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\begin{center}{\huge MATH 1190 - EXAM 4 FALL 2016 \\
SOLUTION}
\end{center}

\begin{flushleft}\vspace{0.5in}
Friday 18 November 2016 \\
Instructor: Tom Cuchta
\end{flushleft}

\textbf{Instructions:}
\begin{itemize}
	\item  Show all work, clearly and in order, if you want to get full
	credit. If you claim something is true \textbf{you must show work backing up your claim}. I reserve the right to take off points if I cannot see how you arrived at your answer (even if your final answer is correct).
	\item Justify your answers algebraically whenever possible to ensure full credit. 	
	\item  Circle or otherwise indicate your final answers.
	\item  Please keep your written answers brief; be clear and to the point.
	\item  Good luck!
\end{itemize}
\vspace*{10pt}


\end{coverpages}

\begin{questions}

\question[15] Use $n=3$ rectangles to approximate the area under the function $f(x)=x^2+3$ and above the interval $[0,3]$ \textbf{using right endpoints}. \\
\textit{Solution:} Using $n=3$ rectangles requires $\Delta x = \dfrac{3-0}{3}=1$. Therefore the right endpoints are at $x_1=1$, $x_2=2$, and $x_3=3$. Evaluating $f$ at these points yields
$$f(1) = 1^2+3=4,$$
$$f(2)=2^2+3=7,$$
and
$$f(3)=3^2+3=12.$$
Therefore the approximate area is given by
$$\rm{Area} \approx \displaystyle\sum_{k=1}^3 f(x_k) \Delta x = f(x_1)\Delta x + f(x_2) \Delta x + f(x_3) \Delta x = 4(1)+7(1)+12(1)=23.$$
\question[15] Compute the following summation using the properties of sums (after applying the sum formula, you do not have to further simplify!): 
$$\displaystyle\sum_{k=1}^{50} k^3+k.$$
\textit{Solution:} Using the formulas for the sum of cubes and sum of first powers, we get
$$\displaystyle\sum_{k=1}^{50} k^3+k = \left( \displaystyle\sum_{k=1}^{50} k^3 \right) + \left( \displaystyle\sum_{k=1}^{50} k \right) = \dfrac{50^2(51)^2}{4} + \dfrac{50(51)}{2}.$$
\question[24] Find the following integrals any way you wish:
\begin{parts}
\part[6] $\displaystyle\int_0^1 x^2+3x+2 \rm{d}x$ \\
\textit{Solution:} Calculate
$$\displaystyle\int_0^1 x^2+3x+2 \rm{d}x = \left. \dfrac{x^3}{3} + \dfrac{3}{2}x^2 + 2x \right|_0^1 = \dfrac{1}{3} + \dfrac{3}{2} + 2 = \dfrac{23}{6}$$
\part[6] $\displaystyle\int_0^{\frac{\pi}{2}} \cos(x) \rm{d}x$ \\
\textit{Solution:} Calculate 
$$\displaystyle\int_0^{\frac{\pi}{2}} \cos(x) \rm{d}x = \sin(x) \Bigg|_0^{\frac{\pi}{2}} = \sin\left(\dfrac{\pi}{2} \right) - \sin(0) = 1-0 = 1$$
\part[6] $\displaystyle\int \dfrac{x}{\sqrt{x+1}} \rm{d}x$ \\
\textit{Solution:} Let $u=x+1$ then $\rm{d}u = \rm{d}x$ and $x=u-1$ so using substitution, we get
$$\begin{array}{ll}
\displaystyle\int \dfrac{x}{\sqrt{x+1}} \rm{d}x &= \displaystyle\int \dfrac{u-1}{\sqrt{u}} \rm{d}u \\
&= \displaystyle\int u^{\frac{1}{2}} - u^{-\frac{1}{2}} \rm{d}u \\
&= \dfrac{u^{\frac{3}{2}}}{\frac{3}{2}} - \dfrac{u^{\frac{1}{2}}}{\frac{1}{2}} + C \\
&= \dfrac{2}{3} (x+1)^{\frac{3}{2}} - 2 (x+1)^{\frac{1}{2}} + C
\end{array}$$
\part[6] $\displaystyle\int \dfrac{\sin(x)}{\cos^3(x)} \rm{d}x$ \\
\textit{Solution:} Let $u=\cos(x)$ so $\rm{d}u = -\sin(x) \rm{d}x$. Use substitution to get
$$\begin{array}{ll} \displaystyle\int \dfrac{\sin(x)}{\cos^3(x)} \rm{d} x &= -\displaystyle\int \dfrac{1}{u^3} \rm{d} u \\
&= - \displaystyle\int u^{-3} \rm{d} u \\
&= - \dfrac{u^{-2}}{-2} + C \\
&= \dfrac{1}{2\cos^2(x)} + C \\
&= \dfrac{1}{2}\sec^2(x) + C
\end{array}$$
\end{parts}

\question[20] In this problem you will use the \textbf{limit definition} of a definite integral to compute
$$\displaystyle\int_0^4 2x+1 \rm{d}x.$$
\begin{parts}
\part[5] What is $\Delta x$? What is $x_k$?  \\
\textit{Solution:} Here
$$\Delta x = \dfrac{4-0}{n} = \dfrac{4}{n}$$
and
$$x_k = 0+k\Delta x = \dfrac{4k}{n}.$$
\part[5] What is $f(x_k)$? \\
\textit{Solution:} Here $f(x)=2x+1$ so using $x_k = \dfrac{4k}{n}$ we get
$$f(x_k) = 2x_k + 1 = \dfrac{8k}{n}+1.$$
\part[5] What is $\displaystyle\sum_{k=1}^n f(x_k) \Delta x$? \\
\textit{Solution:} Calculate
$$\begin{array}{ll}
\displaystyle\sum_{k=1}^n f(x_k) \Delta x &= \dfrac{4}{n} \displaystyle\sum_{k=1}^n \dfrac{8k}{n} + 1 \\
&= \dfrac{32}{n^2} \left( \displaystyle\sum_{k=1}^n k \right) + \dfrac{4}{n} \left( \displaystyle\sum_{k=1}^n 1 \right) \\
&= \dfrac{32}{n^2} \dfrac{n(n+1)}{2} + \dfrac{4}{n} (n) \\
&=16 \left( 1 + \dfrac{1}{n} \right) + 4.
\end{array}$$
\part[5] Use your answers in parts (a)-(c) to compute $\displaystyle\int_0^4 2x+1 \rm{d}x$ using the limit definition. \\
\textit{Solution:} The limit definition tells us that
$$\begin{array}{ll}
\displaystyle\int_0^4 2x+1 \rm{d}x &= \displaystyle\lim_{n \rightarrow \infty} \displaystyle\sum_{k=1}^n f(x_k) \Delta x \\
&= \displaystyle\lim_{n \rightarrow \infty} 16\left( 1 + \dfrac{1}{n} \right) + 4 \\
&= 16+4 \\
&= 20.
\end{array}$$
\vfill
\end{parts}

\newpage

\question[10] Find the average value of $f(x)=x^2+1$ on the interval $[-2,3]$. \\
\textit{Solution:} The average value is given by
$$\begin{array}{ll}
\rm{AvgVal} &= \dfrac{1}{3-(-2)} \displaystyle\int_{-2}^3 x^2+1 \rm{d}x \\
&= \dfrac{1}{5} \left[ \dfrac{x^3}{3} + x \right|_{-2}^3 \\
&= \dfrac{1}{5} \left[ \left( \dfrac{27}{3} + 3 \right) - \left( \dfrac{-8}{3} - 2 \right) \right]
\end{array}$$
\question[15] 
\begin{parts}
\part[7] Let $F(x)=\displaystyle\int_0^x t^2 \rm{d}t$. Compute $F'(x)$. \\
\textit{Solution:} Using the fundamental theorem of calculus (``second part") we get
$$F'(x) = x^2.$$
\part[8] Let $F(x)= \displaystyle\int_{\sin(x)}^x t^2 \rm{d}t$. Compute $F'(x)$.  \\
\textit{Solution:} First rearrange this integral into two integrals each with the variable $x$ appearing in the top:
$$F(x) = \displaystyle\int_{\sin(x)}^0 t^2 \rm{d} t + \displaystyle\int_0^x t^2 \rm{d} t = -\displaystyle\int_0^{\sin(x)} t^2 \rm{d}t + \displaystyle\int_0^x t^2 \rm{d}t.$$
Letting $G(x)=\displaystyle\int_0^x t^2 \rm{d}t$ we see from the second fundamental theorem of calculus that $G'(x)=x^2$ and also we can write
$$F(x) = -G(\sin(x)) + G(x),$$
so that using the chain rule,
$$\begin{array}{ll}
F'(x) &= -\dfrac{\rm{d}}{\rm{d}x} G(\sin(x)) + G'(x) \\
&= -G'(\sin(x))\cos(x) + G'(x) \\
&= -\sin^2(x)\cos(x) + x^2.
\end{array}$$
\end{parts}
\end{questions}
\newpage
\noindent Summation formulas (ok to tear this page out): \\~\\
for any constant $\alpha$, $\displaystyle\sum_{k=1}^n \alpha = n\alpha$ \\
$\displaystyle\sum_{k=1}^n k = \dfrac{n(n+1)}{2}$ \\
$\displaystyle\sum_{k=1}^n k^2 = \dfrac{n(n+1)(2n+1)}{6}$ \\
$\displaystyle\sum_{k=1}^n k^3 = \dfrac{n^2(n+1)^2}{4}$
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