This is the midterm exam for second semester single variable calculus.
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.
 Use the same materials that you are allowed in your exam (unless the instructions with these exams are more strict).
 Use your calculator as little as possible except for graphing and checking your calculations.
 Work the entire exam before checking any solutions.
 After checking your work, rework any problems you missed and go to the 17calculus page discussing the material to perfect your skills.
 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.
 Show all your work.
 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).
 Give exact, simplified answers.
Evaluate these integrals.
Part A  Questions 15
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 

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

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

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

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

\(\displaystyle{ \frac{5}{8}(2+t^2)^4 + C }\) 
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\(\int{ (2x+1)\ln x ~ dx } \)
Problem Statement 

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

\((x^2+x)\ln x  (1/2)x^2  x + C \)
Problem Statement 

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

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

\((x^2+x)\ln x  (1/2)x^2  x + C \) 
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\( \int{ \sin^3\theta ~ d\theta } \)
Problem Statement 

\( \int{ \sin^3\theta ~ d\theta } \)
Final Answer 

\( (1/3)\cos^3\theta  \cos\theta + C \)
Problem Statement 

\( \int{ \sin^3\theta ~ d\theta } \)
Solution 

\( \int{ \sin^3\theta ~ d\theta } \) 
\( \int{ \sin \theta (\sin^2\theta)~d\theta } \) 
\( \int{ \sin \theta ( 1\cos^2\theta )~d\theta } \) 
\( u=\cos\theta \to du = \sin\theta d\theta \) 
\( \int{ (1u^2)(du) }\) 
\( \int{ u^21 ~du } \) 
\( u^3 / 3  u + C \) 
\( (1/3)\cos^3\theta  \cos\theta + C \) 
Final Answer 

\( (1/3)\cos^3\theta  \cos\theta + C \) 
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\(\displaystyle{ \int{ \frac{dx}{(9+x^2)^{3/2}} } }\)
Problem Statement 

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

\(\displaystyle{ \frac{x}{27\sqrt{9+x^2}} + C }\)
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. 
\(\displaystyle{ \frac{1}{27} \frac{x}{\sqrt{9+x^2}} + C }\) 
Final Answer 

\(\displaystyle{ \frac{x}{27\sqrt{9+x^2}} + C }\) 
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\(\displaystyle{ \int{ \frac{2x^2+x4}{x^3x^22x} ~dx } }\)
Problem Statement 

\(\displaystyle{ \int{ \frac{2x^2+x4}{x^3x^22x} ~dx } }\)
Final Answer 

\(\displaystyle{ \ln\left \frac{x^2(x2)}{x+1} \right + C }\)
Problem Statement 

\(\displaystyle{ \int{ \frac{2x^2+x4}{x^3x^22x} ~dx } }\)
Solution 

Check to see if the denominator factors. If so, then try partial fractions.
\(x^3x^22x = x(x^2x2) = x(x2)(x+1)\) 
\(\displaystyle{ \frac{2x^2+x4}{x^3x^22x} }\) 
\(\displaystyle{ \frac{A}{x} + \frac{B}{x2} + \frac{D}{x+1} }\) 
Solving for A, B and C, we get \(A=2\), \(B=1\) and \(D=1\). Now the integral looks like this. 
\(\displaystyle{ \int{ \frac{2}{x} + \frac{1}{x2}  \frac{1}{x+1} ~ dx } }\) 
\(2\lnx + \lnx2  \lnx+1 + C \) 
\(\displaystyle{ \ln\left \frac{x^2(x2)}{x+1} \right + C }\) 
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.
Final Answer 

\(\displaystyle{ \ln\left \frac{x^2(x2)}{x+1} \right + C }\) 
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Part B  Questions 611
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 68, consider the region \(\mathcal{R}\) bounded by the graph of \(y=4x^2\), the yaxis 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.
Final Answer 

\(A = 2\sqrt{3} \)
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 \(y=4x^2\), the lower function is \(y=1\). We need to find the xvalue where these intersect on the right end of the area.
\(4x^2=1 \to x^2 = 3 \to x=\pm \sqrt{3}\)
Since we are in the first quadrant, x is positive, so \(x=\sqrt{3}\).
\(A = \int_a^b{f(x)g(x)~dx}\) 
\(A = \int_0^{\sqrt{3}}{ (4x^2)  1 ~ dx }\) 
\(A = \int_0^{\sqrt{3}}{ 3x^2 ~ dx }\) 
\(A = [ 3x  x^3/3 ]_0^{\sqrt{3}} \) 
\(A = 2\sqrt{3} \) 
For the ydirection, the right function is \(y=4x^2\) which we need to solve for y.
\(y=4x^2 \to x^2=4y \to x=\pm\sqrt{4y} \)
Since the part of the function that we need is in the first quadrant, x is positive, so our right function is \( x=\sqrt{4y} \).
The left function is \(x=0\). We integrate from 1 to 4.
\(A = \int_c^d{f(y)g(y)~dx}\) 
\(A = \int_1^4{\sqrt{4y}0~dx}\) 
Use integration by substitution. 
\(u = 4y \to du = dy\) 
Convert the limits of integration in terms of u. 
lower limit: \(y=1 \to u=41=3 \) 
upper limit: \(y=4 \to u=44= 0\) 
\(\int_3^0{ u^{1/2}~du } \) 
\( \int_0^3{ u^{1/2}~du } \) 
\(\displaystyle{ \left. \frac{u^{3/2}}{3/2} \right_0^3 }\) 
\(\displaystyle{ \frac{2}{3}3^{3/2}  0 }\) 
\(A = 2\sqrt{3}\) 
Final Answer 

\(A = 2\sqrt{3} \) 
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[5 points] A solid is generated by revolving the region \(\mathcal{R}\) about the xaxis. Set up, but do not evaluate, the definite integral for the volume of the resulting solid of revolution using the method of discwashers.
Problem Statement 

[5 points] A solid is generated by revolving the region \(\mathcal{R}\) about the xaxis. Set up, but do not evaluate, the definite integral for the volume of the resulting solid of revolution using the method of discwashers.
Final Answer 

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

[5 points] A solid is generated by revolving the region \(\mathcal{R}\) about the xaxis. Set up, but do not evaluate, the definite integral for the volume of the resulting solid of revolution using the method of discwashers.
Solution 

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

\( V = \pi \int_0^{\sqrt{3}}{ (4x^2)^2  1 ~ dx } \) 
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[5 points] A solid is generated by revolving the region \(\mathcal{R}\) about the xaxis. Set up, but do not evaluate, the definite integral for the volume of the resulting solid of revolution using the method of shellcylinders.
Problem Statement 

[5 points] A solid is generated by revolving the region \(\mathcal{R}\) about the xaxis. Set up, but do not evaluate, the definite integral for the volume of the resulting solid of revolution using the method of shellcylinders.
Final Answer 

\( V = 2\pi\int_1^4{ y\sqrt{4y}~dy } \)
Problem Statement 

[5 points] A solid is generated by revolving the region \(\mathcal{R}\) about the xaxis. Set up, but do not evaluate, the definite integral for the volume of the resulting solid of revolution using the method of shellcylinders.
Solution 

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

\( V = 2\pi\int_1^4{ y\sqrt{4y}~dy } \) 
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Section 2 

[15 points] Find the centroid of the region bounded by the graph of \(y=\cos x\) and the xaxis, 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 xaxis, between \(x=\pi/2\) and \(x=\pi/2\). Make a sketch and use symmetry where possible.
Final Answer 

centroid \((0,\pi/8)\)
Problem Statement 

[15 points] Find the centroid of the region bounded by the graph of \(y=\cos x\) and the xaxis, 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
\(\displaystyle{ \bar{y} = \frac{\int_c^d{y(x_2x_1)dy}}{\int_c^d{(x_2x_1~dy)}} }\)
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. 
\( u=\arccos(y) \to du=dy/\sqrt{1y^2} \) 
\( dv = y~dy \to v=y^2/2 \) 
\(\displaystyle{ 2\left[ \left. \frac{y^2}{2} \arccos(y)\right_0^1  \int_0^1{ \frac{1}{2}\frac{y^2}{\sqrt{1y^2}} ~dy } \right] }\) 
\(\displaystyle{ \int_0^1{ \frac{y^2}{\sqrt{1y^2}} ~dy } }\) 
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 \((0,\pi/8)\)
Final Answer 

centroid \((0,\pi/8)\) 
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[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/ft^{3} 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/ft^{3} for the (weight) density of the corn.]
Final Answer 

\( 43750\pi\delta \) ftlb
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/ft^{3} 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 \(30y\) if the bottom of the tank is at height zero.
Work is force times distance giving us the integral
\(\displaystyle{ W = \int_0^{25}{ \pi r^2 \delta (30y)~dy } }\) 
\(\displaystyle{ W = 100 \pi \delta \int_0^{25}{ 30y ~dy } }\) 
\(\displaystyle{ W = 100 \pi \delta \left[ 30y  \frac{y^2}{2} \right]_0^{25} }\) 
\( W = 100\pi \delta[ 750  625/2 ] = 43750\pi\delta \) ftlb 
Final Answer 

\( 43750\pi\delta \) ftlb 
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[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 \(x=3t^2, y=2t^3, 0 \leq x \leq 2\).
Final Answer 

\( 2[5\sqrt{5}1] \)
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. 
\( t=0 \to u=1; t=2 \to u=5 \) 
\(\displaystyle{ s = \int_1^5{ 6t u^{1/2} \frac{du}{2t} } }\) 
\(\displaystyle{ 3\int_1^5{u^{1/2}~du} }\) 
\(\displaystyle{ \left. 3\frac{u^{3/2}}{3/2} \right_1^5 }\) 
\( 2[5^{3/2}1] = 2[5\sqrt{5}1]\) 
Final Answer 

\( 2[5\sqrt{5}1] \) 
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You CAN Ace Calculus
Integration by Parts 
Integrals  Trig Substitution 
Integrals  Partial Fractions 
Volume Integrals 
Centroid 
Work 
Parametrics 
other exams from calculus 2 

The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) 
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = \sin(t) }\)  
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) 
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = \csc^2(t) }\)  
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) 
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = \csc(t)\cot(t) }\) 
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
Trig Integrals
\(\int{\sin(x)~dx} = \cos(x)+C\) 
\(\int{\cos(x)~dx} = \sin(x)+C\)  
\(\int{\tan(x)~dx} = \ln\abs{\cos(x)}+C\) 
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)  
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) 
\(\int{\csc(x)~dx} = \) \( \ln\abs{\csc(x)+\cot(x)}+C\) 
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