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 Calculus 2 Exams Exam B1 - Exam B2

### 17Calculus Subjects Listed Alphabetically

Single Variable Calculus

 Absolute Convergence Alternating Series Arc Length Area Under Curves Chain Rule Concavity Conics Conics in Polar Form Conditional Convergence Continuity & Discontinuities Convolution, Laplace Transforms Cosine/Sine Integration Critical Points Cylinder-Shell Method - Volume Integrals Definite Integrals Derivatives Differentials Direct Comparison Test Divergence (nth-Term) Test
 Ellipses (Rectangular Conics) Epsilon-Delta Limit Definition Exponential Derivatives Exponential Growth/Decay Finite Limits First Derivative First Derivative Test Formal Limit Definition Fourier Series Geometric Series Graphing Higher Order Derivatives Hyperbolas (Rectangular Conics) Hyperbolic Derivatives
 Implicit Differentiation Improper Integrals Indeterminate Forms Infinite Limits Infinite Series Infinite Series Table Infinite Series Study Techniques Infinite Series, Choosing a Test Infinite Series Exam Preparation Infinite Series Exam A Inflection Points Initial Value Problems, Laplace Transforms Integral Test Integrals Integration by Partial Fractions Integration By Parts Integration By Substitution Intermediate Value Theorem Interval of Convergence Inverse Function Derivatives Inverse Hyperbolic Derivatives Inverse Trig Derivatives
 Laplace Transforms L'Hôpital's Rule Limit Comparison Test Limits Linear Motion Logarithm Derivatives Logarithmic Differentiation Moments, Center of Mass Mean Value Theorem Normal Lines One-Sided Limits Optimization
 p-Series Parabolas (Rectangular Conics) Parabolas (Polar Conics) Parametric Equations Parametric Curves Parametric Surfaces Pinching Theorem Polar Coordinates Plane Regions, Describing Power Rule Power Series Product Rule
 Quotient Rule Radius of Convergence Ratio Test Related Rates Related Rates Areas Related Rates Distances Related Rates Volumes Remainder & Error Bounds Root Test Secant/Tangent Integration Second Derivative Second Derivative Test Shifting Theorems Sine/Cosine Integration Slope and Tangent Lines Square Wave Surface Area
 Tangent/Secant Integration Taylor/Maclaurin Series Telescoping Series Trig Derivatives Trig Integration Trig Limits Trig Substitution Unit Step Function Unit Impulse Function Volume Integrals Washer-Disc Method - Volume Integrals Work

Multi-Variable Calculus

 Acceleration Vector Arc Length (Vector Functions) Arc Length Function Arc Length Parameter Conservative Vector Fields Cross Product Curl Curvature Cylindrical Coordinates
 Directional Derivatives Divergence (Vector Fields) Divergence Theorem Dot Product Double Integrals - Area & Volume Double Integrals - Polar Coordinates Double Integrals - Rectangular Gradients Green's Theorem
 Lagrange Multipliers Line Integrals Partial Derivatives Partial Integrals Path Integrals Potential Functions Principal Unit Normal Vector
 Spherical Coordinates Stokes' Theorem Surface Integrals Tangent Planes Triple Integrals - Cylindrical Triple Integrals - Rectangular Triple Integrals - Spherical
 Unit Tangent Vector Unit Vectors Vector Fields Vectors Vector Functions Vector Functions Equations

Differential Equations

 Boundary Value Problems Bernoulli Equation Cauchy-Euler Equation Chebyshev's Equation Chemical Concentration Classify Differential Equations Differential Equations Euler's Method Exact Equations Existence and Uniqueness Exponential Growth/Decay
 First Order, Linear Fluids, Mixing Fourier Series Inhomogeneous ODE's Integrating Factors, Exact Integrating Factors, Linear Laplace Transforms, Solve Initial Value Problems Linear, First Order Linear, Second Order Linear Systems
 Partial Differential Equations Polynomial Coefficients Population Dynamics Projectile Motion Reduction of Order Resonance
 Second Order, Linear Separation of Variables Slope Fields Stability Substitution Undetermined Coefficients Variation of Parameters Vibration Wronskian

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17calculus > exam list > calc2 exam B1

This is the midterm exam for second semester single variable calculus.

### Practice Exam Tips

Each exam page contains a full exam with detailed solutions. Most of these are actual exams from previous semesters used in college courses. You may use these as practice problems or as practice exams. Here are some suggestions on how to use these to help you prepare for your exams.

- Set aside a chunk of full, uninterrupted time, usually an hour to two, to work each exam.
- Go to a quiet place where you will not be interrupted that duplicates your exam situation as closely as possible.
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- Work the entire exam before checking any solutions.
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- Work as many practice exams as you have time for. This will give you practice in important techniques, experience in different types of exam problems that you may see on your own exam and help you understand the material better by showing you what you need to study.

IMPORTANT -
Exams can cover only so much material. Instructors will sometimes change exams from one semester to the next to adapt an exam to each class depending on how the class performs during the semester while they are learning the material. So just because you do well (or not) on these practice exams, does not necessarily mean you will do the same on your exam. Your instructor may ask completely different questions from these. That is why working lots of practice problems will prepare you better than working just one or two practice exams.
Calculus is not something that can be learned by reading. You have to work problems on your own and struggle through the material to really know calculus, do well on your exam and be able to use it in the future.

Exam Details

Time

1 hour and 40 minutes

Questions

11

Total Points

100

Tools

Calculator

see instructions

Formula Sheet(s)

none

Other Tools

none

Instructions:
- This exam is in two main parts, labeled parts A and B, with different instructions for each part.
- For each problem, correct answers are worth 1 point. The remaining points are earned by showing calculations and giving reasoning that justify your conclusions.
- Correct notation counts (i.e. points will be taken off for incorrect notation).

Evaluate these integrals.

### Part A - Questions 1-5

Instructions for Part A - - You have 40 minutes to complete this part of the exam. No calculators are allowed. All questions in this section are worth 10 points.

$$\int{ 5t(2 + t^2)^3 ~dt }$$

Problem Statement

$$\int{ 5t(2 + t^2)^3 ~dt }$$

$$\displaystyle{ \frac{5}{8}(2+t^2)^4 + C }$$

Problem Statement

$$\int{ 5t(2 + t^2)^3 ~dt }$$

Solution

 $$\int{ 5t(2 + t^2)^3 ~dt }$$ Use integration by substitution. $$u = 2+t^2 \to du=2t~dt \to dt=du/(2t)$$ $$\displaystyle{ \int{ 5t ~ u^3 \frac{du}{2t} } }$$ $$\displaystyle{ \frac{5}{2}\int{ u^3 ~ du } }$$ $$\displaystyle{ \frac{5}{2} \frac{u^4}{4} + C }$$ $$\displaystyle{ \frac{5}{8}(2+t^2)^4 + C }$$

$$\displaystyle{ \frac{5}{8}(2+t^2)^4 + C }$$

$$\int{ (2x+1)\ln x ~ dx }$$

Problem Statement

$$\int{ (2x+1)\ln x ~ dx }$$

$$(x^2+x)\ln x - (1/2)x^2 - x + C$$

Problem Statement

$$\int{ (2x+1)\ln x ~ dx }$$

Solution

 Use integration by parts. $$u=\ln x \to du=(1/x)dx$$ $$dv = (2x+1)dx \to v = x^2+x$$ $$\int{ u ~dv } = u v - \int{ v ~du }$$ $$\displaystyle{ (x^2+x)\ln x - \int{ (x^2+x)\frac{dx}{x} } }$$ $$(x^2+x)\ln x - \int{ x+1~dx }$$ $$(x^2+x)\ln x - (1/2)x^2 - x + C$$

$$(x^2+x)\ln x - (1/2)x^2 - x + C$$

$$\int{ \sin^3\theta ~ d\theta }$$

Problem Statement

$$\int{ \sin^3\theta ~ d\theta }$$

$$(1/3)\cos^3\theta - \cos\theta + C$$

Problem Statement

$$\int{ \sin^3\theta ~ d\theta }$$

Solution

 $$\int{ \sin^3\theta ~ d\theta }$$ $$\int{ \sin \theta (\sin^2\theta)~d\theta }$$ $$\int{ \sin \theta ( 1-\cos^2\theta )~d\theta }$$ $$u=\cos\theta \to du = -\sin\theta d\theta$$ $$\int{ (1-u^2)(-du) }$$ $$\int{ u^2-1 ~du }$$ $$u^3 / 3 - u + C$$ $$(1/3)\cos^3\theta - \cos\theta + C$$

$$(1/3)\cos^3\theta - \cos\theta + C$$

$$\displaystyle{ \int{ \frac{dx}{(9+x^2)^{3/2}} } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{dx}{(9+x^2)^{3/2}} } }$$

$$\displaystyle{ \frac{x}{27\sqrt{9+x^2}} + C }$$

Problem Statement

$$\displaystyle{ \int{ \frac{dx}{(9+x^2)^{3/2}} } }$$

Solution

 Use trigonometric substitution. $$x=3\tan t \to dx = 3\sec^2 t ~dt$$ Draw a triangle that matches $$\tan t = x/3$$ including the hypotenuse $$\sqrt{9+x^2}$$. $$\displaystyle{ \int{ \frac{dx}{(9+x^2)^{3/2}} } }$$ $$\displaystyle{ \int{ \frac{\sec^2t~dt}{(9+9\tan^2 t)^{3/2}} } }$$ $$\displaystyle{ \int{ \frac{\sec^2t~dt}{27 (\sec^2 t)^{3/2}} } }$$ $$\displaystyle{ \frac{1}{27}\int{ \frac{\sec^2 t}{\sec^3 t}~dt } }$$ $$\displaystyle{ \frac{1}{27}\int{ \frac{1}{\sec t}~dt } }$$ $$\displaystyle{ \frac{1}{27}\int{ \cos t ~dt } }$$ $$\displaystyle{ \frac{1}{27}\sin t + C }$$ Use the triangle that you drew above to write $$\sin t$$ in terms of x. $$\displaystyle{ \frac{1}{27} \frac{x}{\sqrt{9+x^2}} + C }$$

$$\displaystyle{ \frac{x}{27\sqrt{9+x^2}} + C }$$

$$\displaystyle{ \int{ \frac{2x^2+x-4}{x^3-x^2-2x} ~dx } }$$

Problem Statement

$$\displaystyle{ \int{ \frac{2x^2+x-4}{x^3-x^2-2x} ~dx } }$$

$$\displaystyle{ \ln\left| \frac{x^2(x-2)}{x+1} \right| + C }$$

Problem Statement

$$\displaystyle{ \int{ \frac{2x^2+x-4}{x^3-x^2-2x} ~dx } }$$

Solution

Check to see if the denominator factors. If so, then try partial fractions.

 $$x^3-x^2-2x = x(x^2-x-2) = x(x-2)(x+1)$$ $$\displaystyle{ \frac{2x^2+x-4}{x^3-x^2-2x} }$$ $$\displaystyle{ \frac{A}{x} + \frac{B}{x-2} + \frac{D}{x+1} }$$ Solving for A, B and C, we get $$A=2$$, $$B=1$$ and $$D=-1$$. Now the integral looks like this. $$\displaystyle{ \int{ \frac{2}{x} + \frac{1}{x-2} - \frac{1}{x+1} ~ dx } }$$ $$2\ln|x| + \ln|x-2| - \ln|x+1| + C$$ $$\displaystyle{ \ln\left| \frac{x^2(x-2)}{x+1} \right| + C }$$

Notes:
1. The absolute values on each term are required since we are not guaranteed that the factors are positive.
2. Some instructors consider the 2nd to the last expression to be simplified, others require the terms to be combined as in the last expression. Check with your instructor to see what they require.

$$\displaystyle{ \ln\left| \frac{x^2(x-2)}{x+1} \right| + C }$$

### Part B - Questions 6-11

Instructions for Part B - - You may use your calculator. You have one hour to complete this part of the exam. Show all your work. Correct answers are worth 1 point. The remaining points are earned by showing calculations and giving reasoning that justify your conclusions. Give exact, simplified answers.

Section 1

For questions 6-8, consider the region $$\mathcal{R}$$ bounded by the graph of $$y=4-x^2$$, the y-axis and the horizontal line $$y=1$$.

[5 points] Sketch the region $$\mathcal{R}$$ and calculate it's area.

Problem Statement

[5 points] Sketch the region $$\mathcal{R}$$ and calculate it's area.

$$A = 2\sqrt{3}$$

Problem Statement

[5 points] Sketch the region $$\mathcal{R}$$ and calculate it's area.

Solution

built with geoGeobra

The region is show in the plot.
For the area, we will set up and evaluate both integrals, i.e. with respect to x and y. The general equation for area is $$A = \int_a^b{f(x)-g(x)~dx}$$. A similar equation holds for y.
First, with respect to x.
The upper function is $$y=4-x^2$$, the lower function is $$y=1$$. We need to find the x-value where these intersect on the right end of the area.
$$4-x^2=1 \to x^2 = 3 \to x=\pm \sqrt{3}$$
Since we are in the first quadrant, x is positive, so $$x=\sqrt{3}$$.

 $$A = \int_a^b{f(x)-g(x)~dx}$$ $$A = \int_0^{\sqrt{3}}{ (4-x^2) - 1 ~ dx }$$ $$A = \int_0^{\sqrt{3}}{ 3-x^2 ~ dx }$$ $$A = [ 3x - x^3/3 ]_0^{\sqrt{3}}$$ $$A = 2\sqrt{3}$$

For the y-direction, the right function is $$y=4-x^2$$ which we need to solve for y.
$$y=4-x^2 \to x^2=4-y \to x=\pm\sqrt{4-y}$$
Since the part of the function that we need is in the first quadrant, x is positive, so our right function is $$x=\sqrt{4-y}$$.
The left function is $$x=0$$. We integrate from 1 to 4.

 $$A = \int_c^d{f(y)-g(y)~dx}$$ $$A = \int_1^4{\sqrt{4-y}-0~dx}$$ Use integration by substitution. $$u = 4-y \to du = -dy$$ Convert the limits of integration in terms of u. lower limit: $$y=1 \to u=4-1=3$$ upper limit: $$y=4 \to u=4-4= 0$$ $$-\int_3^0{ u^{1/2}~du }$$ $$\int_0^3{ u^{1/2}~du }$$ $$\displaystyle{ \left. \frac{u^{3/2}}{3/2} \right|_0^3 }$$ $$\displaystyle{ \frac{2}{3}3^{3/2} - 0 }$$ $$A = 2\sqrt{3}$$

$$A = 2\sqrt{3}$$

[5 points] A solid is generated by revolving the region $$\mathcal{R}$$ about the x-axis. Set up, but do not evaluate, the definite integral for the volume of the resulting solid of revolution using the method of disc-washers.

Problem Statement

[5 points] A solid is generated by revolving the region $$\mathcal{R}$$ about the x-axis. Set up, but do not evaluate, the definite integral for the volume of the resulting solid of revolution using the method of disc-washers.

$$V = \pi \int_0^{\sqrt{3}}{ (4-x^2)^2 - 1 ~ dx }$$

Problem Statement

[5 points] A solid is generated by revolving the region $$\mathcal{R}$$ about the x-axis. Set up, but do not evaluate, the definite integral for the volume of the resulting solid of revolution using the method of disc-washers.

Solution

 $$V = \pi \int_a^b{ R^2 - r^2 ~dx }$$ $$R=y=4-x^2$$; $$r=1$$ x ranges from 0 to $$\sqrt{3}$$ $$V = \pi \int_0^{\sqrt{3}}{ (4-x^2)^2 - 1 ~ dx }$$

$$V = \pi \int_0^{\sqrt{3}}{ (4-x^2)^2 - 1 ~ dx }$$

[5 points] A solid is generated by revolving the region $$\mathcal{R}$$ about the x-axis. Set up, but do not evaluate, the definite integral for the volume of the resulting solid of revolution using the method of shell-cylinders.

Problem Statement

[5 points] A solid is generated by revolving the region $$\mathcal{R}$$ about the x-axis. Set up, but do not evaluate, the definite integral for the volume of the resulting solid of revolution using the method of shell-cylinders.

$$V = 2\pi\int_1^4{ y\sqrt{4-y}~dy }$$

Problem Statement

[5 points] A solid is generated by revolving the region $$\mathcal{R}$$ about the x-axis. Set up, but do not evaluate, the definite integral for the volume of the resulting solid of revolution using the method of shell-cylinders.

Solution

 $$V = 2\pi\int_c^d{ ph~ dy }$$ $$p=y; h=x = \sqrt{4-y}$$ y ranges from 1 to 4 $$V = 2\pi\int_1^4{ y\sqrt{4-y}~dy }$$

$$V = 2\pi\int_1^4{ y\sqrt{4-y}~dy }$$

Section 2

[15 points] Find the centroid of the region bounded by the graph of $$y=\cos x$$ and the x-axis, between $$x=-\pi/2$$ and $$x=\pi/2$$. Make a sketch and use symmetry where possible.

Problem Statement

[15 points] Find the centroid of the region bounded by the graph of $$y=\cos x$$ and the x-axis, between $$x=-\pi/2$$ and $$x=\pi/2$$. Make a sketch and use symmetry where possible.

centroid $$(0,\pi/8)$$

Problem Statement

[15 points] Find the centroid of the region bounded by the graph of $$y=\cos x$$ and the x-axis, between $$x=-\pi/2$$ and $$x=\pi/2$$. Make a sketch and use symmetry where possible.

Solution

built with geoGeobra

The plot is shown here. Using symmetry, $$\bar{x}=0$$. The centroid is moment/area with the equation
$$\displaystyle{ \bar{y} = \frac{\int_c^d{y(x_2-x_1)dy}}{\int_c^d{(x_2-x_1~dy)}} }$$
First, calculate the area.

 area = $$\int_{-\pi/2}^{\pi/2}{ \cos x ~dx }$$ area = $$\sin x |_{-\pi/2}^{\pi/2} = 2$$

Now, let's calculate the moment.

 $$\int_0^1{ y( \arccos(y) - (-\arccos(y)) ) ~ dy }$$ $$\int_0^1{ 2y \arccos(y) ~ dy }$$ Use integration by parts. $$u=\arccos(y) \to du=-dy/\sqrt{1-y^2}$$ $$dv = y~dy \to v=y^2/2$$ $$\displaystyle{ 2\left[ \left. \frac{y^2}{2} \arccos(y)\right|_0^1 - \int_0^1{ \frac{-1}{2}\frac{y^2}{\sqrt{1-y^2}} ~dy } \right] }$$ $$\displaystyle{ \int_0^1{ \frac{y^2}{\sqrt{1-y^2}} ~dy } }$$ Use trig substitution. $$y=\sin\theta \to dy=\cos\theta~d\theta$$ Convert the limits of integration to $$\theta$$. $$y=0 \to \theta=0$$; $$y=1 \to \theta=\pi/2$$ $$\displaystyle{ \int_{0}^{\pi/2}\frac{\sin^2\theta}{\sqrt{1-\sin^2\theta}}\cos \theta ~d\theta }$$ $$\displaystyle{ \int_0^{\pi/2}{ \sin^2\theta d\theta } }$$ $$\displaystyle{ \int_0^{\pi/2}{ (1/2) (1-\cos(2\theta)) d\theta } }$$ $$\displaystyle{ \frac{1}{2}\left[ \theta - \frac{\sin(2\theta)}{2} \right]_0^{\pi/2} }$$ $$\displaystyle{ \frac{1}{2}\left[ \frac{\pi}{2} \right] = \frac{\pi}{4} }$$

$$\displaystyle{ \bar{y}= \frac{\pi/4}{2} = \frac{\pi}{8} }$$
centroid $$(0,\pi/8)$$

centroid $$(0,\pi/8)$$

[10 points] The corn in a grain silo, which as the shape of a right circular cylinder, is to be lifted out over the top edge of the silo. The silo is 20 feet in diameter and is 30 feet high. If the corn in the silo has a uniform depth of 25 feet, how much work is done in lifting all of the corn out of the silo? [Use $$\delta$$ lb/ft3 for the (weight) density of the corn.]

Problem Statement

[10 points] The corn in a grain silo, which as the shape of a right circular cylinder, is to be lifted out over the top edge of the silo. The silo is 20 feet in diameter and is 30 feet high. If the corn in the silo has a uniform depth of 25 feet, how much work is done in lifting all of the corn out of the silo? [Use $$\delta$$ lb/ft3 for the (weight) density of the corn.]

$$43750\pi\delta$$ ft-lb

Problem Statement

[10 points] The corn in a grain silo, which as the shape of a right circular cylinder, is to be lifted out over the top edge of the silo. The silo is 20 feet in diameter and is 30 feet high. If the corn in the silo has a uniform depth of 25 feet, how much work is done in lifting all of the corn out of the silo? [Use $$\delta$$ lb/ft3 for the (weight) density of the corn.]

Solution

Volume of a slice is $$\pi r^2 ~dy$$ giving a differential force of $$dF = \pi r^2 \delta ~dy$$ lbs.
The distance the slice if moved is $$30-y$$ if the bottom of the tank is at height zero.
Work is force times distance giving us the integral

 $$\displaystyle{ W = \int_0^{25}{ \pi r^2 \delta (30-y)~dy } }$$ $$\displaystyle{ W = 100 \pi \delta \int_0^{25}{ 30-y ~dy } }$$ $$\displaystyle{ W = 100 \pi \delta \left[ 30y - \frac{y^2}{2} \right]_0^{25} }$$ $$W = 100\pi \delta[ 750 - 625/2 ] = 43750\pi\delta$$ ft-lb

$$43750\pi\delta$$ ft-lb

[10 points] Calculate the length of the parametric curve $$x=3t^2, y=2t^3, 0 \leq x \leq 2$$.

Problem Statement

[10 points] Calculate the length of the parametric curve $$x=3t^2, y=2t^3, 0 \leq x \leq 2$$.

$$2[5\sqrt{5}-1]$$

Problem Statement

[10 points] Calculate the length of the parametric curve $$x=3t^2, y=2t^3, 0 \leq x \leq 2$$.

Solution

The length of a parametric curve is given by the equation

 $$\displaystyle{ s = \int_0^2{ \sqrt{ (dx/dt)^2 + (dy/dt)^2 } ~dt } }$$ $$\displaystyle{ x=3t^2 \to dx/dt = 6t }$$ $$\displaystyle{ y=2t^3 \to dy/dt = 6t^2 }$$ $$\displaystyle{ s = \int_0^2{ \sqrt{ 36t^2 + 36t^4 } ~dt } }$$ $$\displaystyle{ s = \int_0^2{ 6t \sqrt{ 1+t^2 } ~dt } }$$ Use integration by substitution. $$u=1+t^2 \to du=2t~dt$$ Change the limits of integration. $$t=0 \to u=1; t=2 \to u=5$$ $$\displaystyle{ s = \int_1^5{ 6t u^{1/2} \frac{du}{2t} } }$$ $$\displaystyle{ 3\int_1^5{u^{1/2}~du} }$$ $$\displaystyle{ \left. 3\frac{u^{3/2}}{3/2} \right|_1^5 }$$ $$2[5^{3/2}-1] = 2[5\sqrt{5}-1]$$

$$2[5\sqrt{5}-1]$$