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\begin{center}\huge MATH 3315 - EXAM 1 - FALL 2017\\
SOLUTION
\end{center}
\begin{flushleft}\vspace{0.5in}
Friday 8 September 2017\\
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[36] Compute the following derivatives. 
\begin{parts}
\part[6] $\dfrac{\mathrm{d}}{\mathrm{d}x} \log(5x^2+3x+1)$. \\
\textit{Solution:} \href{http://www.wolframalpha.com/input/?i=d%2Fdx%5Blog(5x%5E2%2B3x%2B1)%5D}{Compute}
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}x} \log(5x^2+3x+1) &= \dfrac{1}{5x^2+3x+1} \dfrac{\mathrm{d}}{\mathrm{d}x} [5x^2+3x+1] \\
&= \dfrac{10x+3}{5x^2+3x+1}.
\end{array}$$
\part[6] $\dfrac{\mathrm{d}}{\mathrm{d}x} (e^{5x}+e^{7x}+1)^3$ \\
\textit{Solution:} \href{http://www.wolframalpha.com/input/?i=d%2Fdx%5Blog(5x%5E2%2B3x%2B1)%5D}{Compute}
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}x} (e^{5x}+e^{7x}+1)^3 &= 3(e^{5x}+e^{7x}+1)^2 \dfrac{\mathrm{d}}{\mathrm{d}x} [e^{5x}+e^{7x}+1] \\
&= 3(e^{5x}+e^{7x}+1)^2 [5e^{5x}+7e^{7x}].
\end{array}$$
\part[6] $\dfrac{\mathrm{d}}{\mathrm{d}t} \arcsin(7t^3-9)$ \\
\textit{Solution:} \href{http://www.wolframalpha.com/input/?i=d%2Fdt%5Barcsin(7t%5E3-9)%5D}{Compute}
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}t} \arcsin(7t^3-9) &= \dfrac{1}{\sqrt{1-(7t^3-9)^2}} \dfrac{\mathrm{d}}{\mathrm{d}t} [7t^3-9] \\
&=\dfrac{21t^2}{\sqrt{1-(7t^3-9)^2}}.
\end{array}$$
\part[6] $\dfrac{\mathrm{d}}{\mathrm{d}z} 3^{5z-1}$ \\
\textit{Solution:} Note that
$$(*) \hspace{35pt} 3^{5z-1} = e^{\log(3^{5z-1})} = e^{(5z-1)\log(3)},$$
so \href{http://www.wolframalpha.com/input/?i=d%2Fdz%5B3%5E(5z-1)%5D}{compute}
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}z} 3^{5z-1} &= \dfrac{\mathrm{d}}{\mathrm{d}z} e^{(5z-1)\log(3)} \\
&= e^{(5z-1)\log(3)} \dfrac{\mathrm{d}}{\mathrm{d}z} [(5z-1)\log(3)] \\
&\stackrel{(*)}{=} 5\log(3) 3^{5z-1}. 
\end{array}$$
\part[6] $\dfrac{\mathrm{d}}{\mathrm{d}t} \dfrac{\cosh(t)}{\sinh(t)+1}$ \\
\textit{Solution:} \href{http://www.wolframalpha.com/input/?i=d%2Fdt%5Bcosh(t)%2F(sinh(t)%2B1)%5D}{Compute}
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}t} \dfrac{\cosh(t)}{\sinh(t)+1} &= \dfrac{(\sinh(t)+1) \dfrac{\mathrm{d}}{\mathrm{d}t} \cosh(t) - \cosh(t) \dfrac{\mathrm{d}}{\mathrm{d}t}[\sinh(t)+1]}{(\sinh(t)+1)^2} \\
&=\dfrac{\sinh(t)(\sinh(t)+1) - \cosh^2(t)}{(\sinh(t)+1)^2}.
\end{array}$$
\part[6] $\dfrac{\mathrm{d}}{\mathrm{d}w} \log(\log(\log(w)))$ \\
\textit{Solution:} \href{http://www.wolframalpha.com/input/?i=d%2Fdw%5Blog(log(log(w)))%5D}{Calculate}
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}w} \log(\log(\log(w))) &= \dfrac{1}{\log(\log(w))} \dfrac{\mathrm{d}}{\mathrm{d}w} \log(\log(w)) \\
&=\dfrac{1}{\log(\log(w))} \dfrac{1}{\log(w)} \dfrac{\mathrm{d}}{\mathrm{d}w} \log(w) \\
&=\dfrac{1}{\log(\log(w))} \dfrac{1}{\log(w)} \dfrac{1}{w}. \\
\end{array}$$
\end{parts}
\question[36] Compute the following indefinite integrals.
\begin{parts}
\part[6] $\displaystyle\int \dfrac{1}{2+x^2} \mathrm{d}x$ \\
\textit{Solution:} Using the formula
$$(**) \hspace{35pt} \displaystyle\int \dfrac{1}{a^2+x^2} \mathrm{d}x = \dfrac{1}{a} \arctan \left( \dfrac{x}{a} \right) + C,$$
we \href{http://www.wolframalpha.com/input/?i=antiderivative+1%2F(2%2Bx%5E2)+dx}{compute}
$$\begin{array}{ll}
\displaystyle\int \dfrac{1}{2+x^2} \mathrm{d}x &= \displaystyle\int \dfrac{1}{(\sqrt{2})^2+x^2} \mathrm{d}x \\
&\stackrel{(**)}{=} \dfrac{1}{\sqrt{2}} \arctan \left( \dfrac{x}{\sqrt{2}} \right) + C.
\end{array}$$
\part[6] $\displaystyle\int \dfrac{w-5}{w^2+1} \mathrm{d}w$ \\
\textit{Solution:} First note that we may split this into two integrals:
$$\displaystyle\int \dfrac{w-5}{w^2+1} \mathrm{d}w = \displaystyle\int \dfrac{w}{w^2+1} \mathrm{d}w - 5 \displaystyle\int \dfrac{1}{w^2+1} \mathrm{d}w.$$
For the first integral, make the substitution $u = w^2+1$ to arrive at $\dfrac{1}{2} \mathrm{d}u = w \mathrm{d}w$ and for the second integral, use $(**)$ to \href{http://www.wolframalpha.com/input/?i=antiderivative+(w-5)%2F(w%5E2%2B1)+dw}{calculate}
$$\begin{array}{ll}
\displaystyle\int \dfrac{w-5}{w^2+1} \mathrm{d}w &= \displaystyle\int \dfrac{w}{w^2+1} \mathrm{d}w - 5 \displaystyle\int \dfrac{1}{w^2+1} \mathrm{d}w \\
&= \dfrac{1}{2} \displaystyle\int \dfrac{1}{u} \mathrm{d}u - 5 \arctan(w) + C \\
&= \dfrac{1}{2} \log(u) - 5 \arctan(w) + C \\
&= \dfrac{1}{2} \log(w^2+1) - 5 \arctan(w) + C.
\end{array}$$
\part[6] $\displaystyle\int \dfrac{4v+4}{2v^2+4v+18} \mathrm{d}v$ \\
\textit{Solution:} Let $u=2v^2+4v+18$ so that $\mathrm{d}u = 4v+4 \mathrm{d}v$. Now \href{http://www.wolframalpha.com/input/?i=antiderivative+(4v%2B4)%2F(2v%5E2%2B4v%2B18)+dv}{calculate}
$$\begin{array}{ll}
\displaystyle\int \dfrac{4v+4}{2v^2+4v+18} \mathrm{d}v &= \displaystyle\int \dfrac{1}{u} \mathrm{d}u \\
&= \log(u) + C \\
&= \log(2v^2+4v+18) + C.
\end{array}$$
\part[6] $\displaystyle\int \mathrm{coth}(\theta) \mathrm{d}\theta$ $\Bigg($recall: $\mathrm{coth}=\dfrac{\cosh}{\sinh}$ $\Bigg)$ \\
\textit{Solution:} Take note that if $u=\sinh(\theta)$ then $\mathrm{d}u=\cosh(\theta) \mathrm{d}\theta$. Now \href{http://www.wolframalpha.com/input/?i=antiderivative+coth(theta)+dtheta}{calculate}
$$\begin{array}{ll}
\displaystyle\int \mathrm{coth}(\theta) \mathrm{d}\theta &= \displaystyle\int \dfrac{\cosh(\theta)}{\sinh(\theta)} \mathrm{d}\theta \\
&=  \displaystyle\int \dfrac{1}{u} \mathrm{d}u \\
&= \log(u) + C \\
&= \log(\sinh(\theta)) + C.
\end{array}$$
\part[6] $\displaystyle\int \dfrac{\cosh(t)}{\sinh(t)+1} \mathrm{d}t$ \\
\textit{Solution:} Take note that if $u=\sinh(t)+1$ then $\mathrm{d}u=\cosh(t) \mathrm{d}t$. Now \href{http://www.wolframalpha.com/input/?i=antiderivative+cosh(t)%2F(sinh(t)%2B1)+dt}{compute}
$$\begin{array}{ll}
\displaystyle\int \dfrac{\cosh(t)}{\sinh(t)+1} \mathrm{d}t &= \displaystyle\int \dfrac{1}{u} \mathrm{d}u \\
&= \log(u) +C \\
&= \log(\sinh(t)+1)+C.
\end{array}$$
\part[6] $\displaystyle\int \dfrac{17}{\sqrt{1-2w^2}} \mathrm{d}w$ \\
\textit{Solution:} Recall the formula
$$(***) \hspace{35pt} \displaystyle\int \dfrac{1}{\sqrt{a^2-x^2}} \mathrm{d}x = \arcsin \left( \dfrac{x}{a} \right)+C.$$
Let $u=\sqrt{2}w$ so that $\dfrac{1}{\sqrt{2}} \mathrm{d}u = \mathrm{d}w$. Then  \href{http://www.wolframalpha.com/input/?i=antiderivative+17%2Fsqrt(1-2w%5E2)+dw}{compute}
$$\begin{array}{ll}
\displaystyle\int \dfrac{17}{\sqrt{1-2w^2}} \mathrm{d}w &= 17 \displaystyle\int \dfrac{1}{\sqrt{1-(\sqrt{2}w)^2}} \mathrm{d}w \\
&= \dfrac{17}{\sqrt{2}} \displaystyle\int \dfrac{1}{\sqrt{1-u^2}} \mathrm{d}u \\
&\stackrel{(***)}{=} \dfrac{17}{\sqrt{2}} \arcsin(u) \mathrm{d}u + C\\
&= \dfrac{17}{\sqrt{2}} \arcsin \left( \sqrt{2}w \right) + C.
\end{array}$$
\end{parts}
\question[28] Compute the following definite integrals.
\begin{parts}
\part[7] $\displaystyle\int_0^1 7^x - 11^x \mathrm{d}x$ \\
\textit{Solution:} Note that $7^x=e^{x\log(7)}$ and $11^x=e^{x\log(11)}$. We will use the substitutions $u=x\log(7)$ so $\dfrac{1}{\log(7)} \mathrm{d}u=\mathrm{d}x$ and $w=x\log(11)$ so that $\dfrac{1}{\log(11)}\mathrm{d}w = \mathrm{d}x$. Now \href{http://www.wolframalpha.com/input/?i=integral+from+x%3D0+to+x%3D1+of+7%5Ex+-+11%5Ex+dx}{compute}
$$\begin{array}{ll}
\displaystyle\int_0^1 7^x - 11^x \mathrm{d}x &= \displaystyle\int_0^1 7^x \mathrm{d}x - \displaystyle\int_0^1 11^x \mathrm{d}x \\
&= \displaystyle\int_0^1 e^{x\log(7)} \mathrm{d}x - \displaystyle\int_0^1 e^{x\log(11)} \mathrm{d}x \\
&= \dfrac{1}{\log(7)} \displaystyle\int_{0}^{\log(7)} e^u \mathrm{d}u - \dfrac{1}{\log(11)} \displaystyle\int_0^{\log(11)} e^w \mathrm{d}w \\
&= \dfrac{1}{\log(7)} \Bigg[ e^u \Bigg|_0^{\log(7)} - \dfrac{1}{\log(11)} \Bigg[ e^u \Bigg|_0^{\log(11)} \\
&= \dfrac{1}{\log(7)} (e^{\log(7)}-1) - \dfrac{1}{\log(11)} (e^{\log(11)}-1) \\
&= \dfrac{6}{\log(7)} - \dfrac{10}{\log(11)}.
\end{array}$$
\part[7] $\displaystyle\int_2^3 \dfrac{1}{\sqrt{16y^2+1}} \mathrm{d}y$ \\
\textit{Solution:} Recall the formula
$$(****) \hspace{35pt} \displaystyle\int \dfrac{1}{\sqrt{a^2+x^2}} \mathrm{d}x = \mathrm{arcsinh} \left( \dfrac{x}{a} \right) +C.$$
We shall use the substitution $u=4y$, hence $\dfrac{1}{4} \mathrm{d}u = \mathrm{d}y$. Now \href{http://www.wolframalpha.com/input/?i=integral+from+y%3D2+to+y%3D3+of+1%2Fsqrt(16y%5E2%2B1)+dy}{compute}
$$\begin{array}{ll}
\displaystyle\int_2^3 \dfrac{1}{\sqrt{16y^2+1}} \mathrm{d}y &= \displaystyle\int_2^3 \dfrac{1}{\sqrt{(4y)^2+1}} \mathrm{d}y \\
&= \dfrac{1}{4} \displaystyle\int_8^12 \dfrac{1}{\sqrt{u^2+1}} \mathrm{d}u \\
&\stackrel{(****)}{=} \mathrm{arcsinh}(u) \Bigg|_2^3 \\
&= \mathrm{arcsinh}(3) - \mathrm{arcsinh}(2).
\end{array}$$
\part[7] $\displaystyle\int_0^3 \xi^4+3\xi^2+\xi+1 \mathrm{d}\xi$ \\
\textit{Solution}: This is just the integral of a polynomial. So \href{http://www.wolframalpha.com/input/?i=integral+from+xi%3D0+to+xi%3D3+of+xi%5E4+%2B+3xi%5E2+%2B+xi+%2B+1}{compute}
$$\begin{array}{ll}
\displaystyle\int_0^3 \xi^4 + 3\xi^2 + \xi + 1 \mathrm{d}\xi &= \dfrac{\xi^5}{5} + \xi^3 + \dfrac{\xi^2}{2} + \xi \Bigg|_0^3 \\
&= \dfrac{3^5}{5} + 3^3 + \dfrac{3^2}{2} + 3.
\end{array}$$
\part[7] $\displaystyle\int_1^2 \dfrac{1}{57-4z^2} \mathrm{d}z$ \\
\textit{Solution}: Recall that
$$(*****) \displaystyle\int\dfrac{1}{a^2-x^2} \mathrm{d}x = \dfrac{1}{a} \mathrm{arctanh} \left( \dfrac{x}{a} \right) +C.$$
We will use the substitution $u=2z$ so that $\dfrac{1}{2} \mathrm{d}u = \mathrm{d}z$. Now \href{http://www.wolframalpha.com/input/?i=integral+from+z%3D1+to+z%3D2+of+1%2F(57-4z^2)+dz}{compute}
$$\begin{array}{ll}
\displaystyle\int_1^2 \dfrac{1}{57-4z^2} \mathrm{d}z &= \displaystyle\int_1^2 \dfrac{1}{(\sqrt{57})^2-(2z)^2} \mathrm{d}z \\
&= \dfrac{1}{2} \displaystyle\int_2^4 \dfrac{1}{(\sqrt{57})^2 - u^2} \mathrm{d}u \\
&\stackrel{(*****)}{=} \dfrac{1}{2\sqrt{57}} \mathrm{arctanh} \left( \dfrac{u}{\sqrt{57}} \right) \Bigg|_2^4 \\
&= \dfrac{\mathrm{arctanh} \left( \frac{4}{\sqrt{57}} \right) - \mathrm{arctanh}\left( \frac{2}{\sqrt{57}} \right)}{2\sqrt{57}}.
\end{array}$$
\end{parts}
\end{questions}
\end{document}