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\centerline{Mathematics 132 -- Calculus for Physical and Life Sciences 2}
\centerline{Exam 1 -- Review Sheet}
\centerline{February 18, 2008}
\bigskip


\noindent
{\it Sample Exam Questions- Solutions}
\bigskip
\noindent I.  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 $y = f(x)$.

\begin{center}
\includegraphics[width=2in]{exam1fig1}
\end{center}


\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}

$\displaystyle F(7)=\int_0^6 f(x)\, dx + \int_6^{13/2} f(x)\, dx +\int_{13/2}{7}= 9 + \frac{1}{4} -\frac{1}{4}=9$   \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)  Sketch the graph $y = F(x)$ if $F(0) = 0$,
and also if $F(0) = 2$.
\bigskip

\vspace{.1in} 

If $F(0)=0$ we have 

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\includegraphics[width=2in]{exam1fig2}
\end{center}
If $F(0)=2$ we have 

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\includegraphics[width=2in]{exam1fig3}
\end{center}

Ignore the vertical lines in the graph. 

\vspace{.1in}


\noindent
II.  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 III. 
\bigskip\vspace{.1in}


\noindent (A) Compute $\int 5x^4 - 3\sqrt{x} + e^x + {2\over x}\ dx$
\medskip\vspace{.1in}

$\displaystyle \int 5x^4 - 3\sqrt{x} + e^x + {2\over x}\ dx = x^5-2x^{3/2}+e^x+2 \ln|x|+C$\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) Apply a $u$-substitution to compute $\int_1^2 e^{\sin(\pi x)}\cos(\pi x)\ dx$
\medskip\vspace{.1in}



$u=\sin(\pi x), \ du= \pi \cos(\pi x)$. Then $\displaystyle \int_1^2 e^{\sin(\pi x)}\cos(\pi x)\ dx= \frac{1}{\pi} \int _0^0 e^u \, du = 0$\vspace{.1in}

\noindent (D)  Do you need partial fractions to 
compute $$\int {t^2 + 1\over t^3 + 3t + 3}\ dt?$$
Explain, and give a simpler method.\vspace{.1in}

No. Just do a $u$-substitution. $u=t^3+3t+3, \ du= (3t^2+3) dt= 3(t^2+1) dt$. Then $\displaystyle \int {t^2 + 1\over t^3 + 3t + 3}\ dt = \frac{1}{3} \int \frac{1}{u}\, du = \frac{1}{3} \ln|u| +C=\frac{1}{3} \ln|t^3-3t+3| + C.$

\vspace{.1in}

\noindent (E) Apply integration by parts to compute $\int x^2 e^{-2x}\ dx$
\medskip\vspace{.1in}

$u=x^2\ \ \ \ \ \ du=2x\, dx$

$\displaystyle dv=e^{-2x} \ \ \ v=-\frac{1}{2}e^{-2x}$

Then $\displaystyle \int x^2 e^{-2x}\ dx = -\frac{1}{2}x^2 e^{-2x} + \int x e^{-2x} \, dx$. Integration by parts again. 

$u=x\ \ \ \ \ \ du=dx$

$\displaystyle dv=e^{-2x} \ \ \ v=-\frac{1}{2}e^{-2x}$ 

Then $\displaystyle \int x^2 e^{-2x}\ dx = -\frac{1}{2}x^2 e^{-2x} -\frac{1}{2}x e^{-2x} + \frac{1}{2}\int e^{-2x}\, dx =  -\frac{1}{2}x^2 e^{-2x} -\frac{1}{2}x e^{-2x} -\frac{1}{4}e^{-2x} + C$

\vspace{.1in}

\noindent (F)  Apply partial fraction decomposition to 
compute $$\int {1\over x(x - 1)(x + 2)}\ dx$$\vspace{.1in}


$\displaystyle \frac{1}{x(x-1)(x+2)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+2}$. Then $A(x-1)(x+2)+Bx(x+2)+Cx(x-1)=1$. If $x=0$, we get $-2A=1$. Thus, $\displaystyle A=-\frac{1}{2}$. If $x=1$, we get $3B=1$. Thus $\displaystyle B=\frac{1}{3}$. If $x=-2$, we get $6C=1$. Thus $\displaystyle C=\frac{1}{6}$. Therefore 

$\displaystyle \int {1\over x(x - 1)(x + 2)}\ dx =-\frac{1}{2} \int \frac{1}{x}\, dx + \frac{1}{3} \int \frac{1}{x-1}\, dx + \frac{1}{6} \int \frac{1}{x+2}\, dx = $ 

\vspace{.1in}

 $-\frac{1}{2} \ln|x| + \frac{1}{3} \ln|x-1| + \frac{1}{6} \ln|x+2| +C.$

\vspace{.1in}

\noindent (G)  Which trigonometric substitution would you apply
to compute $\int {1\over u \sqrt{a^2 - u^2}}\ du$?
What trigonometric integral do you get after making the substitution?
Complete the derivation of the integral.
\bigskip\vspace{.1in} 

Use the trigonometric substitution $u=a \sin \theta, \ du = a \cos \theta \, d\theta$. Then 

$\displaystyle \int {1\over u \sqrt{a^2 - u^2}}\ du = \int \frac{1}{a \sin \theta \sqrt{a^2 - a^2 \sin^2 \theta} } a \cos \theta \, d\theta = \int \frac{\cos \theta}{\sin\theta a \sqrt{\cos^2 \theta }}\, d \theta =$ \vspace{.1in}

$ \displaystyle \frac{1}{a} \int \frac{\cos \theta}{\sin\theta \cos \theta} \, d \theta = \frac{1}{a} \int \frac{1}{\sin \theta}\, d \theta= \frac{1}{a} \int \csc \theta \, d \theta$. Now we use the table of integrals (formula $15$) to obtain that the integral equals $\displaystyle \frac{1}{a} \ln|\csc \theta - \cot \theta| + C$. If $u=a \sin \theta$, then $\csc \theta=a/u$ and $\displaystyle \cot \theta = \frac{\sqrt{a^2-u^2}}{u}$. Therefore $$\int {1\over u \sqrt{a^2 - u^2}}\ du= \frac{1}{a} \ln|\frac{a}{u}-\frac{\sqrt{a^2-u^2}}{u}|+C.$$

\vspace{.1in}
\noindent
IV.  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.  
\bigskip


\noindent (A) $$\int {x^3 \over x^2 + 4x}\ dx$$
\vspace{.1in}

First simplify and do long division to get $\displaystyle \frac{x^3}{x^2+4x}=\frac{x^2}{x+4}=x-4+\frac{16}{x+4}$. Therefore $\displaystyle \int {x^3 \over x^2 + 4x}\ dx= \int (x-4)\, dx + \int \frac{16}{x+4}\, dx= \frac{x^2}{2}-4x + 16 \ln|x+4| + C$.

\vspace{.1in}


\noindent (B) $$\int (x^2 + 2x)^{3/2}\ dx$$\vspace{.1in}
First complete the square to get $\displaystyle \int (x^2 + 2x)^{3/2}\ dx= \int ((x+1)^2-1)^{3/2}\, dx$. 
Now do the substitution $u=x+1, \ du = dx$. The integral equals $\int (u^2-1)^{3/2}\, du$.
We can split this up as 
\begin{eqnarray*}
\int (u^2-1)^{3/2}\, du &=& \int (u^2 - 1)\sqrt{u^2 - 1}\ du\\
&=& \int u^2 \sqrt{u^2 - 1}\ du - \int \sqrt{u^2 - 1}\ du
\end{eqnarray*}
The first integral is \#40 in the table, and the second is \# 39, so we obtain
\begin{eqnarray*}
&=& \frac{u}{8}(2u^2 - 1)\sqrt{u^2 - 1} - \frac{1}{8}\ln|u + \sqrt{u^2 - 1}| 
- \frac{u}{2}\sqrt{u^2 - 1} + \frac{1}{2} \ln|u + \sqrt{u^2-1}| + C\\
&=& \frac{1}{4}(x+1)^3\sqrt{x^2 + 2x} - \frac{5}{8}(x+1)\sqrt{x^2 + 2x} + \frac{3}{8}\ln|x+1 + \sqrt{x^2+2x}| + C
\end{eqnarray*}


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

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

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

\vspace{.1in}
\noindent (D)  Note:  There was a typo in the statement of this one.  The intended integral was
\[
\int \frac{dz}{(z+3)^2\sqrt{z^2 + 6z + 10}}
\]
Completing the square we obtain 
\[
\int \frac{dz}{(z+3)^2\sqrt{z^2 + 6z + 10}}= \int \frac{1}{(z+3)^2\sqrt{(z+3)^2+1}}\, dz.
\]
Now we perform a $u$-substitution: $u=z+3, \ du=dz$ and the integral becomes
\[
\int \frac{1}{u^2 \sqrt{u^2+1}}\, du
\]
which is in the form of \#28 in the table.  The answer is
\[
-\frac{\sqrt{u^2 + 1}}{u} + C = -\frac{\sqrt{z^2 + 6z + 10}}{z + 3} + C.
\] 

\vspace{.1in}
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