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\flushleft\underline{Quiz 1 --- 25 January 2018} \\
\begin{enumerate}
\item Calculate $\displaystyle\int x^2 + 5x + \dfrac{3}{x} \mathrm{d}x$. \\
\textit{Solution}: Compute
$$\begin{array}{ll}
\displaystyle\int x^2 + 5x + \dfrac{3}{x} \mathrm{d}x &= \displaystyle\int x^2  \mathrm{d}x + \displaystyle\int 5x \mathrm{d}x + \displaystyle\int \dfrac{3}{x} \mathrm{d}x \\
&= \dfrac{x^3}{3} + \dfrac{5x^2}{2} + 3\ln(x) + C.
\end{array}$$

\item Calculate $\displaystyle\int_5^6 e^{11t} \mathrm{d}t$.  \\
\textit{Solution}: Let $u=11t$ so that $\mathrm{d}u=11 \mathrm{d}t$. Thus $\dfrac{1}{11} \mathrm{d}u = \mathrm{d}t$. Now calculate
$$\begin{array}{ll}
\displaystyle\int_{5}^{6} e^{11t} \mathrm{d}t &\stackrel{u=11t}{=} \dfrac{1}{11} \displaystyle\int_{t=5}^{t=6} e^u \mathrm{d}u \\
&= \dfrac{1}{11} e^u \Bigg|_{t=5}^{t=6} \\
&=\dfrac{1}{11} e^{11t}\Bigg|_{t=5}^{t=6} \\
&=\left( \dfrac{1}{11} e^{66} \right) - \left( \dfrac{1}{11}e^{55} \right).
\end{array}$$

\item The population of a city is projected to grow at a rate of
$$r(t) = 200 \left( 1 + \dfrac{2t}{15+t^2} \right) \hspace{2pt} \dfrac{\mathrm{people}}{\mathrm{year}}.$$
The current population (at time $t=0$), t measured in years) is $25,000$. What will be the population in $4$ years? \\
\textit{Solution}: To answer this, we must integrate $r(t)$ and then apply the initial condition. First, integrate: we will use the $u$-substitution $u=15+t^2$ so that $\mathrm{d}u=2t \mathrm{d}t$ and now compute
$$\begin{array}{ll}
\displaystyle\int r(t) \mathrm{d}t &= \displaystyle\int 200 \left( 1 + \dfrac{2t}{15+t^2} \right) \mathrm{d}t \\
&= 200 \displaystyle\int 1 \mathrm{d}t + 200 \displaystyle\int \dfrac{2t}{15+t^2} \mathrm{d}t \\
&\stackrel{u=15+t^2}{=} 200t + 200 \displaystyle\int \dfrac{1}{u} \mathrm{d}u \\
&= 200t + 200\ln(u) + C \\
&= 200t + 200 \ln(15+t^2) + C.
\end{array}$$ 
\newpage
\underline{Find $C$} \\
Now we apply the initial condition that, at $t=0$, the value of this integral should be $25,000$:
$$\underbrace{25000}_{\mathrm{given}} = \underbrace{200(0)+200\ln(15+0^2)+C}_{\mathrm{calculated}}$$
this yields
$$25000=200\ln(15)+C.$$
Therefore $C=25000-200\ln(15) \approx 24458.39$. Thus the model for the population is
$$\mathrm{population}(t)=200t+200\ln(15+t^2) + 250000-200\ln(15).$$
In $4$ years, i.e. at $t=4$, the model predicts that the population will be
$$\mathrm{population}(4)=200(4)+200\ln(15+4^2)+25000-200\ln(15)\approx 25945 \hspace{2pt} \mathrm{people}.$$
\item Consider the differential equation $\dfrac{\mathrm{d}y}{\mathrm{d}t} = 7y(t)$. Use the technique of separation of variables to solve the differential equation (your answer should have a ``$C$" in it). \\
\textit{Solution}: Separate variables to get
$$\dfrac{1}{y} \mathrm{d}y = 7 \mathrm{d}t.$$
Place $\displaystyle\int$ on both sides of this equation to get
$$\displaystyle\int \dfrac{1}{y} \mathrm{d}y = \displaystyle\int 7 \mathrm{d}t.$$
Finding these antiderivatives yields
$$\ln(y) = 7t + C.$$
Plug both sides of this equation into the exponential function to get
$$y = e^{7t+C}.$$
\end{enumerate}
\end{document}