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\begin{document}
\begin{center}

\textbf{College of the Holy Cross, Spring 2020} \\
\textbf{Math 136, section 1, Solutions for Midterm Exam 1} \\
\textbf{Friday, February 21}
\end{center}

\vskip 10pt
\noindent
I. 
\begin{enumerate}
\item[A.] (10)  Evaluate the $L_4$ Riemann sum for $f(x) = e^{x^2}$ on the interval $[a,b] = [0,2]$. 
\item[] {\it Solution:}  We have $\Delta x = \frac{2 - 0}{4} = \frac{1}{2}$, so the partition is 
$0 < \frac{1}{2} < 1 < \frac{3}{2} < 2$.  The $L_4$ Riemann sum is 
\begin{align*}
L_4 &= \sum_{j=1}^4 e^{x_{j-1}^2} \Delta x\\
&= e^0 \cdot \frac{1}{2} + e^{1/4} \cdot \frac{1}{2} + e^1\cdot \frac{1}{2} + e^{9/4} \cdot \frac{1}{2}\\
&\doteq 7.245
\end{align*}
\item[B.] (10)  The following limit of a sum would equal the definite integral 
$\int_a^b f(x)\ dx$ for
some function $f(x)$ on some interval $[a,b]$.  What function and what interval?
\[
\lim_{N\to\infty} \sum_{j=1}^N \sqrt{\left( \frac{5(j-1)}{N} \right)^3 + 4} \cdot \frac{5}{N}.
\]
\item[] {\it Solution:} The integral is $\displaystyle \int_0^5 \sqrt{x^3+4}\ dx$.  
(The sum is the $L_N$ Riemann
sum for $f(x) = \sqrt{x^3+4}$ on $[0,5]$ and the limit of the Riemann sums gives the integral.)
\end{enumerate}
\vskip 10pt
\noindent
II.  All parts of this problem refer to 
$$A(x) = \int_0^x t(5-t)\ dt$$ 
\begin{enumerate}
\item[A.]  (10)  Is $A(4)$ a positive or negative number?  How can you tell?
\item[] {\it Solution:}  $A(4)$ is positive since $f(x) = x(5 - x) = 5x - x^2$ is $\ge 0$ for
all $x$ in the interval $[0,4]$--the graph $y = f(x)$ is a parabola opening down 
and crossing the $x$-axis at $x = 0$ and $x = 5$.  
\item[B.]  (10) Where does $A(x)$ have critical points?  Is each of them a local maximum, 
an local minimum, or neither?
\item[] {\it Solution:}  $A'(x) = x(5 - x)$ by the FTC, part II.  This equals $0$
at $x = 0$ and $x = 5$.  Those are the critical points.  Since $f(x) = A'(x)$ changes from
negative to positive at $0$, $A(x)$ has a local minimum there.  Since $f(x) = A'(x)$ 
changes from positive to negative at $5$, $A(x)$ has a local maximum there. 
\item[C.]  (10)  If $B(x) = A(\tan(x)) = \int_0^{\tan(x)} t(5 - t)\ dt$, find $B'(x)$.
\item[] {\it Solution:}  By the FTC part II and the chain rule:
$$B'(x) = A'(\tan(x))\sec^2(x) = \tan(x)(5 - \tan(x))\sec^2(x).$$
\end{enumerate}
\vskip 10pt
\noindent
III.  
\begin{enumerate}
\item[A.] (10)
Integrate with a suitable $u$-substitution:  ${\displaystyle \int (6x^5 + 1)^{2/5} x^4\ dx}$.
\item[] {\it Solution:}  Let $u = 6x^5 + 1$.  Then $du = 30x^4\ dx$.  So $x^4\ dx = \frac{1}{30}\ du$
and the integral becomes
$$\int u^{2/5} \frac{1}{30}\ du = \frac{1}{30}\cdot \frac{5}{7} u^{7/5} + C = 
\frac{1}{42}(6x^5 + 1)^{7/5} + C.$$
\item[B.]  (10) Integrate by parts:  ${\displaystyle \int x^2 \cos(4x)\ dx}$
\item[] {\it Solution:}  We integrate by parts twice, letting $u = x^2$, then $u = x$:
\begin{align*}
\int x^2\cos(4x)\ dx &= \frac{x^2\sin(4x)}{4} - \frac{2}{4}\int x\sin(4x)\ dx\\
&= \frac{x^2\sin(4x)}{4} - \frac{1}{2}\left(\frac{-x\cos(4x)}{4} + \frac{1}{4}\int\cos(4x)\ dx\right)\\
&= \frac{x^2\sin(4x)}{4} + \frac{x\cos(4x)}{8} - \frac{\sin(4x)}{32} + C
\end{align*} 
\item[C.] (10) Integrate:
${\displaystyle \int \sec^5(x) dx}$
\item[] {\it Solution:}  Using the ST3 (twice) and ST4 formulas:
$$\int\sec^5(x)\ dx = \frac{\sec^3(x)\tan(x)}{4} + \frac{3\sec(x)\tan(x)}{8} + \frac{3}{8}\ln|\sec(x) + \tan(x)| + C$$
\end{enumerate}
\vskip 10pt
\noindent
IV.  (20) Integrate ${\displaystyle \int_{-6}^6 x^2\sqrt{36 - x^2}\ dx}$
\vskip 10pt
\noindent
{\it Solution:}  From the form of the integrand, we know we want to use
the trigonometric substitution $x = 6\sin\theta$, with $dx = 6\cos\theta\ d\theta$.
Converting everything, including the limits of integration, we have
$$1296 \int_{-\pi/2}^{\pi/2} \sin^2\theta\cos^2\theta\ d\theta$$
For this integral we can use either the SC 3 or SC 4 reduction formula.  Here's how it turns
out if you use SC 3 to reduce the power of sine, then SC 2 to reduce the power of cosine:
\begin{align*}
&= 1296\left(\left.\frac{-\sin\theta\cos^3\theta}{4}\right|_{-\pi/2}^{\pi/2} + \frac{1}{4}\int_{-\pi/2}^{\pi/2} \cos^2\theta\ d\theta\right)\\
&= 1296\left( 0 + \left.\frac{\cos\theta\sin\theta}{8}\right|_{-\pi/2}^{\pi/2} + 
\left.\frac{\theta}{8}\right|_{-\pi/2}^{\pi/2}\right)\\
&= 1296\left(0 + 0 + \frac{\pi}{8}\right)\\
&= 162 \pi.
\end{align*}

Note:  You can also convert the indefinite integral back to an equivalent function 
of $x$ and subsitute the original limits of integration $x = -6$ and $x = 6$.    
\end{document}

