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Subsections

Definite Integrals

Introduction

In Exercise 1 we look at the definite integral as the limit of Riemann sums (see Sections 5.3 to 5.5 of the text). In Exercise 2 we show how Maple can, in many cases, give us antiderivatives (see Section 5.1). In Exercise 3 we use Maple to implement the amazing theorem about definite integrals given in Section 5.6 - do you know what theorem this is? Exercise 4 gives you an antiderivative problem with two different looking answers. In Exercise 5 we take a look at the kind of approach that could be used on a problem for which there was no practical way to apply that amazing theorem from Section 5.6.

Background

For Exercise 1 ....

It is not claimed that the following procedure is efficient. Rather, it is presented as an educational demonstration in the hope that you will develop a greater appreciation of how the definite integral arises as a limit of Riemann sums.

We use a right sum with n subintervals to set up a Riemann sum. Then we simplify this sum and use summation techniques to reduce the Riemann sum to a function of n. Finally, the limit as n goes to infinity is taken.

Here are Maple commands to carry out the steps outlined above for the given g on the interval [2,5]. Notice the use of the sum command. More information about sum can be found in Maple's online help.

> with(student):
> g:=x->x^2+2*x-4;
> rightsum(g(x),x=2..5,n);
> expand();
> R:=(12/n)*sum(1,i=1..n)+(54/n^2)*sum(i,i=1..n)+(27/n^3)*sum(i^2,i=1..n);
> expand(R);
> limit(,n=infinity);

For Exercise 2 ....

In Maple the int command can be used to obtain antiderivatives. All you need to do is list the function and the variable of integration. Here is a sample of the syntax

> int((sin(x))^2,x);

For Exercise 3 ....

The int command can also be used to evaluate definite integrals. If possible, Maple will find the needed antiderivative and then apply the Fundamental Theorem of Calculus. Here is a sample of the syntax.

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

For Exercise 4 ....

A theorem mentioned on the first day of class will be useful here.

Exercise 5 ....

When a usable antiderivative cannot be found, a definite integral then has to be approximated by one of what are called numerical integration techniques. Here we will approximate a definite integral by using a midpoint rule Riemann sum (see Lab 1).

An approximation to a quantity is usually not of much use unless we have some idea of by how much the approximation might be off'' from the actual value of the quantity. We now discuss such an error bound for midpoint rule approximations.

The midpoint rule with n subintervals (designated as Mn) usually gives better accuracy than either the left endpoint rule (Ln) or the right endpoint rule (Rn). This means that, for a given n, Mn is generally closer to A, the value of the definite integral under discusion, than either Ln or Rn. In numerical analysis texts it is shown that the error, EMn, in using Mn to approximate the area under y = f(x) on [a,b] satisfies

where B is the global maximum of on [a,b]. In practice, B is often approximated by a number K that is an upper bound for B, that is B < K. For instance, if on might be taken as 4. Do you see why? For more complicated functions, Maple can be used to get a value for K that is close to B. Note that the error bound formula gives a worst case estimate, the accuracy achieved for a given number of subintervals n may be much better than the guarantee given by the formula.

In Section 10.3 (part of MA 1023) you can see two additional numerical integration techniques. Note that when, for a definite integral, you use evalf with int, Maple invokes a powerful numerical approximation routine.

The function g given in this exercise is not the kind of function for which a numerical integration technique is required since it has a nice'' antiderivative. However, it yields a good illustration of the technique under discussion and also permits us to make a useful comparison in part (d) of the exercise.

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Christine Marie Bonini
4/5/1999