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\centerline{Mathematics 134 -- Calculus 2 With Fundamentals}
\centerline{Exam 2 -- Answers/Solutions for Sample Questions}
\centerline{March 2, 2018}

\vskip 10pt
\noindent
{\it Sample Exam Questions}
\vskip 10pt
\noindent
{\it Disclaimer: The actual exam questions may be organized differently and ask questions in 
different ways.  This list is also quite a bit longer than the actual exam will be (to give you some
idea of the range of different questions that might be asked).}
\vskip 10pt
\noindent
I.  Compute each of the integrals below using some combination of
basic rules, substitution, integration by parts.  You must show all work
for full credit. 
\begin{enumerate}
\item[A)]  $$\int \frac{e^{\tan^{-1}(x)}}{1 + x^2}\ dx$$
\item[] {\it Solution:}  By substitution.  Let $u = \tan^{-1}(x)$.  The form is $\int e^u\ du$, so the 
integral is $e^{\tan^{-1}(x)} + C$.
\item[B)]  $$\int e^x \sin(2x) \ dx$$
\item[] {\it Answer:}  Apply integration by parts twice; the second time, the original integral
is returned, but with a numerical coefficient, so you can solve for it algebraically.  The integral
equals:  $\frac{1}{5}e^x \sin(2x) - \frac{2}{5} e^x \cos(2x) + C$.
\item[C)]  $$\int \sin^3(x) \cos(x)\ dx$$
\item[] {\it Solution:} By substitution.  Let $u = \sin(x)$, so $du = \cos(x)\ dx$ and the 
form is 
$$\int u^3\ du = \frac{1}{4} u^4 + C = \frac{1}{4} \sin^4(x) + C.$$
\item[D)]  $$\int_0^{1/2} \sin^{-1}(x)\ dx$$
\item[] {\it Answer:}  Use integration by parts with $u = \sin^{-1}(x)$ and $dv = dx$.  The 
answer is
\begin{align*}
\left. x\sin^{-1}(x)\right|_0^{1/2} - \int_0^{1/2} \frac{x}{\sqrt{1 - x^2}}\ dx 
&= \frac{\pi}{12} + \left. \sqrt{1 - x^2}\right|_0^{1/2}\\
&= \frac{\pi}{12} + \frac{\sqrt{3}}{2} - 1.
\end{align*}
\item[E)]  Use integration by parts to show this reduction formula:
If $n$ is a positive integer, then
$$\int x^n e^{ax}\ dx = \frac{x^n e^{ax}}{a} - \frac{n}{a} \int x^{n-1} e^{ax}\ dx$$
\item[] {\it Solution:}  This comes directly from the integration by parts formula if you 
let $u = x^n$ and $dv = e^{ax}\ dx$, so $du = nx^{n-1}$ and $v = \frac{1}{a} e^{ax}$.
\item[F)]  Apply the result from part (E) (repeatedly) to compute $\int x^4 e^{-2x}\ dx$.  
\item[] {\it Answer:}  $\left(\frac{-x^4}{2} - x^3 - \frac{3x^2}{2} - \frac{3x}{2} - \frac{3}{4}\right)e^{-2x} + C$.
\item[G)]  $$\int x^8 \ln(x)\ dx$$
\item[] {\it Solution:}  Recall that this is the case where we \emph{don't} let
$u$ be the power of $x$ in parts.  Instead, let $u = \ln(x)$ and $dv = x^8\ dx$.  Then
$du = \frac{1}{x}\ dx$ and $v = \frac{x^9}{9}$.  Then by the parts formula,
\begin{align*}
\int x^8\ln(x)\ dx &= \frac{x^9\ln(x)}{9} - \int \frac{x^9}{9}\cdot \frac{1}{x}\ dx\\
&= \frac{x^9\ln(x)}{9} - \int \frac{x^8}{9} dx\\
&= \frac{x^9\ln(x)}{9} - \frac{x^9}{81} + C.
\end{align*}
\end{enumerate}
\vskip 10pt 
\noindent
III.
\begin{enumerate}  
\item[(A)]  Let $R$ be the region in the plane bounded by $y = 3 - x^2$ and the $x$-axis.
\begin{enumerate}
\item[(1)]  Sketch the region $R$.  
\item[] Omitted -- the region is extends from $x = -\sqrt{3}$ to $x = +\sqrt{3}$ under
the parabola $y = 3 - x^2$, which opens down from the vertex at $(0,3)$.
\item[(2)]  Find the area of $R$.
\item[] {\it Answer:}  $\int_{-\sqrt{3}}^{\sqrt{3}} 3 - x^2\ dx = 4\sqrt{3}$.
\item[(3)]  Find the volume of the solid generated by rotating $R$ about the
$x$-axis.
\item[] {\it Answer:}  $V = \int_{-\sqrt{3}}^{\sqrt{3}} \pi (3 - x^2)^2\ dx = \frac{48\pi\sqrt{3}}{5}$.
\item[(4)]  Find the volume of the solid generated by rotating $R$ about the line
$y = 5$.
\item[] {\it Solution:}  The cross-sections are ``washers'' with inner radius
$$r_{in} = 5 - (3 - x^2) = 2 + x^2$$
and outer radius $5$.  So the volume is
\begin{align*}
\int_{-\sqrt{3}}^{\sqrt{3}} \pi(5)^2 - \pi(2 + x^2)^2\ dx 
&= \pi \int_{-\sqrt{3}}^{\sqrt{3}} 21 - 4x^2 - x^4\ dx\\
&= \pi \left(\left. 21 x - \frac{4x^3}{3} - \frac{x^5}{5}\right|_{-\sqrt{3}}^{\sqrt{3}}\right)\\
&= \frac{152\sqrt{3}\pi}{5}.
\end{align*}
\item[(5)] Find the volume of the solid generated by rotating $R$ about the line
$y = -2$.
\item[] {\it Solution:}  The cross-sections are ``washers'' again, but with inner radius
$r_{in} = 0 - (-2) = 2$
and outer radius 
$$3 - x^2 - (-2) = 5 - x^2.$$  So the volume is
\begin{align*}
\int_{-\sqrt{3}}^{\sqrt{3}} \pi(5 - x^2)^2 - \pi(2)^2\ dx 
&= \pi \int_{-\sqrt{3}}^{\sqrt{3}} 21 - 10x^2 + x^4\ dx\\
&= \pi \left(\left. 21 x - \frac{10x^3}{3} + \frac{x^5}{5}\right|_{-\sqrt{3}}^{\sqrt{3}}\right)\\
&= \frac{128\sqrt{3}\pi}{5}.
\end{align*}
\item[(6)] Find the volume of the solid generated by rotating $R$ about the line
$x = -4$.  (Note: This can be done ``by slices'' or by ``shells.''  Try it 
both ways and check that you get the same result.    
\item[] {\it Solution 1:}  By ``slices:''  The cross-sections by planes 
perpendicular to the vertical line $x = -4$ are washers.  The inner and outer
radii come from the difference between $x$ as a function of $y$ on the 
left and right halves of the parabola:  $x = \pm \sqrt{3 - y}$ and $x = -4$.
$$r_{in} = -\sqrt{3-y} - (-4) = 4 - \sqrt{3 - y}$$
and 
$$r_{out} = \sqrt{3 - y} - (-4) = 4 + \sqrt{3 - y}$$
Hence the volume is computed by
\begin{align*}
V &= \int_0^3 \pi(4 + \sqrt{3 - y})^2 - \pi(4 - \sqrt{3 - y})^2\ dy\\
&= 16\pi \int_0^3 \sqrt{3 - y}\ dy\quad ({\rm the\ other\ terms\ cancel})\\
&= 16\pi \cdot \left.\frac{-2}{3}(3 - y)^{3/2}\right|_0^3\\
&= 32\sqrt{3}\pi.
\end{align*}  
\item[] {\it Solution 2:} By ``shells:'' The volume this way is given by 
\begin{align*}
V &= \int_{x=-\sqrt{3}}^{x = \sqrt{3}} 2\pi (x - (-4))(3 - x^2)\ dx\\
&= 2\pi \int_{-\sqrt{3}}^{\sqrt{3}} 12 + 3x - 4x^2 - x^3\ dx\\
&= 2\pi \left. 12x + \frac{3x^2}{2} - \frac{4x^3}{3} - \frac{x^4}{4}\right|_{-\sqrt{3}}^{\sqrt{3}}\\
&= 2\pi (24\sqrt{3} - 8\sqrt{3})\\
&= 32\sqrt{3}\pi.
\end{align*}
(The two methods give the same volume, of course!)  

\end{enumerate}
\item[(B)]  Let $R$ be the region in the plane bounded by $y = 3x$ and $y = x^2$.  
\begin{enumerate}
\item[(1)]  Sketch the region $R$.
\item[] Omitted.  The curves cross at $(0,0)$ and $(3,9)$.
\item[(2)]  Find the area of $R$.
\item[] {\it Answer:} $\int_0^3 3x - x^2\ dx = \frac{9}{2}$.
\item[(3)]  Find the volume of the solid generated by rotating $R$ about the
$x$-axis.
\item[] {\it Answer:}  The cross sections are washers with outer radius 
$3x$ and inner radius $x^2$ so $V = \int_0^3 \pi (3x)^2 - \pi(x^2)^2\ dx = \frac{162\pi}{5}$. 
\item[(4)]  Find the volume of the solid generated by rotating $R$ about the
$y$-axis.  
\item[] {\it Answer:}  The cross-sections are washers again, but the outer radius at $y$
is $x = \sqrt{y}$ and the inner radius is $x = y/3$.  The volume is 
$V = \int_0^9 \pi(\sqrt{y})^2 - \pi(y/3)^2\ dy = \frac{27\pi}{2}$.  
\end{enumerate}
\item[(C)]  Let $R$ be the region in the plane bounded by $y = \cos(\pi x)$, $y = 1/2$,
$x = -1/3$ and $x = 1/3$.
\begin{enumerate}
\item[(1)]  Sketch the region $R$.  
\item[] {\it Answer:}  Graph omitted.  Since $\cos(\pi/3) = \cos(-\pi/3) = 1/2$, this is 
a lens-shaped area thickest at $x = 0$.  
\item[(2)]  Find the area of $R$.
\item[] {\it Answer:}  $A = \int_{-1/3}^{1/3} \cos(\pi x) - \frac{1}{2}\ dx = \frac{\sqrt{3}}{\pi} - \frac{1}{3}.$
\item[(3)]  Find the volume of the solid generated by rotating $R$ about the
$x$-axis.
\item{} {\it Solution:}  The cross-sections are washers with inner radius $1/2$ and outer
radius $\cos(\pi x)$.  The volume is 
$$V = \int_{-1/3}^{1/3} \pi \cos^2(\pi x) - \pi (1/2)^2\ dx$$
Letting $u = \pi x$, we see we have to integrate $\int \cos^2 u\ du = \frac{u}{2} + \frac{1}{4}\sin(2u)$
by the double angle formula.  So the volume is 
$$V = \left.\frac{\pi x}{2} + \frac{1}{4}\sin(2\pi x) - \frac{\pi x}{4}\right|_{-1/3}^{1/3} = \frac{\sqrt{3}}{4} + \frac{\pi}{6}.$$
\item[(4)]  Find the volume of the solid generated by rotating $R$ about the line $y = -1$.  
\item[] {\it Answer:}
$$V = \int_{-1/3}^{1/3} \pi (1 + \cos(\pi x))^2 - \pi (3/2)^2\ dx =  \frac{9\sqrt{3}}{4} - \frac{\pi}{2}.$$
\end{enumerate}
\end{enumerate}
\vskip 10pt
\noindent
III.  The height of a monument is 20m.  The horizontal cross-section of
the monument at $x$ meters from the top is an isosceles right triangle
with legs $x/4$ meters.  Is the given information enough to find the volume of the monument?
If so, find the volume.  If not, say why not.  
\vskip 10pt
\noindent
{\it Solution:}  By Cavalieri's Principle, this \emph{is enough}.  In particular, it doesn't actually
matter how the cross-sections are stacked on top of each other.  The volume is 
$$V = \int_0^{20} \frac{1}{2}\left(\frac{x}{4}\right)^2\ dx = \frac{250}{3}.$$
\end{document}
%\vskip 10pt
%\noindent
%V.  A solid paperweight has a circular base of radius 4 cm.  The cross-sections of the paperweight
%by planes perpendicular to one diameter of the base are equilateral triangles.  Find the volume
%of the paperweight.  
%\vskip 10pt
%\noindent
%{\it Solution:}  Let the given diameter be the segment of the $x$-axis from $x = -4$ to $x = 4$.
%The side of the slice at position $x$ is then $2\sqrt{16 - x^2}$.  The area of an equilateral triangle
%of side $s$ is $\frac{1}{2}\cdot s \cdot \frac{s\sqrt{3}}{2} = \frac{s^2 \sqrt{3}}{4}$.   
%By Cavalieri's Principle, 
%$$V = \int_{-4}^4 \frac{4(16 - x^2)\sqrt{3}}{4}\ dx = \frac{256\sqrt{3}}{3}.$$
%(The units would be ${\rm cm}^3$.)
%\vskip 10pt
%\noindent
%VI.  A \emph{45 rpm single} record was a vinyl disk 7 inches in diameter, with a large central hole 1 
%inch in 
%diameter.  A ring-shaped paper label 1 inch wide was usually glued to the vinyl surface outside the 
%hole, but
%inside the grooves where the sound was recorded.  The vinyl making up the disk, plus the paper of the 
%label, had mass density .1 ounce per square inch at all points outside the hole in the central region.
%But the density then decreased linearly from .1 to .07 ounces per square inch at the outer edge.  
%The density was the same at all points in the region the same distance away from the center of the 
%hole.  
%Set up an integral or integrals to compute the total mass of a 45 rpm single record (including the 
%label)
%and compute the total mass.  
%\vskip 10pt
%\noindent
%{\it Answer:}  The total mass is computed by 
%$$M = \int_0^{1/2} 0\cdot 2\pi x\ dx + \int_{1/2}^{3/2} .1 \cdot 2\pi x\ dx + \int_{3/2}^{7/2} ((-.015)%(x -3/2) + .1)\cdot 2\pi x\ dx \doteq 3.24 {\rm oz}$$  
%\end{document} 
