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\flushleft\underline{Homework 2 --- MATH 1586 Spring 2018} \\
\begin{enumerate}[1.]
\item Compute $\displaystyle\int e^{12x} \mathrm{d}x$. \\
\textit{Solution}: Let $u=12x$, then $\mathrm{d}u=12 \mathrm{d}x$ and so $\dfrac{1}{12} \mathrm{d}u = \mathrm{d}x$. Therefore, compute
\[ \begin{array}{ll}
\displaystyle\int e^{12x} \mathrm{d}x &= \dfrac{1}{12} \displaystyle\int e^u \mathrm{d}u \\
&= \dfrac{1}{12} e^u + C \\
&= \dfrac{1}{12} e^{12x} + C.
\end{array} \]
\item Compute $\displaystyle\int (t^3-2t)^{25}(3t^2-2) \mathrm{d}t$. \\
\textit{Solution}: Let $u=t^3-2t$, then $\mathrm{d}u=3t^2-2 \mathrm{d}t$. Therefore, compute
\[ \begin{array}{ll}
\displaystyle\int (t^3-2t)^{25} (3t^2-2) \mathrm{d}t &= \displaystyle\int u^{25} \mathrm{d}u \\
&= \dfrac{u^{26}}{26} + C \\
&= \dfrac{(t^3-2t)^{26}}{26} + C.
\end{array} \]
\item Compute $\displaystyle\int \dfrac{z^4}{1-z^5} \mathrm{d}z$. \\
\textit{Solution}: Let $u=1-z^5$, then $\mathrm{d}u = -5z^4 \mathrm{d}z$, and so $-\dfrac{1}{5} \mathrm{d}u = z^4 \mathrm{d}z$. Therefore, compute
\[ \begin{array}{ll}
\displaystyle\int \dfrac{z^4}{1-z^5} \mathrm{d}z &= -\dfrac{1}{5} \displaystyle\int \dfrac{1}{u} \mathrm{d}u \\
&= -\dfrac{1}{5} \ln(u) + C \\
&= -\dfrac{1}{5} \ln(1-z^5) + C.
\end{array} \]
\item Compute $\displaystyle\int \dfrac{5}{w-3} \mathrm{d}w$.\\
\textit{Solution}: Let $u=w-3$, then $\mathrm{d}u=\mathrm{d}w$. Therefore, compute
\[ \begin{array}{ll}
\displaystyle\int \dfrac{5}{w-3} \mathrm{d}w &= 5 \displaystyle\int \dfrac{1}{u} \mathrm{d}u \\
&= 5 \ln(u) + C \\
&= 5 \ln(w-3) + C.
\end{array} \]
\item Compute $\displaystyle\int q^2 e^{q^3-1} \mathrm{d}q$. \\
\textit{Solution}: Let $u=q^3-1$, then $\mathrm{d}u=3q^2 \mathrm{d}q$ and so $\dfrac{1}{3} \mathrm{d}u = q^2 \mathrm{d}q$. Therefore, compute
\[ \begin{array}{ll}
\displaystyle\int q^2 e^{q^3-1} \mathrm{d}q &= \dfrac{1}{3}\displaystyle\int e^u \mathrm{d}u \\
&= \dfrac{1}{3} e^u + C \\
&= \dfrac{1}{3} e^{q^3-1} + C.
\end{array} \]
\item In calm waters, the oil spilling from the ruptured hull of a grounded tanker forms an oil slick that is circular in shape. If the radius $r$ of the circle is increasing at a rate of 
$$r'(t) = \dfrac{30}{\sqrt{2t+4}} \hspace{5pt} \dfrac{\mathrm{ft}}{\mathrm{min}}$$
$t$ minutes after the rupture occurs, find an expression for the radius at any time $t$. How large is the polluted area $16$ minutes after the rupture occurred? (\textit{note: $r(0)=0$}) \\
\textit{Solution}: We are told $r'(t)$, so to find $r(t)$, we must integrate (with respect to $t$). Let $u=2t+4$, then $\mathrm{d}u=2 \mathrm{d}t$ and so $\dfrac{1}{2} \mathrm{d}u = \mathrm{d}t$. Therefore, compute
\[ \begin{array}{ll}
r(t) &= \displaystyle\int \dfrac{30}{\sqrt{2t+4}} \mathrm{d}t \\
&= \dfrac{30}{2} \displaystyle\int \dfrac{1}{\sqrt{u}} \mathrm{d}u \\
&= 15 \displaystyle\int u^{-\frac{1}{2}} \mathrm{d}u \\
&= 15 \dfrac{u^{\frac{1}{2}}}{\frac{1}{2}} + C \\
&= 30 \sqrt{2t+4} + C.
\end{array} \]
\underline{Find $C$} \\
Using the given information that $r(0)=0$, we see
\[ \underbrace{0}_{\mathrm{given}} = r(0) = \underbrace{30\sqrt{2(0)+4}+C}_{\mathrm{calculated}}.\]
This yields $0=30\sqrt{4}+C=60+C$. Therefore solve for $C$ to get $C=-60$. Thus our model is
\[ r(t) = 30\sqrt{2t+4}-60.\]
\underline{16 minutes after rupture} \\
Here just calculate $r(16)$:
\[ r(16) = 30\sqrt{2(16)+4}-60 = 30\sqrt{36}-60 = 180-60=120 \mathrm{\hspace{2pt} ft}.\]
\item Suppose that in a certain country, the life expectancy at birth of a female is changing at the rate of 
$$g'(t)=\dfrac{5.45218}{(1+1.09t)^{0.9}} \hspace{5pt} \dfrac{\mathrm{years}}{\mathrm{year}}.$$
Here $t$ is measured in years with $t=0$ corresponding to the beginning of the year $1900$. Find an expression $g(t)$ giving the life expectancy at birth (in years) of a female in that country if the life expectancy at the beginning of $1900$ is $50.02$ years. What is the life expectancy at birth of a female born in the year $2000$ according to this model? \\
\textit{Solution}: We are told $g'(t)$, so we must integrate (with respect to $t$) to find $g(t)$. Let $u=1+1.09t$ so $\mathrm{d}u=1.09 \mathrm{d}t$ and thus $\dfrac{1}{1.09} \mathrm{d}u = \mathrm{d}t$. Therefore, calculate
\[ \begin{array}{ll}
g(t) &= \displaystyle\int \dfrac{5.45218}{(1+1.09t)^{0.9}} \mathrm{d}t \\
&= \dfrac{5.45218}{1.09} \displaystyle\int \dfrac{1}{u^{0.9}} \mathrm{d}u \\
&=5.002 \displaystyle\int u^{-0.9} \mathrm{d}u \\
&= 5.002 \dfrac{u^{0.1}}{0.1} + C \\
&= 50.02 (1+1.09t)^{0.1} + C.
\end{array} \]
\underline{Find $C$} \\
Since $t=0$ corresponds to the year $1900$ and we are told that life expectancy in 1900 was $50.02$ years, we see that we were  told that $g(0)=50.02$. Therefore,
\[ \underbrace{50.02}_{\mathrm{given}} = g(0) = \underbrace{50.02(1+1.09(0))^{0.1} + C}_{\mathrm{calculated}}.\]
This equation yields $50.02 = 50.02+C$, or in other words, $C=0$. Therefore our model is 
\[ g(t) = 50.02(1+1.09t)^{0.1}.\]
\underline{Life expectancy in year $2000$} \\
Since $t=0$ corresponds to $1900$, then $t=100$ corresponds to $2000$. Plug this into the model to get
\[ g(100)=50.02(1+1.09(100))^{0.1} = 80.0356 \mathrm{\hspace{2pt} years}.\]
\end{enumerate}
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