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\begin{document}
\begin{center}

\textbf{College of the Holy Cross, Spring 2014} \\
\textbf{Math 136, section 1, Midterm Exam 1} \\
\textbf{Friday, February 21}
\end{center}

\vskip 10pt

\noindent
I. Let $f(x) = x^3 + 1$ on the interval $[a,b] = [1,3]$. 
\begin{enumerate}
\item[A.] (10)  Evaluate the Riemann sum for $f$ on this interval using $n = 4$ and 
$x_i^* = $ right endpoints.
\item[] {\it Solution:}  With $n = 4$, we have $\Delta x = \frac{3 - 1}{3} = \frac{1}{2}$.
The end points of the intervals are $x_0 = 1$, $x_1 = \frac{3}{2}$, $x_2 = 2$, 
$x_3 = \frac{5}{2}$, and $x_4 = 3$.  The right-hand Riemann sum equals:
$$f(3/2) \Delta x + f(2)\Delta x + f(5/2) \Delta x + f(3) \Delta x,$$
which equals
$$(1/2)[((3/2)^3 + 1) + (2^3 + 1) + ((5/2)^3 + 1) + (3^3 + 1)] = 29$$ 
\item[B.] (10) Now repeat part A, but using the left endpoints.
\item[] {\it Solution:}  Similar to part A, but using the left endpoints:
$$f(1) \Delta x + f(3/2)\Delta x + f(2) \Delta x + f(5/2) \Delta x,$$
which equals
$$(1/2)[(1^3 + 1) + ((3/2)^3 + 1) + (2^3 + 1) + ((5/2)^3 + 1)] = 16$$ 
\item[C.] (5) One of your answers in parts A and B is definitely an 
overestimate for the value of $\int_1^3 x^3 + 1\ dx$.  Which is it?
\item[] {\it Solution:}  Since $f(x) = x^3 + 1$ is increasing on its whole
domain, the right-hand sum (part A) gives an overestimate of the integral.
(The exact value is 
$$\int_1^3 x^3 + 1\ dx = \left. \frac{x^4}{4} + x\right|_1^3 = 
\frac{81}{4} + 3 - \frac{1}{4} - 1 = 22.)$$
\end{enumerate}
\noindent
II.  Compute the derivatives of each of the following functions defined by integrals.
\begin{enumerate}
\item[A.]  (5) ${\displaystyle f(x) = \int^x_1 e^{t^2}\ dt}$
\item[] {\it Solution:} By the FTC, part 1, $f'(x) = e^{x^2}$
\item[B.] (5) ${\displaystyle g(x) = \int_{x^2}^4 \frac{\cos(t)}{t^2}\ dt}$
\item[] {\it Solution:} We must reverse the order of the limits, then use
the FTC part 1 and the Chain Rule on this one:
$$g(x) = - \int_4^{x^2} \frac{\cos(t)}{t^2}\ dt \Rightarrow g'(x) = \frac{-\cos(x^2)}{x^4}\cdot 2x
= \frac{-2\cos(x^2)}{x^3}.$$
\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} 
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\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.
\item[] {\it Solution:}  By finding areas of trapezoids and triangles, 
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
   $x$ & 0 & 1 & 2 & 3 & 4\\ \hline 
   $F(x)$    &     0      &  3/2         &    3/2      &    0       &   -1/2        \\  \hline 
\end{tabular}
\end{center}

\item[B.]  (5)  Does $F(x)$ have any critical points?  If so, say where.  If not say why not.
\item[] {\it Solution:} $F'(x) = f(x)$ and this equals $0$ when $x = 3/2$ and 
(in a sense -- see below) again when $x = 4$.  
Of these $x = 3/2$ is definitely a critical point.  Whether or not you call $x = 4$
a critical point is sort of a matter of taste (i.e. of the precise way critical points are 
defined.)  Note that $F(x)$ is not defined for $x > 4$, so to define $F'(4)$, one would
have to consider just a one-sided limit of the difference quotient.  That does exist
and equals zero here.  So it is OK to say $F'(4) = 0$ in that sense. 
\item[C.]  (5)  Over which interval(s) is $F(x)$ concave down?
\item[] {\it Solution:}  This is true on intervals where $F'(x) = f(x)$ is decreasing, 
so $(1,2)$
\end{enumerate}
\noindent
IV.  
\begin{enumerate}
\item[A.] (5)
Integrate with a suitable $u$-substitution:  ${\displaystyle \int_0^1 (4x^3 + 1)^{3/5} x^2\ dx}$.
\item[] {\it Solution:}  Let $u = 4x^3 + 1$.  Then $du = 12x^2\ dx$, so $\frac{1}{12} du = x^2\ dx$.
When $x = 0$, $u = 1$ and when $x = 1$, $u = 5$.  The definite integral
goes over to 
$$\frac{1}{12} \int_1^5 u^{3/5}\ du = \frac{1}{12} \frac{5}{8} \left. u^{8/5} \right|_1^5 = 
\frac{5}{96}[5^{8/5} - 1] \doteq .6319$$
\item[B.]  (10)
Integrate with a suitable $u$-substitution:  ${\displaystyle \int \frac{x\sin(3x^2)}{\cos(3x^2)+1}\ dx}$.
\item[] {\it Solution:}  Let $u = \cos(3x^2) + 1$.  Then $du = -6x \sin(3x^2)\ dx$.  So 
$x\sin(3x^2)\ dx = \frac{-1}{6}\ du$ and the integral goes over to 
$$\frac{-1}{6} \int \frac{du}{u} = \frac{-1}{6}\ln|u| + C = \frac{-1}{6} \ln|\cos(3x^2) + 1| + C.$$  
\item[C.] (15)  Integrate by parts:
${\displaystyle \int x^2 \sin(5x)\ dx}$
\item[] {\it Solution:}  We integrate by parts twice, letting $u = $ power of $x$ each time:
\begin{eqnarray*}
\int x^2\sin(5x)\ dx &=& \frac{-x^2 \cos(5x)}{5} + \frac{2}{5} \int x\cos(5x)\ dx\\
 &=& \frac{-x^2 \cos(5x)}{5} + \frac{2}{5} \left( \frac{x \sin(5x)}{5} - \frac{1}{5} \int \sin(5x)\ dx\right)\\
&=& \frac{-x^2\cos(5x)}{5} + \frac{2x \sin(5x)}{25} + \frac{2\cos(5x)}{125} + C.
\end{eqnarray*}
\item[D.] (10) Integrate with any applicable method we have discussed: 
${\displaystyle \int_0^1 \frac{x}{\sqrt{x^2 + 1}} dx}$
\item[] {\it Solution:}  The $x\ dx$ in the numerator is, up to a constant, the 
derivative of $u = x^2 + 1$ in the radical:
$$= \frac{1}{2} \int_{u=1}^{u=2} u^{-1/2}\ du = \left. u^{1/2}\right|_1^2= \sqrt{2} - 1.$$
\item[E.] (10)  Integrate with any applicable method we have discussed:
${\displaystyle \int e^x \cos(4x)\ dx}$
\item[] {\it Solution:} Use parts twice (letting $dv = e^x\ dx$ each time), 
then solve for the integral:
\begin{eqnarray*}
\int e^x \cos(4x)\ dx &=& e^x \cos(4x) + 4 \int e^x \sin(4x)\ dx\\
&=& e^x \cos(4x) + 4\left( e^x \sin(4x) - 4\int e^x \cos(4x)\ dx\right)\\ 
{\rm so}\ \int e^x \cos(4x)\ dx &=& \frac{e^x}{17}(\cos(4x) + 4\sin(4x)) + C.
\end{eqnarray*}
\end{enumerate}
\end{document}

