
\documentclass[12pt]{article}
\usepackage{latexsym,amsmath,epsfig}
\newcommand{\ds}{\displaystyle}
\newcommand{\vp}{$\vphantom{\big|}$}

\newcommand{\R}{\textbf{R}}
\newcommand{\Item}{\item{$\Box$}\quad}

\newcommand{\Answer}[1]{$\phantom{.}$\hfill
        #1\quad\fboxsep 20pt \fbox{$\phantom{\Big|Answer}$}} 

\newcommand{\MyTallAnswer}[2]{$\phantom{.}$\hfill
        #1\quad\fboxsep 20pt\fbox{$\vphantom{\Big|}#2$}}

\newcommand{\MyAnswer}[2]{$\phantom{.}$\hfill
        #1\quad\fboxsep 5pt\fbox{$\vphantom{\Big|}#2$}}

\newcommand{\soln}{\paragraph{Solution}}
\newcommand{\sep}{{\medskip\hrule\bigskip}}
\newcommand{\blank}{\underline{$\phantom{A blank}$}}

\setlength{\textwidth}{6.5in}
\setlength{\topmargin}{-0.5in}
\setlength{\oddsidemargin}{0in}
\setlength{\textheight}{9.0in}
\begin{document}
\begin{center}

\textbf{College of the Holy Cross, Spring 2018} \\
\textbf{Math 134 Midterm Exam 1 Solutions} \\
\textbf{Friday, February 16}
\end{center}



\noindent
I. Both parts of this problem refer to $f(x) = -x^2 + 4$ on the interval $[a,b] = [0,2]$. 
\begin{enumerate}
\item[A.] (15)  Evaluate the $R_4$ Riemann sum for $f$ on this interval.
\item[] {\it Solution:}  We have $\Delta x = \frac{2 - 0}{4} = \frac{1}{2}$, so 
$x_1 = \frac{1}{2}, x_2 = 1, x_3 = \frac{3}{2}$, and $x_4 = 2$.  The right hand sum is
\begin{align*}
R_4 &= (-(1/2)^2 + 4)(1/2) + (-(1)^2 + 4)(1/2) + (-(3/2)^2 + 4)(1/2) + (-(2)^2 + 4)(1/2)\\
&= (15/4)(1/2) + (3)(1/2) + (7/4)(1/2) + 0\\
&= 34/8 = 4.25.
\end{align*}
\item[B.] (10) Use Part I of the Fundamental Theorem of Calculus to 
evaluate $\int_0^2 -x^2 + 4\ dx$.   
\item[] {\it Solution:}  We have 
\begin{align*}
\int_0^2 -x^2 + 4\ dx &= \left. \frac{-x^3}{3} + 4x\right|_0^2\\
&= \frac{-8}{3} + 8 - 0\\
&= \frac{16}{3}.
\end{align*}
\item[C.] (5) Your answer to part B should be larger than your answer to part A.  Explain
how you could know it would turn out that way even without computing the numerical values.  
\item[] {\it Solution:}  We can tell the integral will be less than the right hand sum
because the function $-x^2 + 4$ is \emph{decreasing} on $[0,2]$ (note, for instance, that
$f'(x) = -2x < 0$ for all $x$ in $[0,2]$).  This implies that the actual value of the
integral is larger than all right-hand sums.  
\end{enumerate}
\vskip 10pt
\noindent
II.  (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?
\[
\lim_{N\to\infty} \sum_{j=1}^N \left(2 + \left(\frac{2j}{N}\right)^4\right)\cdot \frac{2}{N}.
\]
\vskip 10pt
\noindent
{\it Solution:}  The sum is the $R_N$ sum for $f(x) = 2 + x^4$ on $[0,2]$, so the integral
is 
$$\int_0^2 2 + x^4\ dx.$$
\noindent
III.  Both parts of this problem refer to $f(x) = (x - 2)^2 (x - 4)$.  
\begin{enumerate}
\item[A.]  (10) Let $\displaystyle A(x) = \int_0^x f(t)\ dt$. What is $A'(x)$?
\item[] {\it Solution:}  By the FTC Part II, 
$$A'(x) = f(x) = (x - 2)^2 (x - 4).$$
\item[B.]  (10) Now consider $\displaystyle B(x) = \int_1^{x^3} f(t)\ dt$.  What is $B'(x)$?
\item[] {\it Solution:}  By the FTC Part II and the Chain Rule, 
$$B'(x) = (x^3 - 2)^2 (x^3 - 4)\cdot 3x^2.$$
\end{enumerate}
\vskip 10pt
\noindent
IV.  Compute the following integrals.  
\begin{enumerate}
\item[A.] (10) $\displaystyle \int \sqrt{x} + \frac{1}{x} - e^x\ dx$
\item[] {\it Solution:} 
$$\int \sqrt{x} + \frac{1}{x} - e^x\ dx = \frac{2}{3} x^{3/2} + \ln|x| - e^x + C$$
\item[B.] (10) $\displaystyle $
\item[] {\it Solution:}
$$\int \cos(x) + \frac{1}{\sqrt{1 - x^2}}\ dx = \sin(x) + \sin^{-1}(x) + C$$
\item[C.] (10)
Integrate with a suitable $u$-substitution:  ${\displaystyle \int_0^1 (2x^4 + 1)^{6/5} x^3\ dx}$
\item[] {\it Solution:}  Let $u(x) = 2x^4 + 1$.  Then $du = 8x^3\ dx$, so $x^3\ dx = \frac{1}{8}\ du$.
When $x = 0$, $u(0) = 1$ and when $x = 1$, $u(1) = 3$.  
In terms of $u$ the integral becomes
$$\frac{1}{8} \cdot \int_{u = 1}^{u = 3} u^{6/5}\ du = \frac{1}{8}\cdot \frac{5}{11}\left. u^{11/5} \right|_1^3
=\frac{5}{88} (3^{11/5} - 1) \doteq .5802$$
\item[D.]  (10)
Integrate with a suitable $u$-substitution:  $\displaystyle \int (\tan(x))^3 \sec^2(x)\ dx$.
\item[] {\it Solution:}  Let $u = \tan(x)$.  Then $du = \sec^2(x)\ dx$ and the integral
becomes
$$\int u^3\ du = \frac{u^4}{4} + C = \frac{(\tan(x))^4}{4} + C.$$
\end{enumerate}
\end{document}

