## 17Calculus - Calculus 2 - Practice Exam 1 (Midterm) (Semester B)

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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.

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

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

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

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

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

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

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

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

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

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

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. $$\displaystyle{ \frac{1}{27} \frac{x}{\sqrt{9+x^2}} + C }$$

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

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

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.

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

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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: $$y=4 \to u=4-4= 0$$ $$-\int_3^0{ u^{1/2}~du }$$ $$\int_0^3{ u^{1/2}~du }$$ $$\displaystyle{ \left. \frac{u^{3/2}}{3/2} \right|_0^3 }$$ $$\displaystyle{ \frac{2}{3}3^{3/2} - 0 }$$ $$A = 2\sqrt{3}$$

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

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

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

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

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

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

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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/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

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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 $$x=3t^2, y=2t^3, 0 \leq x \leq 2$$.

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

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. $$t=0 \to u=1; t=2 \to u=5$$ $$\displaystyle{ s = \int_1^5{ 6t u^{1/2} \frac{du}{2t} } }$$ $$\displaystyle{ 3\int_1^5{u^{1/2}~du} }$$ $$\displaystyle{ \left. 3\frac{u^{3/2}}{3/2} \right|_1^5 }$$ $$2[5^{3/2}-1] = 2[5\sqrt{5}-1]$$

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

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You CAN Ace Calculus

 Integration by Parts Integrals - Trig Substitution Integrals - Partial Fractions Volume Integrals Centroid Work Parametrics

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other exams from calculus 2

### Trig Formulas

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 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle 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

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

### Topics Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

Precalculus

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