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- Computing net or total distance traveled by a moving object.
- Computing average values, e.g. centroids and centers of mass, moments of inertia, and averages of probability distributions.

Of course, when we use a definite integral to compute an area or a volume, we are adding up area or volume. So you might ask why make a distinction? The answer is that the notion of an integral as a means of computing an area or volume is much more concrete and is easier to understand.

You have learned in class that the definite integral is actually defined as a (complicated) limit of sums, so it makes sense that the integral should be thought of as a sum. You have also learned in class that the indefinite integral, or anti-derivative, can be used to evaluate definite integrals. Students often concentrate on techniques for evaluating integrals, and ingnore the definition of the integral as a sum. This is a mistake, for the following reasons.

- 1.
- Many functions don't have anti-deriviatives that can be written down as formulas. Definite integrals of such function must be done using numerical techniques, which always depend on the definition of the integral as a sum.
- 2.
- In many applications of the integral in engineering and science, you aren't given the function which is to be integrated and must derive it. The derivation always depends on the definition of the integral as a sum. You will see examples of this later on in the course.

The command to use is shown below.

> int(x^2,x=0..4);

Notice that Maple gives an exact answer, as a fraction. If you want a decimal approximation to an integral, you just put an

> evalf(int(x^2,x=0..4));

To compute an indefinite integral with Maple, you just leave out the range for the limits of integration, as shown below.

> int(x^2,x);

Note that Maple does not include a constant of integration.

You can also use the Maple `int` command with functions or
expressions you have defined in Maple.
For
example, suppose you wanted to find area under the curve of the
function on the
interval . Then you can define this function in Maple with
the command

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

and then use this definition as shown below.

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

You can also simply give the expression corresponding to *f*(*x*) a
label in Maple, and then use that label in subsequent commands as
shown below. However, notice the difference between the two
methods. You are urged you to choose one or the other, so you don't
mix the syntax up.

> p := x*sin(x);

> int(p,x=0..Pi);

> 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

> int(2-x^2,x=-5..1)+int(x,x=1..5);

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);

- 1.
- Use Maple to compute the following definite integrals.
- (a)
- (b)
- Explain why your answer to this part is exactly twice your answer to the first part of this exercise.

- 2.
- It is easy to see that if
*f*(*x*)=0, then However, it is not true that having implies*g*(*x*) is identically zero on the interval [*a*,*b*]. Explain why this is so, and come up with a function and an interval of your own that illustrates this. That is, find a function*g*(*x*), and an interval [*a*,*b*] such that but it is not true that*g*(*x*) is identically zero on the interval [*a*,*b*]. Include the integral and a plot of your function in your worksheet. - 3.
- Use Maple to find the area of the region bounded above by the
curve
*y*=*x*+2 and below by the curve*y*=*x*-3^{2}*x*-3. Include a plot of the region. - 4.
- Consider the function Compute the following integral.
- 5.
- Suppose that the velocity of a particle moving in one dimension
is given by
*v*(*t*) = -32*t*+ 15

3/31/2000