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Calculus 1  Exam B3 (Final Exam) 
This is the final exam for first semester single variable calculus. The topics covered are integration by substitution, both indefinite and definite, solution of separable differential equations, optimization, area between curves, exponential growth/decay and volumes of revolution using both the disc/washer method and the shell/cylinder method. 

Exam Details 

Tools 
Time  2.5 hours  
Calculators  not allowed 
Questions  10  
Formula Sheet(s) 
one page 
Total Points  135  
Other Tools  ruler 


Calculus 1  Exam B3 
This is the final exam for first semester single variable calculus. The topics covered are integration by substitution, both indefinite and definite, solution of separable differential equations, optimization, area between curves, exponential growth/decay and volumes of revolution using both the disc/washer method and the shell/cylinder method. 

Exam Details 
Time  2.5 hours 
Questions  10 
Total Points  135 

Tools 
Calculators  not allowed 
Formula Sheet(s)  one page 
Other Tools  ruler 


Instructions:
 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 your answers in exact, simplified form.
Question 1 
[10 points] Evaluate \(\displaystyle{ \int{ \frac{e^{1/x}}{x^2}~dx } }\) 


\(\displaystyle{ e^{1/x}+C }\) 
Use integration by substitution.
Let \(u=1/x = x^{1} ~~ \to ~~ du=x^{2}dx\)
Move the negative sign to the left in the last equation to get \(du=x^{2}dx\).
Now the integral is
\(\displaystyle{ \int{e^u~du} = }\) \(\displaystyle{ e^u+C = }\) \(\displaystyle{ e^{1/x}+C }\)
Question 1 Final Answer 
\(\displaystyle{ e^{1/x}+C }\) 
Question 2 
[10 points] Evaluate \(\displaystyle{ \int_{1}^{e^3}{ \frac{\ln\sqrt{x}}{x}~dx } }\) 


\(\displaystyle{ 9/4 }\) 
Before integrating, we simplify the integrand a bit, i.e. \(\ln\sqrt{x} = \ln(x^{1/2}) = (1/2)\ln x\). 
\(\displaystyle{ \int_{1}^{e^3}{ \frac{\ln\sqrt{x}}{x}~dx } = }\)
\(\displaystyle{ \frac{1}{2}\int_{1}^{e^3}{ \frac{\ln x}{x}~dx } }\) 
Use integration by substitution. Let \(u=\ln x \to \) \(du=dx/x\).
We need to change the limits of integration. \(x=e^3 \to u=3\) and \(x=1 \to u=0\)

\(\displaystyle{ \frac{1}{2}\int_{1}^{e^3}{ \frac{\ln x}{x}~dx } = }\)
\(\displaystyle{ \frac{1}{2}\int_{0}^{3}{ u~du } = }\) \(\displaystyle{ \left. \frac{1}{2} \frac{u^2}{2} \right_0^3 =}\) \(\displaystyle{ \frac{1}{4}[ 3^2  0^2] =}\) \(\displaystyle{ \frac{9}{4} }\)

Another way
If you don't notice the initial simplification, this integral can still be solved with a different (more difficult) substitution, i.e. \(u=\ln x^{1/2}\) as follows.
Let \(u=\ln x^{1/2} \to \) \(du=(1/x^{1/2})(1/2)x^{1/2}dx =1/(2x)dx \to \) \(2du=dx/x\)
Change the limits of integration.
\(x=e^3 \to u=3/2\) and \(x=1 \to u=0\)
So the integral in terms of \(u\) is
\(\displaystyle{ \int_0^{3/2}{ 2u~du} = }\) \(\displaystyle{ \left. \frac{2u^2}{2}\right_0^{3/2} = }\) \(\displaystyle{ \left( \frac{3}{2}\right)^2  0^2 = \frac{9}{4} }\)
Question 2 Final Answer 
\(\displaystyle{ 9/4 }\) 
Question 3 
[15 points] Evaluate \(\displaystyle{ \int{ x\sqrt{2x+3}~dx } }\) 


\(\displaystyle{ \frac{1}{5}(2x+3)^{3/2}(x1) + C }\) 
Use integration by substitution. So let \(u=2x+3 \to \) \(du=2dx\) 
When we substitute, the integrand becomes \(x\sqrt{u}\). We need another factor to get rid of the \(x\). To do so, we solve our choice of \(u\) for \(x\) to get \(x=(u3)/2\). 
\(\displaystyle{ \int{ x\sqrt{2x+3}~dx } = }\)
\(\displaystyle{ \int{ \left( \frac{u3}{2} \right) u^{1/2} \frac{du}{2} } }\) 
Now we combine the \(1/2\)'s and pull the constant outside the integral. 
\(\displaystyle{ \frac{1}{4}\int{ (u3)u^{1/2}~du } }\) 
Since there is no product rule in integration, we need to multiply the \(u^{1/2}\) term through the \((u3)\) term. 
\(\displaystyle{ \frac{1}{4}\int{ u^{3/2}3u^{1/2}~du } }\) 
Now integrate. 
\(\displaystyle{ \frac{1}{4} \left[ \frac{u^{5/2}}{5/2}  \frac{3u^{3/2}}{3/2} \right] + C }\) 
Finally, we replace \(u\) to get an answer in terms of \(x\). 
\(\displaystyle{ \frac{1}{2}\left[ \frac{1}{5}(2x+3)^{5/2}  (2x+3)^{3/2} \right] + C }\) 
Depending on how much simplification your instructor requires, this can be reduced to 
\(\displaystyle{ \frac{1}{5}(2x+3)^{3/2}(x1) + C }\) 
There is another substitution that works here.
If we let \(u=\sqrt{2x+3}\) then \(\displaystyle{ du=\frac{1}{2}(2x+3)^{1/2}(2)dx }\). This gives us the integral in terms of \(u\) as \((1/2)\int{u^43u^2~du}\) and after integrating we get the same answer.
Question 3 Final Answer 
\(\displaystyle{ \frac{1}{5}(2x+3)^{3/2}(x1) + C }\) 
Question 4 
[10 points] Evaluate \(\displaystyle{ \int_2^5{ x^2\sqrt{x^34}~dx } }\) 


\(\displaystyle{ 294 }\) 
Use integration by substitution. So, let \(u=x^34 \to\) \(du=3x^2dx\) 
We also need to change our limits of integration. \(x=5 \to u=121\) and \(x=2 \to u=4\) 
\(\displaystyle{ \int_2^5{ x^2\sqrt{x^34}~dx } = }\)
\(\displaystyle{ \int_4^{121}{ u^{1/2} \frac{du}{3} } }\) 
Now we integrate. 
\(\displaystyle{ \left. \frac{1}{3} \frac{u^{3/2}}{3/2} \right_4^{121} }\) 
Now substitute the limits of integration. 
\(\displaystyle{ \frac{2}{9} \left[ (121)^{3/2}  4^{3/2} \right] }\) 
To simplify, we recognize that \((121)^{3/2} = 121(121)^{1/2}\) and \(121=11^2\) to get \((121)^{3/2} = 11^3\). Similarly, \(4^{3/2} = 4(4)^{1/2}\) which gives us \(4^{3/2}=2^3\). 
\(\displaystyle{ \frac{2}{9}[11^38] }\) 
Without a calculator, this answer is probably sufficient. However, if your instructor requires a more simplified answer, we get 
\(\displaystyle{ \frac{2}{9}(13318) = }\) \(\displaystyle{ \frac{2}{9}(1323) = }\) \(2(147) = 294\) 
Question 4 Final Answer 
\(\displaystyle{ 294 }\) 
Question 5 
[10 points] \(\int{ \text{_________}~dx} = (ax^2a^2)^4 + C \) If \(a\) is a constant, fill in the blank. 


\(\displaystyle{ 8ax(ax^2a^2)^3 }\) 
The key to this problem is to remember that integration is the opposite of differentiation. So all you need to do is to take the derivative of \((ax^2a^2)^4 + C\) to get the integrand of the integral. As stated in the problem, \(a\) is just a constant.
\(\displaystyle{ \frac{d}{dx}[(ax^2a^2)^4 + C] = }\) \(\displaystyle{ 4(ax^2a^2)^3 \frac{d}{dx}[ax^2a^2] = }\) \( 4(ax^2a^2)^3 (2ax) = 8ax(ax^2a^2)^3 \)
Question 5 Final Answer 
\(\displaystyle{ 8ax(ax^2a^2)^3 }\) 
Question 6 
[20 points] Find the particular solution of the differential equation \(f''(x)=e^{3x}+\cos(x)\) that satisfies the initial conditions \(f(0)=10/9\) and \(f'(0)=4/3\). 


\(\displaystyle{ f(x) = \frac{e^{3x}}{9}\cos x+x+2 }\) 
Usually, with separable differential equations, we have to do some work to separate the variables. However, in this problem we already have all the x's on the right and the function on the left. Also, it helps to think back about position, velocity and acceleration to realize that you need to integrate twice and, for each integration, determine the values of the constants.
Starting with \(f''(x)=e^{3x}+\cos(x)\) we integrate to get 
\(f'(x) = \int{e^{3x}+\cos x~dx} = \) \(\displaystyle{ \frac{e^{3x}}{3} + \sin x + C }\) 
Now it is time to use one of the initial conditions, \(f'(0)=4/3\) to find \(C\). When \(f'(0)=4/3\), we know that \(x=0\). 
\(\displaystyle{ f'(0) = \frac{1}{3} + 0 + C = \frac{4}{3} \to C=1}\) 
So, now we have \(\displaystyle{ f'(x)= \frac{e^{3x}}{3} + \sin x + 1 }\) 
To find the general solution \(f(x)\), we integrate again. 
\(\displaystyle{ f(x) = \int{ \frac{e^{3x}}{3} + \sin x + 1~dx} = }\) \(\displaystyle{ \frac{e^{3x}}{9}  \cos x + x +C }\) 
Now we use the other initial condition to find \(C\). 
\(\displaystyle{ \frac{1}{9}  \cos 0 + 0 + C = \frac{10}{9} \to C=2 }\) 
Question 6 Final Answer 
\(\displaystyle{ f(x) = \frac{e^{3x}}{9}\cos x+x+2 }\) 
Question 7 
[15 points] For the region bounded by the graphs \(y=9x^2\), \(x=2\) and \(x=3\) in the first quadrant, a. accurately sketch the region; b. using calculus, calculate the area of the region that you sketched in part a. 


\(8/3\) 
a. Here is a plot. We need to find the area of the shaded region.
\(
\begin{array}{rcl}
A & = & \displaystyle{ \int_2^3{ 9x^2~dx } } \\
& = & \displaystyle{ \left[ 9x\frac{x^3}{3} \right]_2^3 } \\
& = & \displaystyle{ \left[ 27  \frac{27}{3} \right]  \left[ 18  \frac{8}{3} \right] } \\
& = & \displaystyle{ 18  18 + \frac{8}{3} = \frac{8}{3}}
\end{array}
\)
Question 7 Final Answer 
\(8/3\) 
Question 8 
[10 points] The mass of radioactive material in a sample has decreased by \(30%\) since the decay began. Assuming a halflife of \(1500\) years, how long ago did the decay begin? 


\(\displaystyle{ t=\frac{1500\ln(0.70)}{\ln 2} }\) 
This is an exponential decay problem. So our general equation is \(y(t)=y_0 e^{kt}\) where \(k<0\).
We can also think about this as \(y(t)=y_0 e^{kt}\) where \(k>0\). We will use the first equation. 
First, we are going to use the information about the halflife at \(t=1500\) to get a value for \(k\). If the initial amount is \(y_0\), then half of the initial amount is \(y_0/2\). 
\(
\begin{array}{rcl}
\displaystyle{ \frac{y_0}{2} } & = & y_0 e^{k(1500)} \\
\displaystyle{ \frac{1}{2} } & = & e^{k(1500)} \\
\ln 2 & = & k(1500) \\
k & = & \displaystyle{ \frac{\ln 2}{1500} }
\end{array}
\) 
Now, we know that after \(t\) years, the initial amount has decreased \(30%\). This tells us that \(70%\) remains. So know we need to find \(t\) when we have \(70%\) of the initial amount. 
\(\displaystyle{ 0.70 y_0 = y_0 exp\left[ \frac{t\ln 2}{1500} \right] }\) 
In this last equation, we have used the alternate notation for the exponential function, i.e. \(e^t = exp[t]\). This makes it easier to see the exponent of \(e\). 

Next, we take the natural log of both sides and solve for \(t\). 
\(\displaystyle{ \ln(0.70) = \frac{t\ln 2}{1500} \to }\) \(\displaystyle{ t=\frac{1500\ln(0.70)}{\ln 2} }\) 
Question 8 Final Answer 
\(\displaystyle{ t=\frac{1500\ln(0.70)}{\ln 2} }\) 
Question 9 
[10 points] A zookeeper needs to add a rectangular outdoor pen to an animal house with a corner notch, as shown in the figure. If \(85\)m of new fence is available, what dimensions of the pen will maximize its area? No fence will be used along the walls of the animal house.



\(25\)m X \(25\)m 
For this problem, we need to maximize the area of a rectangle, so we need to assign some variables in order to come up with an area equation. We will call the bottom length \(x\) and the right side length \(y\). Therefore our area equation is \(A=xy\).
Now we need to use the information that we have \(85\)m of fence to come up with an equation for the perimeter of the rectangle including only the lengths where we will be using fence. So, the right and the bottom side area pretty obvious, they are \(y\) and \(x\). However, the top side that requires fence is \(x10\) and the left side is \(y5\). So the length of fence we need is \(x+(x10)+y+(y5)\) and this is equal to how much fence we have, \(85\)m. So we have \(x+(x10)+y+(y5)=85\). After simplification we have \(x+y=50\). We will solve this for \(y\) to be substituted back into the area equation before taking the derivative.
\(y=50x\) and since \(A=xy\), then \(A=x(50x)\).
Now we can take the derivative of the area equation with respect to \(x\) since we have only one variable there now.
\(\displaystyle{ \frac{dA}{dx} = 502x }\)
Setting this last equation to zero and solving for \(x\) gives us \(x=25\).
Question 9 Final Answer 
\(25\)m X \(25\)m 
Question 10 
[25 points] The region bounded by the graphs of \(y=2x\), \(y=6x\) and \(y=0\) is revolved about the line \(y=2\) to produce a volume. Calculate the volume that is produced. Set up the integrals for both the disc/washer and shell/cylinder methods and evaluate one of them. Notes: Make sure you plot and shade the region and show the example rectangles, as well as the distances R, r, p and h. Each volume should have its own plot, i.e. you should have one plot for the disc/washer method and another separate plot for the shell/cylinder method. [10 points extra credit] Evaluate the other integral. 


\( 80\pi \) 

disc/washer method 
basic integral \(\displaystyle{ V=\pi\int_a^b{ R^2r^2~dx} }\) 
We need two integrals with a break at \(x=2\). 
\([0,2]\): \(R=2x+2, ~~ r=2\) 
\([2,6]\): \(R=6x+2, ~~ r=2\) 
\(
\begin{array}{rcl}
V & = & \displaystyle{ \pi\int_0^2{ (2x+2)^2  2^2~dx} + \pi\int_2^6{ (8x)^2  2^2~dx} } \\
& = & \displaystyle{ \pi\int_0^2{ 4x^2+8x~dx } + \pi\int_2^6{ 6016x+x^2~dx } } \\
& = & \displaystyle{ \pi\left[ \frac{4x^3}{3} + \frac{8x^2}{2} \right]_0^2 + \pi\left[ 60x\frac{16x^2}{2} + \frac{x^3}{3} \right]_2^6 } \\
& = & 80\pi
\end{array}
\)


shell/cylinder method 
basic integral \(\displaystyle{ V=2\pi\int_c^d{ ph~dy} }\) 
\(p=y+2\) and \(h=(6y) (y/2)\) 
\(
\begin{array}{rcl}
V & = & \displaystyle{ 2\pi \int_0^4{ (y+2)(63y/2)~dy } } \\
& = & \displaystyle{ 2\pi \int_0^4{ 3y+(3/2)y^2+12~dy } } \\
& = & \displaystyle{ 2\pi \left[ \frac{3y^2}{2}  \frac{3}{2}\frac{y^3}{3} + 12y \right]_0^4 } \\
& = & 80\pi
\end{array}
\)

Question 10 Final Answer 
\( 80\pi \) 