Substitute $t = 10$: $y(10) = \frac{2000}{1 + 3e^{-0.1(10)}} = \frac{2000}{1 + 3e^{-1}}$.
4. Calculate $e^{-1} \approx 0.3679$, so $y(10) \approx \frac{2000}{1 + 1.1037} \approx \frac{2000}{2.1037} \approx 950.7$.

Answer: $y(10) \approx 951$

1. Compare the given equation $\frac{dy}{dt} = 0.02y(1 - \frac{y}{100000})$ to the standard form $\frac{dy}{dt} = ky(1 - \frac{y}{L})$.
2. The coefficient $k$ is the value multiplying the $y$ term outside the parentheses, so $k = 0.02$.
3. The value $L$ is the denominator in the fraction inside the parentheses, so $L = 100,000$.

Answer: $k = 0.02, L = 100,000$

1. The growth rate is given by the function $f(y) = ky - \frac{k}{L}y^2$.
2. To find the maximum of $f(y)$, take the derivative with respect to $y$ and set it to zero: $f'(y) = k - \frac{2k}{L}y$.
3. Set $k - \frac{2k}{L}y = 0$.
4. Divide by $k$ (assuming $k \neq 0$): $1 - \frac{2y}{L} = 0 \implies \frac{2y}{L} = 1 \implies y = L/2$.

Answer: $y = L/2$