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MATH 180\hfill SPRING, 1996

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\centerline{ASSIGNMENT \#18}

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3.3, \#2. The units of $\int_a^b f(t)\,dt$ are
$(\hbox{meters/second}^2) \times \hbox{seconds}=\hbox{meters/second}$.


\bigskip\noindent
3.3, \#5.
The area is given by $\int_0^\pi \sin x\,dx$.
We estimate this integral by taking averages of left and right
sums for large $n$.  For example, when $n=200$, the left and
right sums are $1.992$ and $2.008$.
The area under one  arch of $\sin x$ is then equal to 2.
Note that $\sin x$ is not monotonic on $[0,\pi]$, so
it does not follow from general principles that the
integral is between the left and right sums for given $n$.
However, $\sin x$ is increasing on $[0,\pi/2]$ and
decreasing on $[\pi/2,\pi]$, so the integrals
$\int_0^{\pi/2}\sin x\,dx$ and $\int_{\pi/2}^\pi\sin x\,dx$
can be estimated to any prescribed degree of accurancy.  Note also
that the arch is symmetric about the line $x=\pi/2$ so that
$\int_0^{\pi}\sin x\,dx=2\int_0^{\pi/2}\sin x\,dx$.

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3.3, \#16.
a) The average of a function over an interval from $a$ to $b$ 
is defined to be $1/(b-a)
\int^b_af(x)\,dx.$ If we graph $f(x)$, we note that
it is the unit circle restricted to the first quadrant.
The area under the graph is $1/4$ of the area of the 
unit circle, thus, $\pi/4$, or approximately 0.79.

b) The area is certainly bigger than 1/2.  Draw a line segment connecting
(0,1) to (1,0).  The area under the line segment is 1/2, and the 
area under $f(x)$ is greater than 1/2.

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3.4, \#1.
$F(b)=F(0)+\int^b_0f(x)\,dx=\int^b_0f(x)\,dx$ by the
Fundamental Theorem since $F(0)=0$.
The integral $\int_0^bf(x)\,dx$ can be interpreted as the sum of
{$\pm$}-areas
of regions bounded by the graph of $f(x)$, the $x$-axis, and the
lines $y=i$ for $i=0, 1,\ldots,b$.
We find
$$\vbox{
\halign{\hfil\qquad#&\hfil\qquad#&\hfil\qquad#&\hfil\qquad#&
     \hfil\qquad#&\hfil\qquad#&\quad\hfil#\quad\cr
  $b$& 1 & $2$ & 3 & 4 & 5 & 6\cr
  $F(b)$& 1 & $1.5$ & 1 & 0 & $-1$ & $-1.5$\cr}}$$

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3.4, \#2. 
a) $\int^0_{-3}f(x)\,dx=-2$.

b) $\int^4_{-3}f(x)\,dx= \int^0_{-3}f(x)\,dx+
\int^3_0f(x)\,dx.+
\int^4_3f(x)\,dx=  -2 + 2 -A/2 = -A/2$.

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3.4, \#4.  $F$ is increasing when $F'$ is positive, decreasing
when $F'$ is negative.  So $F$ is increasing from 0 to about 1.77
and decreasing from about 1.77 to 2.5.






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