As we stated above, definite integrals are often used to find sums. As
an example, suppose that *v*(*t*) represents the velocity of an object
that moves in one dimension. At any instant of *t*, the value of
*v*(*t*) can be positive, negative, or zero. If it is zero, the object
is not moving at that instant. If it is not zero, then the sign of
*v*(*t*) determines the direction of motion.

Now, suppose we were given a function *v*(*t*) for ,
and we wanted to find the net distance traveled by the object over
this time interval. If *v*(*t*) is constant, then this is easy to
determine: distance is velocity times elapsed time. If the velocity is
not constant, then it is more difficult to find the answer. The idea
of using the integral is based on breaking the elapsed time up into
subintervals. We approximate the distance traveled over each
subinterval by picking an average velocity over that subinterval and
using our formula distance equals velocity times elapsed time. Adding
up the approximations for each subinterval gives an approximation to
the net distance traveled. Intuitively, the more subintervals we use,
the better the approximation will be. Note also that how we pick the
average velocity over each subinterval can affect the accuracy of our
approximation.

For example, suppose *v*(*t*)=3-*t* for and we want the
approximate the net distance traveled. Suppose we use five
subintervals and choose the velocity at the left endpoint of each
subinterval for our average velocity. Then we can use the Maple
`leftsum` command to calculate an approximate distance as shown
below. A different approximation is obtained by using the velocity at
the right endpoint of each subinterval, using the Maple
`rightsum` command, and this is also shown below.

> v := t -> 3-t;

> plot(v(t),t=0..4);

> evalf(leftsum(v(t),t=0..4,5));

> evalf(rightsum(v(t),t=0..4,5));

Notice how different the two approximations are. However, if we increase the number of subintervals to 100, they are much closer.

> evalf(leftsum(v(t),t=0..4,100));

> evalf(rightsum(v(t),t=0..4,100));

These sums can also be interpreted geometrically, using the Maple
`leftbox` and `rightbox` commands.

> leftbox(v(t),t=0..4,5);

> rightbox(v(t),t=0..4,5);

The sums we computed above with the `leftbox` and
`rightbox` commands were obtained by adding up the areas of the
rectangles with the following important convention: if a rectangle
lies below the *t* axis, it appears in the sum with a minus sign. That
is, we interpret a rectangle above the axis as having positive area
and a rectangle below the axis as having negative area.

In part 1 of this lab, we learned how to use the Maple `int`
command to compute areas. The same command works to compute any
definite integral, whether it corresponds to an area or not. For
example, to compute the definite integral

you could use the following Maple command.

> int((2*x-3)^5,x=-2..4);

Sometimes you need to compute a definite integral involving a
piecewise-defined function. For example, suppose you have a function
*f*(*x*) defined as follows

and you needed to compute the definite integral

The best way to do this in Maple is to split it up into two integrals and use the appropriate formula, as shown below. How you split the integral up is determined by where the formula defining the function changes.

If a function *f* is integrable over an interval [*a*,*b*], then we
define the average value of *f*, which we'll denote as ,
on this interval to be

Note that the average value is just a number. Note furthermore that we can rearrange the definition to give

If on [*a*,*b*], then the average value has the following
geometrical interpretation: is the height of a
rectangle of width *b*-*a* such that the area of this rectangle is equal
to the area under the graph of *f* from *a* to *b*. The following
example shows you how to compute an average. The last plot command
shows the function and the top of this rectangle.

> f :=x -> x*sin(x) ;

> plot(f(x),x=0..Pi);

> f_ave := int(f(x),x=0..Pi)/Pi;

> plot(f(x),f_ave,x=0..Pi);

- Consider the function on the interval
[-1,3]. Use the command
`leftsum`to approximate the definite integralto two decimal places. Then explain why the

`leftsum`and`rightsum`commands give the same numerical values for the same number of subintervals. (Hint - use the`leftbox`and`rightbox`commands to see what is going on.) - Suppose that the velocity of a particle moving in one dimension
is given by
Find the average value of

*v*(*t*) on the following time intervals.**0,5****0,10****5,10**

- Consider the function F(x) defined for x between zero and one given by:
Compute the function for

*x*=.25,*x*=.5,*x*=.75, and*x*=1. Does this function have a derivatave when the argument ``*x*" is between zero and one. If so what is it. Please write out all your steps to get full credit for this problem. (Would you use maple to get the answer or can you figure it out without using maple?)

Tue Sep 9 11:18:41 EDT 1997