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\begin{center}
  Math 108\\
  Exam I\vspace{3mm}\\
 February 3, 1998
\end{center}

\vspace{3mm}
\noindent
{\bf 3.}  To find the points where the two curves intersect we set $4 -
x^2 = -2x + 1$.  Then we obtain $x^2 - 2x - 3 = 0$ or $(x+1)(x-3) = 0$. 
Thus the two curves intersect for $x = -1$ and $x = -3$.  Their graphs
look like the following picture.

\vspace{5mm}
{\bf Insert Graph}

The area of the enclosed region is given by
\[
\int^3_{-1} [4 - x^2 - (-2x+1)]dx = \int^3_{-1} (-x^2 + 2x + 3)dx=
\]
\[
= \left (- \frac{x^3}{3} + x^2 + 3x \right )\big |^3_{-1} = \left (-9 +9
+9\right ) -
\left (\frac{1}{3} + 1 -3\right ) = \fbox{$\displaystyle{\frac{32}{3}}$}.
\]

\vspace{3mm}
\noindent
{\bf 4. d  }  14/3

We have area
\begin{eqnarray*}
&= & \int_1^4 x^{\frac{1}{2}} dx = \frac{x^{\frac{1}{2} +
1}}{\frac{1}{2} + 1} \bigg |^4_1 = \frac{2}{3} x^{\frac{3}{2}} \bigg |^4_1
=\\
 &= & \frac{2}{3} \left [4^{\frac{3}{2}} - 1^{\frac{3}{2}} \right
] =\frac{2}{3} [8-1) = \fbox{$\displaystyle{\frac{14}{3}}$}
\end{eqnarray*}

\vspace{3mm}
\noindent
{\bf 5.a  }  11/8

The four subintervals are $\left [ 0, \frac{1}{4}\right ], \left [
\frac{1}{4}, \frac{1}{2}\right ], \left [ \frac{1}{2}, \frac{3}{4}\right ],
\left [ \frac{3}{4}, 1\right ],$ and $\Delta x = \frac{1}{4}$.  Thus the
desired Riemann sum is $\left ( 2 - \frac{1}{4} \right ) \cdot
\frac{1}{4} + \left ( 2 - \frac{1}{2}\right ) \cdot \frac{1}{4} + \left ( 2
- \frac{3}{4}\right ) \cdot \frac{1}{4} + (2-1) \cdot \frac{1}{4} =
$\fbox{$\displaystyle{\frac{11}{8}} $}.

\vspace{3mm}
\noindent
{\bf 6.b  }  3

We have $y(x) = \int (2x^3 - x + 1) dx = \frac{1}{2} x^4 - \frac{1}{2} x^2
+ x + c.$.  Since $0 = y(1) = \frac{1}{2} - \frac{1}{2} + 1 + c$ we have
$c = -1$.  Thus $y(x) = \frac{1}{4} x^4 - \frac{1}{2} x^2 + x-1$ and $F(2)
= y(2) = \frac{1}{4} \cdot 2^4 - \frac{1}{2} \cdot 2^2 + 2 -1 = \ $\fbox{
3 }.

\vspace{3mm}
\noindent
{\bf 7.e  }  $\displaystyle{\frac{1}{5} + \frac{1}{6} + \frac{1}{7} +
\frac{1}{8} + \frac{1}{9}}$

The five equal subintervals are $\left [1, \frac{6}{5} \right ], \left [
\frac{6}{5}, \frac{7}{5}\right ], \left [ \frac{7}{5}, \frac{8}{5} \right ]
\left [ \frac{8}{5}, \frac{9}{5} \right ], \left [ \frac{9}{5}, 2\right ]$,
and
$\Delta x = \frac{1}{5}$.  Thus the Riemann sum corresponding to the
left-hand endpoints is
\[
1 \cdot \frac{1}{5} + \frac{5}{6} \cdot \frac{1}{5} + \frac{5}{7} \cdot
\frac{1}{5} + \frac{5}{8} \cdot \frac{1}{5} + \frac{5}{9} \cdot
\frac{1}{5} = \fbox{$\displaystyle{\frac{1}{5} + \frac{1}{6} +
\frac{1}{7} +
\frac{1}{8} + \frac{1}{9}}.$}
\]

\vspace{3mm}
\noindent
{\bf 8.d  }  $\displaystyle{\frac{x^2}{2} \ln x - \frac{x^2}{4} + C}$

We apply the IBP formula $\int udv = uv - \int v du$ with $u = \ell n x$
and $dv = xdx$.  Then $du = \frac{1}{x}dx$ and $v = \frac{x^2}{2}$.Thus
we obtain
\[
\int x\ell nx dx = \frac{x^2}{2} \ell nx - \int \frac{x^2}{2} \frac{1}{x}
dx = \frac{x^2}{2} \ell nx - \frac{1}{2} \int x dx =
\fbox{$\displaystyle{\frac{x^2}{2} \ell nx - \frac{x^2}{4} + c}$}.
\]


\vspace{3mm}
\noindent
{\bf 9.b  }  $2\int e^u du$

If we let $u = \sqrt{x} = x^{\frac{1}{2}}$ then $du = \frac{1}{2} x^{-
\frac{1}{2}} dx  = \frac{1}{2} \frac{dx}{\sqrt{x}}.$  Therefore

\[
\int \frac{1}{\sqrt{x}} e^{\sqrt{x}} dx = \fbox{$\displaystyle{ 2 \int e^u
du}$.}
\]

\vspace{3mm}
\noindent
{\bf 10.c  }  $\frac{1}{1 + t^2}$

We have $F(t) = \int^t_0 \frac{1}{1 + x^2} \ dt$ and by the FTC
$F^\prime(t) = \frac{1}{1+t^2}.$

\vspace{3mm}
\noindent
{\bf 11.b  }  -8

The definite integral is equal to $-8 -2 + 2 = -8$.

\vspace{3mm}
\noindent
{\bf 12.d  }  3/20

Let $u = x^2 + 1$.  Then $du = 2xdx$ and the integral becomes
\[
\frac{1}{2} \int^5_2 u^{-2} du = \frac{1}{2} \frac{u^{-1}}{-1} \bigg |^5_2
= - \frac{1}{2} \left ( \frac{1}{5} - \frac{1}{2}\right ) = \frac{3}{20}.
\]

\vspace{3mm}
\noindent
{\bf 13.e  }  1

Let $u = x$ and $dv = e^xdx$.  Then $du = dx$ and $v = e^x$.  By applying
the IBP formula we obtain
\[
\int^1_0 x e^x dx = xe^x \bigg |^1_0 - \int^1_0 e^x dx = e - e^x \bigg
|^1_0 = e - e + 1 = 1.
\]








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