Step-by-Step Calculus II Midterm Study Guidance

Q1. Water flows into a pool at a rate of $r(t) = 100 + 20 \sin \pi t$ liters per minute. If the pool started at 1000L, how much will it hold after 30 minutes?

Background

Topic: Applications of Definite Integrals

This question tests your ability to use definite integrals to find the total accumulation of water over a time interval, given a rate function.

Key Terms and Formulas:

• Rate function: $r(t)$ gives the rate at which water flows in (liters/minute).

• Total volume added: $\int_{a}^{b} r(t) \, dt$

• Initial volume: $V_0$

• Final volume: $V = V_0 + \int_{a}^{b} r(t) \, dt$

Step-by-Step Guidance

1. Identify the initial volume: $V_0 = 1000$ liters.

2. Set up the definite integral for the total water added: $\int_{0}^{30} (100 + 20 \sin \pi t) \, dt$.

3. Break the integral into two parts: $\int_{0}^{30} 100 \, dt$ and $\int_{0}^{30} 20 \sin \pi t \, dt$.

4. Recall the antiderivative of $\sin \pi t$ is $-\frac{1}{\pi} \cos \pi t$.

5. Set up the expressions for each part, but do not compute the final values yet.

Try solving on your own before revealing the answer!

Final Answer: 4000 L

The total water added is $\int_{0}^{30} (100 + 20 \sin \pi t) \, dt = 100 \times 30 + 0 = 3000$ liters. Adding the initial 1000L gives $4000$ liters.

The sine term integrates to zero over 30 full periods, so only the constant rate contributes.

Q2. Find the shaded area between the curves $y = \sin 2x$ and $y = \cos x$ from $x = -1$ to $x = 1$.

Background

Topic: Area Between Curves

This question tests your ability to set up and compute the area between two curves using definite integrals.

Key Terms and Formulas:

• Area between curves: $A = \int_{a}^{b} [f(x) - g(x)] \, dx$ where $f(x)$ is the upper curve.

Step-by-Step Guidance

1. Determine which function is on top between $x = -1$ and $x = 1$ by comparing $\sin 2x$ and $\cos x$.

2. Set up the integral: $A = \int_{-1}^{1} [\cos x - \sin 2x] \, dx$ (assuming $\cos x$ is above $\sin 2x$).

3. Recall the antiderivatives: $\int \cos x \, dx = \sin x$, $\int \sin 2x \, dx = -\frac{1}{2} \cos 2x$.

4. Write the expression for the area using these antiderivatives, but do not evaluate at the endpoints yet.

Try solving on your own before revealing the answer!

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

Evaluating the definite integral gives $A = \left[\sin x + \frac{1}{2} \cos 2x\right]_{-1}^{1} = \frac{1}{2}$.

The area is exactly $\frac{1}{2}$ square units.

Q3. A solid is formed so that the cross-section in the $x$-$y$ plane is bounded by $y = \pm e^x$ for $0 \leq x \leq \ln 3$. The cross-sections perpendicular to the $x$-axis are squares. Find the volume.

Background

Topic: Volumes by Slicing (Cross-Sections)

This question tests your ability to find the volume of a solid with known cross-sectional area using definite integrals.

Key Terms and Formulas:

• Cross-sectional area: For squares, $A(x) = [\text{side length}]^2$.

• Side length: $y = e^x - (-e^x) = 2e^x$.

• Volume: $V = \int_{a}^{b} A(x) \, dx$.

Step-by-Step Guidance

1. Find the side length of the square at each $x$: $2e^x$.

2. Compute the area of the square: $A(x) = (2e^x)^2 = 4e^{2x}$.

3. Set up the volume integral: $V = \int_{0}^{\ln 3} 4e^{2x} \, dx$.

4. Recall the antiderivative: $\int e^{2x} \, dx = \frac{1}{2} e^{2x}$.

5. Write the expression for the volume, but do not evaluate at the endpoints yet.

Try solving on your own before revealing the answer!

Final Answer: 16

Evaluating the integral gives $V = 4 \times \frac{1}{2} [e^{2x}]_{0}^{\ln 3} = 2(e^{2\ln 3} - e^{0}) = 2(9 - 1) = 16$.

The volume is 16 cubic units.

Q4. A solid is formed by revolving the region bounded by $y = x$ and $y = x^2$ about the $x$-axis for $x$ in $[0,1]$. Find the volume.

Background

Topic: Volumes of Revolution (Washer/Disc Method)

This question tests your ability to set up and compute the volume of a solid formed by revolving a region about an axis.

Key Terms and Formulas:

• Washer method: $V = \pi \int_{a}^{b} [R^2(x) - r^2(x)] \, dx$

• Outer radius: $R(x) = x$

• Inner radius: $r(x) = x^2$

Step-by-Step Guidance

1. Set up the volume integral: $V = \pi \int_{0}^{1} [x^2 - (x^2)^2] \, dx$

2. Simplify the integrand: $x^2 - x^4$

3. Recall the antiderivatives: $\int x^2 \, dx = \frac{1}{3} x^3$, $\int x^4 \, dx = \frac{1}{5} x^5$

4. Write the expression for the volume, but do not evaluate at the endpoints yet.

Try solving on your own before revealing the answer!

Final Answer: $\frac{2\pi}{15}$

Evaluating the integral gives $V = \pi \left( \frac{1}{3} - \frac{1}{5} \right ) = \frac{2\pi}{15}$.

This is the exact volume of the solid of revolution.

Q5. Find the length of $y = \frac{e^x + e^{-x}}{2}$ on $[-\ln 3, \ln 3]$.

Background

Topic: Arc Length of a Curve

This question tests your ability to compute the arc length of a function over a given interval.

Key Terms and Formulas:

• Arc length formula: $L = \int_{a}^{b} \sqrt{1 + [y'(x)]^2} \, dx$

• Derivative: $y'(x)$

Step-by-Step Guidance

1. Compute $y'(x)$ for $y = \frac{e^x + e^{-x}}{2}$: $y'(x) = \frac{e^x - e^{-x}}{2}$

2. Square $y'(x)$: $[y'(x)]^2 = \left( \frac{e^x - e^{-x}}{2} \right )^2$

3. Add 1: $1 + [y'(x)]^2$

4. Simplify the expression under the square root.

5. Set up the arc length integral, but do not evaluate it yet.

Try solving on your own before revealing the answer!

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

The arc length is $L = \int_{-\ln 3}^{\ln 3} e^x \, dx = [e^x]_{-\ln 3}^{\ln 3} = 3 - \frac{1}{3} = \frac{8}{3}$.

The simplification comes from recognizing the expression under the square root simplifies to $e^{2x}$.

Q6. Set up (but do not evaluate) the integral for the surface area of revolution for $y = 1 - x^2$ with $0 \leq x \leq 1$ about the $x$-axis.

Background

Topic: Surface Area of Revolution

This question tests your ability to set up the integral for the surface area generated by revolving a curve about the $x$-axis.

Key Terms and Formulas:

• Surface area formula: $S = 2\pi \int_{a}^{b} y(x) \sqrt{1 + [y'(x)]^2} \, dx$

• Derivative: $y'(x)$

Step-by-Step Guidance

1. Compute $y'(x)$ for $y = 1 - x^2$: $y'(x) = -2x$

2. Square $y'(x)$: $[y'(x)]^2 = 4x^2$

3. Set up the surface area integral: $S = 2\pi \int_{0}^{1} (1 - x^2) \sqrt{1 + 4x^2} \, dx$

4. Do not evaluate the integral; just write it in its final setup form.

Try solving on your own before revealing the answer!

Final Answer:

$S = 2\pi \int_{0}^{1} (1 - x^2) \sqrt{1 + 4x^2} \, dx$

This is the correct setup for the surface area integral.

Q7. A 10m length of chain weighs 100kg. It is laying on the ground and a hoist picks it up until the bottom end is 2m off the ground. Find the work done.

Background

Topic: Work as an Integral (Physics Application)

This question tests your ability to compute work done in lifting a variable mass using integrals.

Key Terms and Formulas:

• Work: $W = \int_{a}^{b} F(x) \, dx$

• Chain density: $\frac{100}{10} = 10$ kg/m

• Gravity: $g = 9.8$ m/s$^2$

Step-by-Step Guidance

1. Let $x$ be the distance from the top of the chain; set up the integral for work as $W = \int_{0}^{10} (\text{weight per meter}) \times (\text{distance lifted}) \, dx$

2. Each segment at position $x$ is lifted $10 - x - 2$ meters (since the bottom is 2m off the ground).

3. Set up the integral: $W = \int_{0}^{10} 10 \times 9.8 \times (10 - x - 2) \, dx$

4. Simplify the integrand and set up the limits, but do not evaluate yet.

Try solving on your own before revealing the answer!

Final Answer: 6860 N·m

Evaluating the integral gives $W = 6860$ N·m (Joules).

This is the total work required to lift the chain so the bottom is 2m off the ground.

Q8. Evaluate the indefinite integral $\int \ln x \, dx$.

Background

Topic: Integration by Parts

This question tests your ability to use integration by parts to evaluate an integral involving $\ln x$.

Key Terms and Formulas:

• Integration by parts: $\int u \, dv = u v - \int v \, du$

• Let $u = \ln x$, $dv = dx$

Step-by-Step Guidance

1. Let $u = \ln x$, $dv = dx$

2. Compute $du = \frac{1}{x} dx$, $v = x$

3. Apply integration by parts: $\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} dx$

4. Simplify the remaining integral, but do not write the final answer yet.

Try solving on your own before revealing the answer!

Final Answer: $x \ln x - x + C$

Using integration by parts, $\int \ln x \, dx = x \ln x - x + C$.

This is a standard result for integrating $\ln x$.

Q9. Evaluate the definite integral $\int_{0}^{\pi/4} \tan x \sec^2 x \, dx$.

Background

Topic: Substitution in Definite Integrals

This question tests your ability to use substitution to evaluate a definite integral involving trigonometric functions.

Key Terms and Formulas:

• $\frac{d}{dx} \tan x = \sec^2 x$

• Substitution: Let $u = \tan x$

Step-by-Step Guidance

1. Let $u = \tan x$, then $du = \sec^2 x \, dx$

2. Rewrite the integral in terms of $u$: $\int u \, du$

3. Find the new limits: when $x = 0$, $u = 0$; when $x = \pi/4$, $u = 1$

4. Set up the definite integral, but do not evaluate yet.

Try solving on your own before revealing the answer!

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

Evaluating the integral gives $\int_{0}^{1} u \, du = \frac{1}{2}$.

This is the value of the definite integral.

Q10. Evaluate the indefinite integral $\int \frac{dx}{x^2 + 6x + 25}$.

Background

Topic: Integrals Involving Quadratic Denominators

This question tests your ability to integrate functions with quadratic denominators, often using completing the square and arctangent formulas.

Key Terms and Formulas:

• Complete the square: $x^2 + 6x + 25 = (x + 3)^2 + 16$

• Arctangent formula: $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right ) + C$

Step-by-Step Guidance

1. Rewrite the denominator by completing the square: $(x + 3)^2 + 16$

2. Let $u = x + 3$, so $du = dx$

3. Set up the integral: $\int \frac{du}{u^2 + 16}$

4. Recall the arctangent formula, but do not write the final answer yet.

Try solving on your own before revealing the answer!

Final Answer: $\frac{1}{4} \tan^{-1} \left( \frac{x + 3}{4} \right ) + C$

Using the arctangent formula, the integral evaluates to $\frac{1}{4} \tan^{-1} \left( \frac{x + 3}{4} \right ) + C$.

Q11. Evaluate the definite integral $\int_{1}^{e} \frac{dx}{x(x + 1)^2}$.

Background

Topic: Partial Fraction Decomposition

This question tests your ability to decompose a rational function and integrate each part over a definite interval.

Key Terms and Formulas:

• Partial fractions: $\frac{1}{x(x + 1)^2} = \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{(x + 1)^2}$

• Integrate each term separately.

Step-by-Step Guidance

1. Decompose $\frac{1}{x(x + 1)^2}$ into partial fractions.

2. Find constants $A$, $B$, $C$ by solving the system.

3. Integrate each term over $[1, e]$.

4. Set up the sum of the integrals, but do not evaluate yet.

Try solving on your own before revealing the answer!

Final Answer: $\frac{1}{2} + \frac{1}{e + 1} + \ln \frac{2}{e + 1}$

After decomposing and integrating, the definite integral evaluates to $\frac{1}{2} + \frac{1}{e + 1} + \ln \frac{2}{e + 1}$.

Q12. When the region bounded by $x = 1$, $y = 0$, and $y = \frac{1}{x}$ is revolved about the $x$-axis, the resulting volume is $\pi \int_{1}^{\infty} \frac{dx}{x^2}$. Find the value of the integral, if it converges.

Background

Topic: Improper Integrals and Volumes of Revolution

This question tests your ability to evaluate improper integrals and determine convergence.

Key Terms and Formulas:

• Improper integral: $\int_{1}^{\infty} \frac{dx}{x^2}$

• Limit definition: $\lim_{a \to \infty} \int_{1}^{a} \frac{dx}{x^2}$

• Antiderivative: $\int x^{-2} dx = -x^{-1}$

Step-by-Step Guidance

1. Set up the improper integral as a limit: $\lim_{a \to \infty} \int_{1}^{a} x^{-2} dx$

2. Find the antiderivative: $-x^{-1}$

3. Write the expression for the definite integral, but do not evaluate the limit yet.

Try solving on your own before revealing the answer!

Final Answer: $\pi$

The improper integral converges to $\pi$.

Evaluating the limit gives $\pi \left( 0 + 1 \right ) = \pi$.