Calculus 2  Exam A1
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This is the first exam for second semester single variable calculus.
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 answers.
Part A  Questions 14
Instructions for Part A   You have 30 minutes to complete this part of the exam. No calculators are allowed. Show all your work, justify your conclusions and give exact, completely factored answers.
Evaluate the following integrals. Each question in this section is worth 9 points.
Question 1 
\(\displaystyle{ \int_{0}^{1}{x^2e^{4x}~dx} }\)



\(\displaystyle{ \int_{0}^{1}{x^2e^{4x}~dx} =\frac{e^413}{32e^4} }\)

Use integration by parts: \( \int{u~dv} = uv\int{v~du}\)
\(u=x^2\)  \(\to\)  \(du=2x~dx\) 
\(dv=e^{4x}dx\)  \(\to\)  \(\displaystyle{ \frac{e^{4x}}{4} }\) 
\(\displaystyle{
\left. \frac{x^2}{4}e^{4x} \right_0^1  \int_{0}^{1}{\left(\frac{1}{4} e^{4x} \right) 2x~dx} = }\) \(\displaystyle{
\frac{1}{4}e^{4} + 0 + \frac{1}{2} \int_{0}^{1}{xe^{4x}dx}
}\)
Use integration by parts again.
\(u=x\)  \(\to\)  \(du=dx\) 
\(dv = e^{4x}dx\)  \(\to\)  \(\displaystyle{ v=\frac{e^{4x}}{4} }\) 
\(\displaystyle{
\frac{1}{4e^4} + \frac{1}{2}\left[ \left. \frac{x}{4}e^{4x} \right_0^1  \int_{0}^{1}{\frac{1}{4}e^{4x}dx} \right]
}\)
\(\displaystyle{
\frac{1}{4e^4} + \frac{1}{2}\left[ \frac{1}{4}e^{4} +0 \right] + \frac{1}{8} \int_{0}^{1}{e^{4x}dx}
}\)
\(\displaystyle{
\frac{3}{8e^4} + \left. \frac{1}{8}\frac{e^{4x}}{4} \right_0^1
}\)
\(\displaystyle{
\frac{3}{8e^4}  \frac{1}{32}\left[ e^{4}  1 \right] = \frac{e^413}{32e^4}
}\)
Question 1 Final Answer 
\(\displaystyle{ \int_{0}^{1}{x^2e^{4x}~dx} =\frac{e^413}{32e^4} }\)

Question 2 
\( \int{\sin^3(\pi t) \cos^3 (\pi t)~dt}\)



\(\displaystyle{ \int{\sin^3(\pi t) \cos^3 (\pi t)~dt} = }\) \(\displaystyle{ \frac{\cos^4 (\pi t )}{12} (2\cos^2 (\pi t)  3) + C}\)

\( \int{\sin(\pi t) \sin^2(\pi t) \cos^3 (\pi t)~dt}\)
\( \int{\sin(\pi t) (1\cos^2 (\pi t) ) \cos^3 (\pi t)~dt}\)
\( u = \cos(\pi t) ~~ \to ~~ du = \sin(\pi t)dt\)
\(\displaystyle{
\begin{array}{rcl}
\int{(1u^2)u^3du} & = & \int{u^5u^3 du} \\
& = & \frac{u^6}{6}  \frac{u^4}{4} +C \\
& = & \frac{u^4}{12}(2u^23) + C \\
& = & \frac{\cos^4 (\pi t )}{12} (2\cos^2 (\pi t)  3) + C
\end{array}
}\)
Question 2 Final Answer 
\(\displaystyle{ \int{\sin^3(\pi t) \cos^3 (\pi t)~dt} = }\) \(\displaystyle{ \frac{\cos^4 (\pi t )}{12} (2\cos^2 (\pi t)  3) + C}\)

Question 3 
\(\displaystyle{ \int_{0}^{2}{\frac{3}{(r^2+4)^{3/2}}dr} }\)



\(\displaystyle{ \int_{0}^{2}{\frac{3}{(r^2+4)^{3/2}}dr} = \frac{3\sqrt{2}}{8} }\)

Use \( 1+\tan^2(\theta) = \sec^2(\theta) \) and let \( r = 2\tan(\theta) ~~ \to ~~ dr = 2\sec^2(\theta)~d\theta \).
The triangle based on \( r = 2\tan(\theta) \) is shown to the right. Using the Pythagorean Theorem, the hypotenuse is
\(\displaystyle{ \sqrt{r^2+4} }\).
First, we will evaluate the integral without the limits of integration (the indefinite integral).
\(\displaystyle{
\begin{array}{rcl}
\int{\frac{3}{(r^2+4)^{3/2}}dr} & = & \int{\frac{3}{(4\tan^2(\theta) + 4)^{3/2}}2\sec^2(\theta)d\theta} \\
& = & \int{\frac{6\sec^2(\theta)}{4^{3/2}(1+\tan^2(\theta))^{3/2}}d\theta} \\
& = & \frac{3}{4} \int{\frac{\sec^2(\theta)}{(\sec^2(\theta))^{3/2}}d\theta} \\
& = & \frac{3}{4} \int{\frac{1}{\sec(\theta)}d\theta} \\
& = & \frac{3}{4} \int{\cos(\theta)~d\theta} \\
& = & \frac{3}{4} \sin(\theta) + C \\
& = & \frac{3}{4} \frac{r}{\sqrt{r^2+4}} + C
\end{array}
}\)
In the last step above, we used the triangle to convert from \(\theta\) to \(r\).
Now we will evaluate the integral using the limits of integration.
\(\displaystyle{
\left. \frac{3}{4} \frac{r}{\sqrt{r^2+4}} \right_0^2 = \frac{3}{4} \frac{2}{\sqrt{8}}  0 =
\frac{3}{2} \frac{1}{2\sqrt{2}} = \frac{3\sqrt{2}}{8}
}\)
Question 3 Final Answer 
\(\displaystyle{ \int_{0}^{2}{\frac{3}{(r^2+4)^{3/2}}dr} = \frac{3\sqrt{2}}{8} }\)

Question 4 
\(\displaystyle{ \int{\frac{9x^2+6x+12}{(x+1)(x^2+4)} dx} }\)



\(\displaystyle{ \int{\frac{9x^2+6x+12}{(x+1)(x^2+4)} dx} = }\) \(\displaystyle{ \ln \abs{(x+1)^3 (x^2+4)^3} + k }\)

Use partial fraction expansion.
\(\displaystyle{ \frac{9x^2+6x+12}{(x+1)(x^2+4)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+4} }\)
\( 9x^2+6x+12 = A(x^2+4) + (Bx+C)(x+1) \)
\( x = 1 \): \( ~~ 96+12 = A(1+4) \) \(\to\) \( 15 = A(5) \) \(\to\) \( A=3 \)
\( x=0 \): \( ~~ 12 = A(4)+C(1) \) \(\to\) \( 12 = 12+C \) \(\to\) \( C=0 \)
\( x=1 \): \( ~~ 9+6+12 = A(5) + B(1+1) \) \(\to\) \( 27 = 15 +2B \) \(\to\) \( (2715)/2 = B \) \(\to\) \( B = 6 \)
The integral can now be written
\(\displaystyle{ \int{\frac{3}{x+1} + \frac{6x}{x^2+4} dx} }\)
The first fraction is easily integrated.
\(\displaystyle{ \int{\frac{3}{x+1}dx} = 3\ln\abs{x+1} + c_1 }\)
Use substitution for the second fraction.
\( u = x^2+4 ~~ \to ~~ du = 2x~dx \)
\(\displaystyle{ \int{\frac{6x}{x^2+4}dx} = }\) \(\displaystyle{ \int{\frac{3~du}{u} dx} = }\) \(\displaystyle{3\ln\abs{u} + c_2 = }\) \(\displaystyle{ 3\ln\abs{x^2+4} + c_2 }\)
Putting the results together, we get
\( 3 \ln\abs{x+1} + 3 \ln\abs{x^2+4} + k = \) \( 3\ln\abs{(x+1)(x^2+4)} + k = \) \( \ln \abs{(x+1)^3 (x^2+4)^3} +k \)
We combined the constants \( k = c_1 + c_2 \).
Question 4 Final Answer 
\(\displaystyle{ \int{\frac{9x^2+6x+12}{(x+1)(x^2+4)} dx} = }\) \(\displaystyle{ \ln \abs{(x+1)^3 (x^2+4)^3} + k }\)

Part B  Questions 510
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 answers.
A region \(R\) is bounded by the graph of \( y = e^x \), the line \( y=1 \) and the line \(x=3\).
Sketch the region \(R\). [ click here to show/hide the plot ]
Question 5 
A solid is generated by revolving the region \(R\) about the line \( x=1 \). Set up, but do not evaluate, a definite integral for the volume of the resulting solid of revolution.




Since the problem statement did not specify which method, we can use either the disc or shell method. Both solutions are shown here.
Disc Method 
\(\displaystyle{ V = \pi \int_c^d{R^2r^2~dy} }\)
\( R = 4 ~~~~ r = x+1 \)
\( y = e^x ~~ \to ~~ \ln(y)=x ~~ \to ~~ r = \ln(y)+1 \)
\(\displaystyle{ V = \pi \int_1^{e^3}{16  [1+\ln(y)]^2~dy} }\) 
Shell Method 
\(\displaystyle{ V = 2\pi \int_a^b{rh~dx} }\)
\( r = 1+x ~~~~ h = y1 = e^x  1 \)
\(\displaystyle{ V = 2\pi \int_0^3{(1+x)(e^x1)~dx} }\) 
Question 6 
A second solid has base \(R\) and crosssections perpendicular to the xaxis are disks with a diameter in \(R\). Set up, but do not evaluate, a definite integral for the volume of this region.




Disks mean that we have circles with diameter \(d\), so the radius is \(r=d/2\). The area of a disk is \(\pi r^2\). The radius is determined by crosssections perpendicular to the xaxis, which are dependent on \(x\).
\(d = y1 = e^x1 ~~ \to ~~ r = (e^x1)/2 \)
\(\displaystyle{ V = \int_0^3{\pi \left( \frac{e^x1}{2} \right)^2 dx}}\)
Solve the following problems. Make sure your answers have correct units.
Question 7 
(16 points) A thin metal plate in the plane occupies the triangle with vertices \((0,2)\),
\((0,0)\) and \((2,0)\). Find the centroid of the plate. You may use symmetry and known area or volume formulas from geometry to evaluate integrals.



\(\displaystyle{ \left( \frac{2}{3}, \frac{2}{3} \right) }\)

From symmetry we can see that \(\bar{x} = \bar{y} \).
The equation for centroid is
\(\displaystyle{ \bar{x} = \frac{\text{moment about x}}{\text{area}} = \frac{\int_a^b{x(y_2y_1)~dx}}{\text{area}} }\)
where \( y_1 = 0 \) and \( y_2 = 2x \)
The area of the triangle is \( A = bh/2 = (2)(2)/2 = 2 \)
\(\displaystyle{
\begin{array}{rcl}
\int_0^2{x(2x0)~dx} & = & \int_0^2{2xx^2~dx} \\
& = & \left[ \frac{2x^2}{2}  \frac{x^3}{3} \right]_0^2 \\
& = & 4  \frac{8}{3} = \frac{4}{3}
\end{array}
}\)
\( \displaystyle{\bar{x} = \frac{4/3}{2} = \frac{2}{3}}\)
Question 7 Final Answer 
\(\displaystyle{ \left( \frac{2}{3}, \frac{2}{3} \right) }\)

Question 8 
(16 points) An upright water tank has a flat base in the plane and is filled with water. The tank has height \(7\) feet, and, for \( 0 \le y \le 7 \), the crosssection parallel to the base at height \(y\) has area \( y^2+2\). How much work is done in pumping all the water over the top edge of the tank? [Use \( \delta~ lb/ft^3\) for the weightdensity of water.]



\(\displaystyle{ W = \frac{2989\delta}{12}} ~ftlbs \)

The picture to the right is meant to represent an upright tank filled with water. The \(dy\) in the picture is a thin differential slice of water with cross sectional area is given in the problem statement as \( A=y^2+2\).
To calculate the work, we know that the work, \(W\) is force times distance, \( W = F \cdot d \). To get force, we multiply the volume of one small slice of water times the weightdensity.
The volume of one slice of water is \( dV = (y^2+2)dy ~ ft^3 \).
The force of one slice of water is \( dF = \delta \cdot dV = \delta (y^2+2)dy ~ lbs \)
Now, the distance \(d\) is the distance one slice of the water is moved up over the top of the tank. The very top slice of water is moved a distance \( d=0~ft\) while the slice at the bottom of the tank is moved \(d=7~ft\). So our expression for the distance of any given slice of water is dependent on \(y\) and is \( d= 7y \). Now we are ready to set up our integral.
\( \displaystyle{\int_0^7{\delta (y^2+2)(7y)dy}}\)
We integrate from the bottom of the tank (\(y=0)\) to the top of the tank \((y=7)\) since the tank is filled to the top with water.
\( \displaystyle{
\begin{array}{rcl}
\int_0^7{\delta (y^2+2)(7y)dy} & = & \int_0^7{7y^2y^3+142y~dy} \\
& = & \delta \left[ \frac{7y^3}{3}\frac{y^4}{4} + 14y \frac{2y^2}{2} \right]_0^7 \\
& = & \delta \left[ \frac{7}{3}(7^3)  \frac{7^4}{4} + 14(7)  7^2 \right]  0 \\
& = & \frac{49(61)\delta}{12}
\end{array}
}\)
Question 8 Final Answer 
\(\displaystyle{ W = \frac{2989\delta}{12}} ~ftlbs \)

Let G be the portion of the graph with parametric equations
\( x= 1+\sin(t)\) \( y = 3+2\cos(t)\) for \( 0 \le t \le 2\pi\).
Plot G and answer the following questions based on the graph.
[ click here to show/hide the plot of G ]
Question 9 
(8 points) Set up, but do not evaluate, the integral for the length of G.



\(\displaystyle{ s=\int_0^{2\pi}{\sqrt{\cos^2t+4\sin^2t}~dt}}\)

The general equation is \(\displaystyle{ s = \int_a^b{\sqrt{[x'(t)]^2 + [y'(t)]^2}~dt} }\)
In this question
\( x(t) = 1+\sin(t) ~~ \to ~~ x'(t) = \cos(t) \)
\( y(t) = 3+2\cos(t) ~~ \to ~~ y'(t) = 2\sin(t) \)
So our integral is \(\displaystyle{ s=\int_0^{2\pi}{\sqrt{\cos^2t+4\sin^2t}~dt}}\)
Question 9 Final Answer 
\(\displaystyle{ s=\int_0^{2\pi}{\sqrt{\cos^2t+4\sin^2t}~dt}}\)

Question 10 
(8 points) Set up, but do not evaluate, the integral for the area of the surface generated by revolving G about the xaxis.



\(\displaystyle{ A_x = 2\pi \int_0^{2\pi}{(3+2\cos(t)) \sqrt{\cos^2t + 4\sin^2t } ~ dt} }\)

The general equation is \(\displaystyle{ A_x = 2\pi \int_a^b{y(t) \sqrt{(dx/dt)^2 + (dy/dt)^2} ~ dt} }\)
From the previous question, we know \( dx/dt\) and \(dy/dt\). So we just need to write down the integral.
\(\displaystyle{ A_x = 2\pi \int_0^{2\pi}{(3+2\cos(t)) \sqrt{\cos^2t + 4\sin^2t } ~ dt} }\)
Question 10 Final Answer 
\(\displaystyle{ A_x = 2\pi \int_0^{2\pi}{(3+2\cos(t)) \sqrt{\cos^2t + 4\sin^2t } ~ dt} }\)
