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\preface
\magnification=1200

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\title
\noindent
{\bf Math 120: Calculus}\hfill
Name:\hbox to 2.5truein{\hrulefill}\medskip\noindent
{\bf Exam III}\hfill
Tutorial Instructor:\hbox to 2.5truein{\hrulefill}\medskip\noindent
{\sl April 16, 1996}\hfill
Tutorial Section:\hbox to 2.5truein{\hrulefill}

\bigskip\noindent
Calculators are not allowed.  Hand in this answer page only.  Record your answers 
to the multiple choice problems by  placing an $\times$ through one letter for
each problem on this answer sheet. There are  16 multiple choice questions, worth
6 points each.  An additional 4 points will be given for your correct tutorial
section number.  \vskip .25cm \noindent {\bf You are taking this exam under the
honor code.}

\def\answerskip{\vskip0.25in}

\def\problemskip{\vskip1in}

\note %1
\problem  Evaluate $\displaystyle \int x e^{2x} dx$

\correct ${1 \over 2} xe^{2x} - {1 \over 4} e^{2x} +C$
\wrong ${1 \over 2} xe^{2x} -  e^{2x} +C$
\wrong $ xe^{2x} -  e^{2x} +C$
\wrong $2 xe^{2x} - 4 e^{2x} +C$
\wrong $2 xe^{2x} -  e^{2x} +C$

\note %2
\problem Evaluate $\displaystyle \int_0^{\pi/2} \cos^2 x \ dx$

\correct $\displaystyle {{\pi} \over 4}$
\wrong $\displaystyle {{\pi} \over 4} + {1 \over 4}$
\wrong $0$
\wrong $\displaystyle {{\pi} \over 4} - {1 \over 4}$
\wrong $\pi$

\note %3
\problem 
In the partial fraction decomposition $$\displaystyle {{2x +1} \over {x^3 - x}}
= {A \over x} + {B \over {x-1}} + {C \over {x+1}}$$
find the value of $B$.  (You don't have to find the values of $A$ or $C$.)
\correct$3/2$
\wrong $1$
\wrong $-1$
\wrong $2/3$
\wrong $-2/3$

\note %4
\problem Use long division to divide
$\displaystyle {{x^3 + x^2 +x+1} \over {x^2 -1}}$
\correct $\displaystyle x+1 + {{2x+2} \over {x^2 -1}}$
\wrong $\displaystyle x+1 + {{2} \over {x^2 -1}}$
\wrong $\displaystyle x+1 + {{2x} \over {x^2 -1}}$
\wrong $\displaystyle x + {{2x+1} \over {x^2 -1}}$
\wrong $\displaystyle x + {{x+1} \over {x^2 -1}}$

\note %5
\problem Evaluate $\displaystyle \int_{1/2}^{\sqrt{3}/2} \left ( {{dx} \over {x^2
\sqrt{1-x^2}}} \right )$

\correct $\displaystyle \sqrt{3} - {1 \over {\sqrt{3}}}$
\wrong $\displaystyle {{\pi} \over 3} - {\pi \over 6}$
\wrong $\displaystyle {\sqrt{3} \over 2} - {1 \over 2}$
\wrong $\displaystyle {1 \over 2} - \sqrt{3}$
\wrong $\displaystyle {1 \over \sqrt{3}} - {1 \over 2}$


\note %6
\problem  Evaluate $\displaystyle \int {x \over {x^2  +1}} dx$
\correct $\displaystyle {1 \over 2} \ln |1+x^2 | +C$
\wrong $\displaystyle \ln |1+x^2 | +C$
\wrong $\displaystyle -{1 \over 2} \ln |1+x^2 | +C$
\wrong $\displaystyle {1 \over 2} |1+x^2 |^2 +C$
\wrong $\displaystyle -2\ln  |1+x^2 | +C$

\note %7
\problem Evaluate $\displaystyle \int_1^{\hbox{$e$}} \sqrt{x} \ln x \ dx$
 
\correct $\displaystyle {2 \over 9} e^{3/2} + {4 \over 9}$
\wrong $\displaystyle {2 \over 9} e^{3/2} - {2 \over 9}$
\wrong $\displaystyle {2 \over 9} e^{3/2} + {2 \over 3}$
\wrong $\displaystyle {2 \over 9} e^{3/2} - {2 \over 3}$
\wrong $\displaystyle {2 \over 9} e^{3/2} - 1$

\note %8
\problem If you were to use integration by parts to evaluate $\displaystyle \int
x^3 \sin x \ dx$, you would get
$$
\int x^3 \sin x \ dx = -x^3 \cos x - _{\hbox{??} 
\over \phantom{12222222222222222222222222 \over 2}}.
$$
What should be in the blank? (Notice that it's a minus sign in front of the
blank.)

\correct $\displaystyle -3 \int x^2 \cos x \ dx$
\wrong $\displaystyle 3 \int x^2 \cos x \ dx$
\wrong $\displaystyle -{1 \over 3} \int x^2 \cos x \ dx$
\wrong $\displaystyle +{1 \over 3} \int x^2 \cos x \ dx$
\wrong $\displaystyle -{1 \over 3} \int  \cos x^2 \ dx$

\note %9
\problem Evaluate $\displaystyle \int_0^{\pi/2} \sin^3 x \
dx$.  

\correct $2/3$
\wrong $-1$
\wrong $0$
\wrong $1$
\wrong $1/3$

\note %10
\problem Suppose you want to evaluate $\displaystyle \int_1^4 {{x^3} \over
{\sqrt{2x^2 +7}}}\ dx$ using a trig substitution.  What trig substitution should
you use?

\correct $\displaystyle x = {{\sqrt{7}} \over {\sqrt{2}}} \tan \theta$
\wrong $\displaystyle x = {{\sqrt{2}} \over {\sqrt{7}}} \tan \theta$
\wrong $\displaystyle x = {2 \over 7} \tan \theta$
\wrong $\displaystyle x = {7 \over 2} \tan \theta$
\wrong  $\displaystyle x = -{7 \over 2} \tan \theta$

\note %11
\problem Evaluate $\displaystyle \int {x \over {x^2 - 2x +1}} \ dx$

\correct $\displaystyle \ln |x-1| - {1 \over {x-1}} +C$
\wrong $\displaystyle \ln |x-1| + {1 \over {x-1}} +C$
\wrong $\displaystyle \ln |x-1| + {1 \over {(x-1)^2}} +C$
\wrong $\displaystyle \ln |x-1| - {1 \over {(x-1)^2}} +C$
\wrong $\displaystyle -\ln |x-1| + {1 \over {x-1}} +C$

\note %12
\problem Evaluate $\displaystyle \int {{(\sin x)(\cos x)} \over {1+\sin
x}} \ dx$

\correct $\displaystyle \sin x - \ln (1+\sin x) +C$
\wrong $\displaystyle \ln (1+\sin x) +C$
\wrong $\displaystyle {{(1/2) \sin^2 x}\over {\ln (1+\sin x)}} +C$
\wrong $\displaystyle {{(1/2) \sin^2 x} - {\ln (1+\sin x)}} +C$
\wrong $\displaystyle (\cos x)\cdot \ln (1+\sin x) +C$

\note %13
\problem Suppose you want to evaluate $\displaystyle \int_0^1 x^3 \sqrt{4-x^2} \
dx$ using the trig substitution $x = 2 \sin \theta$.  What new integal do you
then get?

\correct $\displaystyle 32 \int_0^{\pi/6}\sin^3 \theta \cos^2 \theta\ d \theta$
\wrong $\displaystyle 32 \int_0^{\pi/2}\sin^3 \theta \cos^2 \theta\ d \theta$
\wrong $\displaystyle 32 \int_0^1\sin^3 \theta \cos^2 \theta\ d \theta$
\wrong $\displaystyle 32 \int_0^{\pi/3}\sin^3 \theta \cos^2 \theta\ d \theta$
\wrong $\displaystyle 32 \int_0^{\pi}\sin^3 \theta \cos^2 \theta\ d \theta$

\note %14
\problem Suppose you want to evaluate $\displaystyle \int_{\qed}^{\qed} (1-4x^2
)^{15} dx$ using a trig substitution.  (I don't care about the limits of
integration in this problem.)  What integral do you then get?

\correct $\displaystyle {1 \over 2}\int_{\qed}^{\qed} \cos^{31} \theta \ d\theta$
\wrong $\displaystyle {1 \over 2}\int_{\qed}^{\qed} \sin^{30} \theta \cdot \cos
\theta
\ d\theta$
\wrong $\displaystyle {1 \over 2}\int_{\qed}^{\qed} \cos^{16} \theta \ d\theta$
\wrong $\displaystyle {1 \over 2}\int_{\qed}^{\qed} \sin^{15} \theta \cdot \cos
\theta
\ d\theta$
\wrong $\displaystyle {1 \over 2}\int_{\qed}^{\qed} \cos^{18} \theta \ d\theta$

\note %15
\problem Evaluate $\displaystyle \int \cos^5 x \ dx$.  Hint: use the fact that
$\displaystyle \cos^5 x =
\cos^4 x \cdot \cos x$.

\correct $\displaystyle \sin x - {2 \over 3} \sin^3 x + {1 \over 5} \sin^5 x + C$
\wrong $\displaystyle {1 \over 6} \cos^6 x + C$
\wrong $\displaystyle \left ( x - {1 \over 3} \sin^3 x \right )^2 +C$
\wrong $\displaystyle {1 \over 3} \left ( 1 - \sin^2 x  \right )^3 +C$
\wrong $\displaystyle \sin x + \sin^2 x + \sin^3 x + \sin^4 x + \sin^5 x +C$

\note %16
\problem Evaluate $\displaystyle \int \ln x \ dx$

\correct $x \ln x - x + C$
\wrong $\displaystyle {1 \over 2} \left ( \ln x \right )^2 +C$
\wrong $x \ln x +C$
\wrong $\displaystyle {1 \over x} \ln x +x +C$
\wrong $\displaystyle e^x +C$

\appendix
%That's all folks!

\end
