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Projects Math 112 \hfill Prelim 1 \hfill October 2, 2001 \hfill Fall
2001\\[.5cm]

\bf SOLUTIONS TO 2, 3, and 4a\\[.5cm]
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\noindent{\bf 2.} (20  points) The following is a sketch of a function
$f(t)$, for $0 \leq t \leq 3$.\\

\begin{figure}[here]
\begin{center}
{\scalebox{.8}{\includegraphics{pre1-graph}}}
\end{center}
\end{figure}

Let $\displaystyle g(x) = \int_{0}^{x} f(t)\, dt$, for $0\leq x \leq 3$.\\

There are two ways to do this problem.  One is to think of
$g(x)$ as the area under the curve $f(t)$ from $0$ to $x$. The other is
to use the FTC to realize $g'(x) = f(x)$.  \\[.5cm]

\noindent a) Find the interval(s) where $g(x)$ is increasing.  \\
Either way you think about the problem, you see $g(x)$ is
increasing when $f(t)$ is positive, so on the interval $2 \leq x \leq 3$.
Why? In the area interpretation, this is because the area from $0$ to $x$
is increasing as $x$ moves from $2$ to $3$, since you are adding on
positive area.  If you use the FTC to think of $f(x)=g'(x)$, then you
realize immediately that $g$ is increasing when $f$ is positive.\\[.5cm]

\noindent b) Find the interval(s) where $g(x)$ is decreasing.\\
Using the above reasoning, $g(x)$ is decreasing on $0 \leq x
\leq 2$.  For, $f(x)=g'(x)$ is negative there.  Also, since $f$ is
negative, as $x$ moves from $0$ toward $2$, you are adding on more
negative area, so $g(x)$ is becoming more and more negative, thus
decreasing.\\[.5cm]

\noindent c) Find the $x$-value(s) where $g(x)$ has a global minimum.\\
The candidates for extrema of $g$ on the interval $0 \leq x
\leq 3$ are critical points (i.e. places where $f=g'$ is zero) or
endpoints.  These are $x=0,2,$ and $3$.  Note $g(0)=0$, since the integral
from $0$ to $0$ of any function is $0$.  Then $g$ decreases til $x=2$, and
there $g(2)$ is some large negative number.  Then all the way from $x=2$
to $x=3$, $g$ is increasing.  Thus $g$ had to have its global minimum at
$x=2$.\\[.5cm]

\noindent d) Find the $x$-value(s) where $g(x)$ has a global maximum.\\
This is trickier.  Notice the positive area under the
curve from $2$ to $3$ is much smaller than the negative area from $0$ to
$2$.  Thus $g(x)$ is negative for every $x > 0$.  Thus the global max.
occurs when $g(0)=0$, at $x=0$.\\[.5cm]

\noindent e) Find $g'(1)$.\\
Here we need the FTC.  Since $g' = f$, we know $g'(1)=f(1)$,
and we estimate that $f(1)= -1$.  Thus, $g'(1) = -1$.\\[.5cm]


\noindent{\bf 3.} (15 points) a) Estimate
$\displaystyle\int_{0}^{4}x^{2}\, dx$ using the left and right endpoint
approximations with $n = 4$. \\

Using $n=4$ rectangles, we get endpoints $x_0=0, x_1=1, x_2=2, x_3=3,
x_4=4$, and a rectangle width $\Delta x = 1$.  So, the left hand endpoint
approximation is \[ L_4 = (1)*(0^2 + 1^2 + 2^2 + 3^2) = 14, \]  and the
right hand endpoint approximation is \[R_4 = (1)*(1^2+2^2+3^2+4^2) = 30. \]
\\[.5cm]

\noindent b) Compute $\displaystyle\int_{0}^{4}x^{2}\, dx$.\\
Since the antiderivative of $x^2$ is $x^3/3$, we use the Evaluation
Theorem to compute:
\[
\int_0^4 x^2 \ dx = \frac{x^3}{3} |_0^4 = \frac{4^3}{3} -
\frac{0^3}{3} =
\frac{64}{3},
\]
which is about 21.33333333333. \\[.5cm]

\noindent c) Which choice of endpoints (left or right) in part a) gave a
more accurate approximation?  Explain why that same choice will give a
more accurate approximation for every value of $n$.  Give specific
reasons.  (Hint: Draw a picture.) \\ 
Left endpoints will always give a better approximation in this case. 
Basically, this is because the function is concave up and increasing, so
the underestimate error from left-hand endpoints is less than the
overestimate from right hand endpoints, because the function lies below
the diagonal line connecting the endpoints. See the figure; it shows the
rectangle which is the difference between left-hand and right-hand
rectangles over one interval. \\

\begin{figure}[here]
\begin{center}
{\scalebox{.8}{\includegraphics{pre1-soln2pic2}}}
\hspace{2cm}
{\scalebox{.8}{\includegraphics{pre1-soln2pic1}}}
\end{center}
\end{figure}

The above explanation, together with the picture, are a solution. But, if
you want to see how one might make the argument really sound, read the
following.  Consider the picture of the box above.  As marked in the
figure, let $A$ be the area of the upper left, above the diagonal, let $C$
be the area under the curve, and let $B$ be the area in between.  Now
$A+B$ is the error in using right-hand endpoints, and $C$ is the error in
using left-hand endpoints, so we want to determine which of the areas
$A+B$ and $C$ is bigger.  It's pretty clear that $A+B$ is bigger.  To make
that precise, notice that $A =B+C$.  Hence, $A+B = (B+C)+B$.  Next, we
know $B>0$, because the function is concave up and increasing (look at
the picture). Hence, $(B+C)+B = 2B+C > C$.  This gives us what we desired,
that $A+B$, the right-hand error, is bigger than $C$, the left-hand error.  
Thus left-hand endpoints give a better approximation.\\[.5cm]

\noindent{\bf 4.} (14 points) Determine by comparison whether the
following improper integrals are convergent or divergent.\\

\noindent a) $\displaystyle\int_{1}^{\infty}\dfrac{dx}{x^{2}+\sqrt{x}}$\\
Here the first obvious choice of what to compare the integrand to is
$1/x^2$.  Notice that for $x \geq 1$ (which is true over the entire
integral), we have $x^2 + \sqrt(x) > x^2$.  Hence, $1/(x^2+\sqrt(x)) <
1/x^2$.  Thus by the comparison theorem, since we know that the larger
integral $\int_1^\infty dx / (x^2)$ converges, then so
does the smaller $\int_1^\infty dx / (x^2 + \sqrt(x))$. (We are allowed to
use the fact that $\int_1^\infty dx / (x^2)$ converges, without proof).


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