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\begin{document}
\begin{center}

\textbf{College of the Holy Cross, Spring 2008} \\
\textbf{Math 132, Solutions for Midterm Exam 1} (All Sections) \\
\textbf{Wednesday, February 20}
\end{center}

\noindent
I. (10) State the definition of the definite integral $\displaystyle \int_a^b f(x)\ dx$.  

\soln
If $f(x)$ is continuous on $[a,b]$ and $\Delta x = \frac{b - a}{n}$, then 
the definite integral is the limit of the Riemann sums
\[
\int_a^b f(x)\ dx = \lim_{n\to\infty} \sum_{i=1}^n f(x_i^*)\ \Delta x
\]
for all choices of $x_i^* \in [x_{i-1},x_i]$.  

\vskip 10pt
\noindent
II.  Compute the derivatives of each of the following functions defined by integrals.
\begin{enumerate}
\item[A.]  (5) ${\displaystyle f(x) = \int^x_1 e^{t^2}\ dt}$

\soln
By the first part of the Fundamental Theorem of Calculus,
\[
f'(x) = e^{x^2}.
\] 
\item[B.] (5) ${\displaystyle g(x) = \int_{x^2}^4 \frac{\cos(t)}{t^2}\ dt}$

\soln
Reversing the limits of integration, 
\[
g(x) = -\int_4^{x^2} \frac{\cos(t)}{t^2}\ dt.
\]
Then by the first part of the Fundamental Theorem and the Chain Rule,
\[
g'(x) = -\frac{\cos(x^2)}{x^4}\cdot 2x = -\frac{2\cos(x^2)}{x^3}.
\]
\end{enumerate}

\noindent
III.  The following graph (made up of straight line segments)
shows $y = f(t)$ for $0 \le t \le 4$.

\begin{center}
\includegraphics[width=2.3truein,height=2.3truein]{132P1F1.ps} 
\end{center}

\noindent
Given: $f(1) = 2$, $f(2) = -2$, $f(3) = -1$, and $f(4) = 0$,  
The function $F$ is defined by $F(x) = \int_0^x f(t)\ dt$.  
\begin{enumerate}
\item[A.] (5) Determine the values $F(x)$ for $x = 0,1,2,3,4$ and enter them in the following table.

\soln
By computing areas of trapezoids, we have:
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
   $x$ & 0 & 1 & 2 & 3 & 4\\ \hline
       &   &   &   &   &     \\
$F(x)$ & $\qquad$ 0 $\qquad$  & $\qquad$ 3/2 $\qquad$  & $\qquad$ 3/2 $\qquad$  & $\qquad$ 0 $\qquad$  & $\qquad$ -1/2 $\qquad$  \\ 
       &           &           &           &           &           \\  \hline 
\end{tabular}
\end{center}

\item[B.]  (5)  Does $F(x)$ have any critical points?  If so, say where.  If not say why not.
\vskip 10pt
\soln
By the first part of the Fundamental Theorem, $F'(x) = f(x) = 0$ when $x = 3/2$ and $x = 4$.
(Since $x = 4$ is an endpoint of the interval, we will also accept just $x = 3/2$.)  
\vskip 15 pt
\item[C.]  (5)  Over which interval(s) is $F(x)$ concave down?
\soln
$F(x)$ is concave down on intervals where $F''(x) < 0$, or $F'(x) = f(x)$
is \emph{decreasing}.  This occurs for $1 < x < 2$ here.   
\end{enumerate}
\noindent
IV.  
\begin{enumerate}
\item[A.] (5)
Integrate with a suitable $u$-substitution:  ${\displaystyle \int_0^1 (4x^3 + 1)^{3/5} x^2\ dx}$.

\soln
Note that the $x^2$ outside the power is $du$ up to a constant for $u = 4x^3 + 1$. 
So do this substitution where $du = 12x^2\, dx$.  So 
\begin{eqnarray*}
\int_0^1 (4x^3 + 1)^{3/5} x^2\ dx &=& \frac{1}{12} \int_{u=1}^{u=5} u^{3/5}\ du\\
&=& \left. \frac{5}{96} u^{8/5} \right|_{u=1}^{u=5}\\
&=& \frac{5}{96}\left(5^{8/5} - 1\right)
\end{eqnarray*} 

\vskip 10pt
\item[B.]  (5)
Integrate with a suitable $u$-substitution:  ${\displaystyle \int \frac{x\sin(3x^2)}{\cos(3x^2)+1}\ dx}$.

\soln
Let $u = \cos(3x^2)+1$.  Then $du = -\sin(3x^2)(6x)\ dx$ by the Chain Rule.  
So 
\begin{eqnarray*}
\int \frac{x\sin(3x^2)}{\cos(3x^2)+1}\ dx &=&
-\frac{1}{6} \int \frac{du}{u} \\
&=& -\frac{1}{6} \ln|u| + C\\
&=& -\frac{1}{6} \ln|\cos(3x^2) + 1| + C.
\end{eqnarray*}
\vskip 10pt
\item[C.] (7.5)  Integrate by parts:
${\displaystyle \int x^2 \sin(5x)\ dx}$

\soln
We let $u = x^2$, $dv = \sin(5x)\ dx$, so $du = 2x\ dx$ and $v = -\frac{1}{5}\cos(5x)$.
Applying the parts formula,
\[
\int x^2 \sin(5x)\ dx = -\frac{x^2}{5} \cos(5x) + \frac{2}{5}\int x\cos(5x)\, dx.
\]
Now do parts again on the remaining integral with $u = x$, $dv = \cos(5x)\ dx$, 
so $du = dx$ and $v = \frac{1}{5}\sin(5x)$.  This gives:
\begin{eqnarray*}
\int x^2 \sin(5x)\ dx &=& -\frac{x^2}{5} \cos(5x) + \frac{2}{5}\left(\frac{x}{5} \sin(5x) - \frac{1}{5}\int\sin(5x)\ dx\right)\\
&=& -\frac{x^2}{5} \cos(5x) + \frac{2x}{25}\sin(5x) + \frac{2}{125} \cos(5x) + C
\end{eqnarray*}
\vskip 10pt
\item[D.] (7.5) Integrate with the partial fraction method:  
${\displaystyle \int \frac{3x^2 + 1}{x^2 + 5x + 6}\ dx}$

\soln
Since the degree of the top is equal to the degree of the bottom, we divide
first:  $3x^2 + 1 = 3(x^2 + 5x + 6) - 15x - 17$ so
\[
\int \frac{3x^2 + 1}{x^2 + 5x + 6}\ dx = \int 3 - \frac{15x + 17}{x^2 + 5x + 6}\ dx.
\]
Now $x^2 + 5x + 6 = (x + 2)(x + 3)$ so the partial fractions are 
\[
\frac{15x + 17}{(x+2)(x+3)}= \frac{A}{x+2} + \frac{B}{x+3}.
\]
Clearing denominators,
\[
15x + 17 = A(x+3) + B(x+2),
\]
so with $x = -2$ we see $A = -13$ and with $x = -3$ we see $B = 28$.  Then 
\[
\int 3 - \frac{15x + 17}{x^2 + 5x + 6}\ dx = \int 3 - \left(\frac{-13}{x+2} + \frac{28}{x+3}\right)\ dx
= 3x + 13\ln|x+2| - 28\ln|x + 3| + C.
\]

\item[E.] (10)  Integrate via trigonometric substitution:
${\displaystyle \int \sqrt{36 - x^2}\ dx}$

\soln
From the form of the integrand, we use $x = 6\sin\theta$, so $dx = 6\cos\theta\, d\theta$ and
the integral becomes
\[
\int \sqrt{36 - x^2}\ dx = \int \sqrt{36(1 - \sin^2\theta)}\, 6\cos\theta\, d\theta = 
36\int \cos^2\theta\, d\theta.
\]
To integrate this, we use the half-angle formula:
\[
36\int \cos^2\theta\, d\theta = 36 \int \frac{1}{2}(1 + \cos(2\theta))\, d\theta = 
18 \theta + 9\sin(2\theta) + C
\]
Now we convert back to expressions in the original variable $x$.  From the 
substitution $x = 6\sin\theta$, we have $\theta = \sin^{-1}\left(\frac{x}{6}\right)$
and $\sin(2\theta) = 2\sin(\theta)\cos(\theta) = 2\frac{x}{6}\frac{\sqrt{36-x^2}}{6} = 
\frac{x\sqrt{36 - x^2}}{18}$.  The integral equals
\[
18 \sin^{-1}\left(\frac{x}{6}\right) + \frac{x \sqrt{36 - x^2}}{2} + C.
\]
(Note that you can check your work using \#39 in the table!)
\end{enumerate}
\vskip 10pt
\noindent
V.  Integrate using any applicable method or the table.  If you 
do use the table, give the number of the entry you are using.

\begin{enumerate}
\item[A.] (10) ${\displaystyle \int \tan x \sec^4 x\ dx}$.

\soln
Method 1:
Since the derivative of $u = \sec x$ is $du = \sec x\tan x$,
the form here is 
\[
\int \sec^3 x (\sec x \tan x)\, dx = \int u^3\ du = \frac{1}{4} u^4 + C = \frac{1}{4}\sec^4 x + C.
\]

\noindent
Method 2:  Since the derivative of $\tan x$ is $\sec^2 x$, this one can 
also be done by splitting off two powers of $\sec x$ and converting
the other $\sec^2 x$ to powers of $\tan x$:
\begin{eqnarray*}
\int \tan x \sec^4 x\ dx &=& \int \tan x (\sec^2 x) \sec^2 x\, dx\\
&=& \int \tan x (1 + \tan^2 x) \sec^2 x\, dx\\
&=& \int \tan x \sec^2 x\, dx + \int \tan^3 x \sec^2 x\, dx\\
&=& \int u\, du + \int u^3\, du \text{ $\qquad$ (for $u = \tan x$) }\\
&=& \frac{u^2}{2} + \frac{u^4}{4} + C\\
&=& \frac{\tan^2 x}{2} + \frac{\tan^4 x}{4} + C.
\end{eqnarray*}
This is equivalent to the answer above via the trig identity 
$1 + \tan^2 x = \sec^2 x$. 

Method 3:  A third way to do this one is to convert everything to sines and
cosines.  The integral obtained that way can be done by a $u$-substitution
letting $u = \cos x$, $du = -\sin x\, dx$:
\[
\int \frac{\sin x}{\cos^5 x}\ dx = -\int u^{-5}\, du = \frac{u^{-4}}{4} + C = \frac{1}{4\cos^4 x} + C.
\]
This is also equivalent to the answer in Method 1 above since $\sec x = \frac{1}{\cos x}$.  
\vskip 10pt
\item[B.] (10) ${\displaystyle \int \frac{\sqrt{9 + e^{2x}}}{e^x}\ dx}$

\soln
Substitute $u = e^x$, then $du = e^x dx$, so $dx = \frac{du}{e^x} = \frac{du}{u}$.
Making these substitutions yields
\[
\int \frac{\sqrt{9 + u^2}}{u^2}\ du,
\]
which is \# 24 in the table with
$a = 3$.  We apply the table entry:
\[
= -\frac{\sqrt{9 + u^2}}{u} + \ln\left|u + \sqrt{9 + u^2}\right| + C 
= -\frac{\sqrt{9 + e^{2x}}}{e^x} + \ln\left|e^x + \sqrt{9 + e^{2x}}\right|+C.
\]
Note that this is \emph{not} \# 23 because of the ``extra'' $u$
in the denominator from $dx$.  
\vskip 10pt
\item[C.] (10) ${\displaystyle \int\frac{dx}{x^2 + 8x + 17}}$.

\soln
Complete the square in the denominator:  $x^2 + 8x + 17 = (x + 4)^2 + 1$.
Therefore, the substitution $u = x + 4$, $du = dx$ takes this to the 
basic form 
\[
\int \frac{1}{u^2 + 1}\ du = \tan^{-1}(u) + C = \tan^{-1}(x + 4) + C.
\] 
\end{enumerate}
\end{document}

