\documentclass[12pt]{article}

\usepackage{latexsym,amsmath,epsfig,graphicx,amssymb}
\setlength{\textwidth}{6.5in} 
\setlength{\textheight}{8.5in} 
\oddsidemargin = -.1 cm
\evensidemargin = -.1 cm 
%\topmargin=-2cm


\begin{document}

\centerline{Mathematics 134 -- Calculus With Fundamentals 2}
\centerline{Exam 1 -- Review Sheet}
\centerline{February 9, 2018}
\bigskip


\noindent
{\it Sample Exam Questions- Solutions}
\vskip 10pt
\noindent
I.  \vskip 10pt
\noindent
(A)  Since the interval is $[0,1]$ and $n = 4$, the sums are
\begin{enumerate}
\item[]
$$L_4 = f(0)(.25) + f(.25)(.25) + f(.5)(.25) + f(.75)(.25) = .96875$$
\item[] 
$$R_4 = f(.25)(.25) + f(.5)(.25) + f(.75)(.25) + f(1)(.25) = 1.71875$$
\item[] 
$$M_4 = f(.125)(.25) + f(.375)(.25) + f(.625)(.25) + f(.875)(.25) = 1.328125$$
\end{enumerate}
\noindent
(B)  $f'(x) = 2x + 2 > 0$ for all $x \in [0,1]$.  Hence $f$ is \emph{increasing}
on this interval.  This implies that the left-hand Riemann sum is less than
$\int_0^1 x^2 + 2x\ dx$, and the 
right-hand Riemann sum is greater than the value of the integral.
Using the Evaluation Theorem, we can check this:
$$\int_0^1 x^2 + 2x\ dx = \left. \frac{x^3}{3} + x^2\right|_0^1 = \frac{4}{3} = 1.\overline{3}.$$
\noindent
(C) The sum is the $R_N$ sum for $f(x) = \frac{\cos(x)}{1 + x^2}$ on $[0,\pi]$, so the 
limit is 
$$\int_0^\pi \frac{\cos(x)}{1 + x^2}\ dx.$$
\vskip 10pt
\noindent 
II.  Let 
$$f(x) = \left\{ \begin{array}{ll} 1 & \mbox{\   if \  } 0 \le x \le 3 \\
                x - 2 & \mbox{\ if \ } 3\le x \le 5\\ 
                13 - 2x & \mbox{\ if \ } 5 \le x \le 8\end{array} \right. $$
\noindent (A) Sketch the graph $y = f(x)$.  The graph is made up of 
segments of three different straight lines.  See figure at top of next page.


\begin{figure}
\begin{center}
\includegraphics[width=3in]{exam1fig1.png}
\caption{Figure for Problem II.}
\end{center}
\end{figure}

\vspace{.1in}
\noindent
In the rest of the parts, $F(x) = \int_0^x f(t)\ dt$, where $f$ is the 
function from part A.  \vspace{.1in}

\noindent (B)  Assuming $F(0) = 0$, Compute $F(1), F(2), F(3), F(4), F(5), F(6), F(7), F(8)$ given 
the information in the graph of $f$. \vspace{.1in}

\noindent Using the area interpretation of the definite integral we have  \vspace{.1in}

$\displaystyle F(1)=\int_0^1 f(x)\, dx= 1$   \vspace{.1in}

$\displaystyle F(2)=\int_0^2 f(x)\, dx= 2$   \vspace{.1in}

$\displaystyle F(3)=\int_0^3 f(x)\, dx= 3$   \vspace{.1in}

$\displaystyle F(4)=\int_0^3 f(x)\, dx + \int_3^4 f(x)\, dx = 3+ \frac{3}{2}= \frac{9}{2}$   \vspace{.1in}

$\displaystyle F(5)=\int_0^4 f(x)\, dx  + \int_4^5 f(x) \, dx =  \frac{9}{2} + \frac{5}{2}= 7$   \vspace{.1in}

$\displaystyle F(6)=\int_0^5 f(x)\, dx + \int_5^6 f(x)\, dx = 7+ 2=9$   \vspace{.1in}

$\displaystyle F(7)=\int_0^6 f(x)\, dx + \int_6^{13/2} f(x)\, dx +\int_{13/2}^{7} f(x)\ dx = 9 + \frac{1}{4} -\frac{1}{4}=9$   \vspace{.1in}

$\displaystyle F(8)=\int_0^5 f(x)\, dx + \int_5^{13/2} f(x)\, dx + \int_{13/2}^{8} f(x) \, dx = \int_0^5 f(x)\, dx=7 $    (the last two integrals cancel since they represent equal areas with opposite signs). \vspace{.1in}

\noindent (C) By the Fundamental Theorem of Calculus, $F'(x)=f(x)$. Since $f(13/2)=0$, 
the point $x=13/2$ is a critical point. Since $F'=f$ changes sign from positive to negative 
at the critical point, $x=13/2$ is a local maximum. 

\vspace{.1in}


\noindent (D)  Since $G(x) = \int_2^x f(t)\ dt = \int_0^x f(t)\ dt - \int_0^2 f(t)\ dt = F(x) - 2$, 
the graph $y = G(x)$ is obtained from $y = F(x)$ by shifting down 2 units along the $y$-axis.
\bigskip

%\vspace{.1in} 
%
%If $F(0)=0$ we have 
%
%\begin{center}
%\includegraphics[width=2in]{exam1fig2}
%\end{center}
%If $F(0)=2$ we have 
%
%\begin{center}
%\includegraphics[width=2in]{exam1fig3}
%\end{center}
%
%Ignore the vertical lines in the graph. 

\vspace{.1in}


\noindent
III.  Find the derivatives of the following functions
\vskip 10pt
\noindent (A)  $f(x) = \int_0^x \sin(t)/t\, dt$. 

\vspace{.1in}

$\displaystyle f'(x)=\frac{\sin x}{x}$ \vspace{.1in}

\noindent (B)  $g(x) = \int_5^{x^3} \tan^4(t)\ dt$.\vspace{.1in}

$g(x)=m(x^3)$, where $\displaystyle m(x)= \int_5^{x} \tan^4(t)\ dt$. Then, $\displaystyle g'(x)=m'(x^3) \cdot 3x^2= \tan^4(x^3) \cdot 3x^2$. 

\noindent (C))  $h(x) = \int_{-3x}^{5x} e^{t^2} \sin(t)\ dt$. \vspace{.1in}

$\displaystyle h(x) = n(x) + l(x)$, where $\displaystyle n(x)=  \int_{-3x}^{0} e^{t^2} \sin(t)\ dt$ and  $\displaystyle l(x)=  \int_{0}^{5x} e^{t^2} \sin(t)\ dt$. Then $\displaystyle h'(x)=n'(x)+l'(x)= -(e^{(-3x)^2} \sin(-3x))\cdot(-3)+5 \cdot  e^{(5x)^2} \sin(5x)= 3 e^{9x^2} \sin(-3x) +5 e^{25x^2} \sin(5x)$.
\bigskip

\vspace{.1in}

\noindent IV. 
\bigskip\vspace{.1in}


\noindent (A) Compute $\int 5x^4 - 3\sqrt{x} + e^x + \frac{2}{x}\ dx$
\medskip\vspace{.1in}

$\displaystyle \int 5x^4 - 3\sqrt{x} + e^x + \frac{2}{x}\ dx = x^5-2x^{3/2}+e^x+2 \ln|x|+C$\vspace{.1in}

\noindent (B) Apply a $u$-substitution to compute $\int x(4x^2 - 3)^{3/5}\ dx$
\medskip\vspace{.1in}

$u=4x^2-3, \ du=8x \, dx$. Then $\displaystyle \int x(4x^2 - 3)^{3/5}\ dx= \int  \frac{1}{8} u^{3/5}\, dx = \frac{1}{8} \frac{u^{8/5}}{8/5} + C = \frac{5}{64} (4x^2-3)^{8/5} + C$
\vspace{.1in}

\noindent (C) Apply a $u$-substitution to compute $\int_1^2 e^{\sin(\pi x)}\cos(\pi x)\ dx$
\medskip\vspace{.1in}


$u=\sin(\pi x), \ du= \pi \cos(\pi x)$. Then $\displaystyle \int_1^2 e^{\sin(\pi x)}\cos(\pi x)\ dx= \frac{1}{\pi} \int _0^0 e^u \, du = 0$\vspace{.1in}

\noindent (D) Let $u = t^3 + 3t + 3$, then $du = (3t^2 + 3)\ dt = 3(t^2 + 1)\ dt$. Then
given integral is
$$\int \frac{t^2 + 1}{t^3 + 3t + 3}\ dt = \int \frac{1}{3u}\ du = \frac{1}{3}\ln|u| + C = 
\frac{1}{3}\ln|t^3 + 3t + 3| + C.$$

\noindent (E) Let $u = -7x$ or ``guess and check.''  {\it Answer}:  $\frac{-1}{7} e^{-7x} + C$.

\noindent (F) One solution:  Let $u = \sin(2x)$.  Then $du = 2\cos(2x)\ dx$, so the given integral is
$$\int u\cdot \frac{1}{2}\ du = \frac{1}{4} u^2 + C = \frac{1}{4}\sin^2(2x) + C.$$
You can equally well let $u = \cos(2x)$ and then $du = -2\sin(2x)\ dx$ and the given integral is
$$\int u \cdot\frac{-1}{2}\ du = -\frac{1}{4} u^2 + C = -\frac{1}{4}\cos^2(2x) + C.$$
Both answers are correct, and each differs from the other by an additive constant because of the 
identity $\cos^2(2x) + \sin^2(2x) = 1$ for all $x$.    
\vspace{.1in}

\end{document}
