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\centerline{Mathematics 132 -- Calculus for Physical and Life Sciences 2}
\centerline{Exam 1 -- Review Sheet}
\centerline{February 14, 2014}
\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[] Left endpoints:
$$f(0)(.25) + f(.25)(.25) + f(.5)(.25) + f(.75)(.25) = -.53125$$
\item[] Right endpoints:
$$f(.25)(.25) + f(.5)(.25) + f(.75)(.25) + f(1)(.25) = -.78125$$
\item[] Midpoints:
$$f(.125)(.25) + f(.375)(.25) + f(.625)(.25) + f(.875)(.25) = -.671875$$
\end{enumerate}
\noindent
B)  $f'(x) = 2x - 2 \le 0$ for all $x \in [0,1]$.  Hence $f$ is \emph{decreasing}
on this interval.  This implies that the left-hand Riemann sum is greater than
$\int_0^1 x^2 - 2x\ dx$ (i.e. less negative than the integral), and the 
right-hand Riemann sum is less (i.e. more negative) than the value of the integral.
Using the Evaluation Theorem, we can check this:
$$\int_0^1 x^2 - 2x\ dx = \left. \frac{x^3}{3} - x^2\right|_0^1 = -\frac{2}{3} \doteq -.66667$$
The left-hand sum is greater (less negative) and the right-hand sum is less (more negative)
than this value. 
\vskip 10pt
\noindent
C) The sum is a right-hand Riemann sum for the function $f(x) = \frac{\cos(x)}{x^2 + 1}$
on the interval $[0,\pi]$.  So it will equal
$$\int_0^\pi \frac{\cos(x)}{x^2 + 1}\ dx.$$ 

\vskip 10pt
II.
\begin{enumerate}
\item[A)]  We have $\Delta t = 2$ and 
$$L_5 = v(0)\cdot 2 + v(2) \cdot 2 + v(4)\cdot 2 + v(6)\cdot 2 + v(8)\cdot 2 \doteq 29.62$$
(meters).
\item[B)]  The exact value is 
$$\int_0^10 \sqrt{t + 5}\ dt = \left. \frac{2}{3}(t + 5)^{3/2}\right|_0^{10} = 
\frac{2}{3}(15\sqrt{15} - 5\sqrt{5}) \doteq 31.28$$
\end{enumerate}
\vskip 10pt
\noindent
III.  
\begin{enumerate}
\item[A)]  We have
\begin{align*}
\sum_{i=1}^N \left(\frac{5i}{N}\right)^2 \cdot \frac{5}{N} &=
\frac{125}{N^3}\sum_{i=1}^N i^2 \\
&= \frac{125}{N^3}\cdot \frac{N(N+1)(2N+1)}{6}\\
&= \frac{125}{3} + \frac{125}{2N} + \frac{125}{6N^2}.
\end{align*}
Hence the limit as $N \to \infty$ is $\frac{125}{3}$.  
\item[B)]  The limit in part A is the limit of a $R_N$ Riemann sum and computes
the integral:
$$\int_0^5 x^2\ dx = \left.\frac{x^3}{3}\right|_0^5 = \frac{125}{3}.$$
\end{enumerate}
\noindent 
IV.  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 $y = f(x)$.  (Omitted -- the graph is made up of 
segments of three different straight lines.)

\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)  Are there any critical points of $F$?  If so, find them and say whether 
they are local maxima, local minima, or neither.  If not, say why not. \vspace{.1in}

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)  How is the graph of $F(x)$ related to the graph of 
$$G(x) = \int_2^x f(t)\ dt?$$
This is another antiderivative of $f$.  It satisfies $F(x) = \int_0^2 f(t)\ dt + G(x)$
by the interval union property for integrals.  So the graph of $F(x)$ would be the graph
of $G(x)$ shifted up by $\int_0^2 f(t)\ dt = F(2) = 2$.  
\bigskip

\vspace{.1in}


\noindent
V.  Find the derivatives of the following functions
\bigskip
\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 VI. 
\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) Integrate by parts to compute $\int x^2 e^{6x}\ dx$.  We let $u = x^2$ and 
$dv = e^{6x}$ the first time, then integrate by parts again with $u = x$, $dv = e^{6x}$.  
The result is 
\begin{align*}
\int x^2 e^{6x}\ dx &= \frac{x^2 e^{6x}}{6} - \frac{1}{3}\int x e^{6x}\ dx \\
&= \frac{x^2 e^{6x}}{6} - \frac{x e^{6x}}{18} + \frac{1}{18} \int e^{6x}\ dx\\
&= \frac{x^2 e^{6x}}{6} - \frac{x e^{6x}}{18} + \frac{e^{6x}}{108} + C
\end{align*}
\medskip\vspace{.1in}


\noindent (D)  Integrate with an appropriate $u$-substitution:
$$\int \frac{t^2 + 1}{t^3 + 3t + 3}\ dt?$$

\vspace{.1in}
Let $u=t^3+3t+3, \ du= (3t^2+3) dt= 3(t^2+1) dt$. Then $\displaystyle \int \frac{t^2 + 1}{t^3 + 3t + 3}\ dt = \frac{1}{3} \int \frac{1}{u}\, du = \frac{1}{3} \ln|u| +C=\frac{1}{3} \ln|t^3-3t+3| + C.$

\vspace{.1in}

\noindent (E) Compute $\int x^4\ln(3x)\ dx$ using integration by parts.
\medskip\vspace{.1in}

Let $u = \ln(3x)$, so $du = \frac{1}{x}\ dx$.  Then $dv = x^4\ dx$, and $v = \frac{x^5}{5}$
Then 
$$\int x^4\ln(3x)\ dx = \frac{x^5\ln(3x)}{5} - \int \frac{x^4}{5}\ dx = \frac{x^5\ln(3x)}{5}
- \frac{x^5}{25} + C.$$

\noindent (F) Using the substitution $u = 3x$ and the SC1 reduction formula twice:
\begin{align*}
\int \sin^4(3x)\ dx &= \frac{1}{3}\int \sin^4(u)\ du\\
&= \frac{1}{3}\left(\frac{-\sin^3(u)\cos(u)}{4} + \frac{3}{4}\int \sin^2(u)\ du\right)\\
&= \frac{1}{3}\left(\frac{-\sin^3(u)\cos(u)}{4} + \frac{3}{4}\left(\frac{-\sin(u)\cos(u)}{2} + \frac{u}{2} \right)\\
&= -\frac{\sin^3(u)\cos(u)}{12} - \frac{\sin(u)\cos(u)}{8} + \frac{u}{8} + C\\
&= -\frac{\sin^3(3x)\cos(3x)}{12} - \frac{\sin(3x)\cos(3x)}{8} + \frac{3x}{8} + C
\end{align*}
\noindent
V.  Compute each of the integrals below using some combination of
basic rules, substitution, integration by parts, the table of integrals, 
partial fractions, and trigonometric substition. You must show all work
for full credit.  
\vskip 10pt

\noindent (A)  $$\int \frac{e^{\sqrt{\sin(x)}} \cos(x)}{\sqrt{\sin(x)}}\ dx$$\vspace{.1in}

Use the substitution $\displaystyle u=\sqrt{\sin x}, du = \frac{1}{2 \sqrt{\sin x}} \cos x\, dx $. Then \vspace{.1in}

$\displaystyle\int \frac{e^{\sqrt{\sin(x)}} \cos(x)}{\sqrt{\sin(x)}}\ dx = 2\int e^u \, du= 2e^u + C = 2e^{\sqrt{\sin x}}+C$.

\vspace{.1in}
\noindent (B)  Use parts twice (with $u = $ the trig function and $dv = e^x\ dx$, 
then solve for the desired integral algebraically:
\begin{eqnarray*}
\int e^x \sin(2x)\ dx &=& e^x \sin(2x) - 2\int e^x \cos(2x)\ dx\\
\int e^x \sin(2x)\ dx &=& e^x \sin(2x) - 2\left(e^x \cos(2x) + 2\int e^x \sin(2x)\ dx\right)\\ 
\int e^x \sin(2x)\ dx &=& e^x \sin(2x) - 2 e^x \cos(2x) - 4\int e^x \sin(2x)\ dx \\
\therefore
5\int e^x \sin(2x)\ dx &=& e^x \sin(2x) - 2 e^x \cos(2x), \quad \text{and}\\
\int e^x \sin(2x)\ dx &=& \frac{e^x \sin(2x) - 2 e^x \cos(2x)}{5} + C.
\end{eqnarray*}

\vspace{.1in}
\noindent
(C) Integrate by parts ($u = \tan^{-1}(x), dv = dx$), the  
resulting integral is $\frac{1}{2}\int \frac{dw}{w}$ for $w =x^2 + 1$, 
and use the Evaluation Theorem:
\begin{eqnarray*}
\int_0^1 \tan^{-1}\ dx &=& 
\left. x\tan^{-1}(x)\right|_0^1 - \int_0^1 \frac{x}{x^2 + 1}\ dx\\
&=& \frac{\pi}{4} - \left. \frac{1}{2} \ln(x^2 + 1)\right|_0^1\\
&=& \frac{\pi}{4} - \frac{\ln(2)}{2}.
\end{eqnarray*}

\vspace{.1in}
\noindent
(D) This reduction formula
falls out immediately from the integration by parts formula, letting $u = x^n$
and $dv = e^{ax}\ dx$.

\vspace{.1in}
\noindent
(E)  We use the reduction formula from (D) with $n = 4$, $a = -2$, then 
again with $n = 3,2,1$:
\begin{eqnarray*}
\int x^4 e^{-2x}\ dx &=& \frac{-x^4 e^{-2x}}{2} + 2\int x^3 e^{-2x}\ dx\\
&=& \frac{-x^4 e^{-2x}}{2} + 2\left(\frac{-x^3 e^{-2x}}{-2} + \frac{3}{2} \int x^2 e^{-2x}\ dx\right)\\
&\vdots & \\
&=& - \left(\frac{x^4}{2} + x^3 + \frac{3}{2} x^2 + \frac{3}{2}x + \frac{3}{4}\right)e^{-2x}
\end{eqnarray*}
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