\def\wsn{25}
\input worksheet.tex


\itemitem{1.\hskip12pt a)}  State the Fundamental Theorem of Calculus.

\medskip
\itemitem{b)}  What is wrong with the following argument?
$$
\int_{-1}^2{dt\over t^2}=\left.{-1\over t}\right|_{-1}^2
=-{1\over2}-\left(-{1\over-1}\right)=-{1\over2}-1=-{3\over2}
$$

\bigskip
\item{2.}  Find all continuous functions $f(x)$ satisfying
$$
\int_0^xf(t)\,dt=\left[f(x)\right]^2+C
$$
(Hint: Differentiate both sides with respect to $x$.)


\bigskip
\item{3.}  Suppose there is a differentiable function $y=f(x)$ on the
interval $[a,b]$.  Let $P_1=(x_k,f(x_k))$ and
$P_2=(x_{k+1},f(x_{k+1}))$ be points on the graph of this function.
Suppose also that $x_{k+1}-x_k=\Delta x$ is a very small quantity.

\medskip
\itemitem{a)}  Show that the length of the line segment
$P_1P_2$ is approximately equal to $\Delta x\sqrt{1+(f'(x))^2}$.

\medskip
\itemitem{b)}  Write a definite integral which expresses the length of
the graph of $f$ on the interval $[a,b]$.

\bigskip
\item{4.}  Find the length of the curve $y=(1-x^{2/3})^{3/2}$ from
$x=1/8$ to $x=1$.

\bigskip
\item{5.}  Graph $f(x)=1/x^2$ on your calculator for $-5\leq x<0$.
Notice that as the graph is drawn, it appears to speed up as it
approaches $x=0$.  This is because (roughly) your calculator plots the
graph by evaluating the function at various values of $x$ evenly
spaced throughout the domain and ``connecting the dots'' as it goes.
For $x$ values close to 0, the values of the function are farther
apart than they are for values of $x$ near $-5$.  Let's figure out how
fast things appear to be moving on your calculator.

\medskip
\itemitem{a)}  Express as an integral the length of the curve from
$(-5,f(-5))$ to $(b,f(b))$.

\medskip
\itemitem{b)}  Suppose that your calculator can plot this graph in 1.5
seconds (You may time your own calculator and change this value if it
does not seem accurate for your calculator.)   Let's assume that the
plotting moves left to right (along the $x$-axis) at a constant rate.
How fast is the length of the curve you see on your calculator
increasing when $(-3,f(-3))$ is being plotted?  when $(-2,f(-2))$ is
being plotted?  when $(-.5,f(-.5))$ is being plotted?


\bigskip
\itemitem{6.\hskip12pt a)}  Sketch the graph of $f(x)=4x-x^2$ on the
interval $[0,4]$.  Estimate a value which you think is roughly the
average height of $f$ over this interval. 

\medskip
\itemitem{b)}  Partition the interval $[0,4]$ into five subintervals
(not all the same size).  Use an actual value of $f$ on each
subinterval to compute an estimate for the average value of $f$ on
$[0,4]$.

\smallskip
\itemitem{}  (Note:  You must take into account that not all the
intervals are the same size, so you must weight each value
appropriately.)

\medskip
\itemitem{c)}  Write down the actual average $A$ as a limit of sums.
Use this to show that
$$
A={1\over4-0}\int_0^44x-x^2\,dx.
$$

\eject
\item{7.}    Average Velocity:  Two methods.
\item{}  Suppose an object travels along a straight wire such that (i)
it starts at one end of the wire, and (ii) the velocity of the object
$t$ seconds later is
$$
v(t)={24\over(t+1)^2}-4
$$

\medskip
\itemitem{a)}  Where is the object at time $t=1$? at $t=4$?  (How did
you get your answers?  Could you have done it another way?) 

\medskip
\itemitem{b)}  Use your answer above to determine the average velocity
of the object over that time. 

\medskip
\itemitem{c)}  Now calculate the average velocity with a definite
integral. 

\medskip
\itemitem{d)}  How do your methods compare?  Why do they give you the
same answer?

\medskip
\itemitem{e)}  Does the object ever return to the end where it
started?  If so, when?




\bye