7.8. Approximate Integration

Transcription

1 7.8. Approimate Integration Recall that when we originally defined the definite integral of a function interval, that we did so with the limit of a Riemann sum: on an where is the number of intervals that is divided into,,, and is between and, for all. If is continuous for all in, then the limit eists and is defined to be. An approimation to this integral can be found by choosing a large value of and appropriate values for. If we choose to be, for all, then the Left-Hand Rule is in use. If we choose to be, for all, then we are using the Right-Hand Rule. If the value of is large, then this sum can best be evaluated using a graphing calculator, or by Maple itself. Maple can not only evaluate these sums for us, it can also provide a graphical representation. Before we get to that, we provide a short description of how Maple treats functions. If you write f() = ^, then Maple will treat this like an equation. If you write f() := ^, then this assigns the value of to, but Maple cannot evaluate. The proper way to assign a function is to use the arrow notation, which indicates that has been transformed into : f := -> ^; f(3); 9 () () Before we continue, we need to load the student package: with (Student[Calculus]); with (plots);

2 (3) Now, let's define a more interesting function for : f := -> ep(sin(*)); (4) We want to approimate the integral of this function from to. The Maple command to approimate an integral is ApproimateInt. It needs the function to integrate, a range for, the approimation technique, and the number of intervals to divide the interval into. If you want to look at the graph of and see what is actually being calculated, you can add the option output=plot. (If you leave it out, Maple will try to find an eact value for the answer.) ApproimateInt(f(), =..Pi, partition=8, method=left, output= plot);

3 An Approimation of the Integral of f() = ep(sin(*)) on the Interval [, Pi] Using a Left-endpoint Riemann Sum Area: Partitions: 8 f() Not surprisingly, if we want to use the Right-hand rule, we change the command to method=right. ApproimateInt(f(), =..Pi, partition=8, method=right, output= plot);

4 An Approimation of the Integral of f() = ep(sin(*)) on the Interval [, Pi] Using a Right-endpoint Riemann Sum Area: Partitions: 8 f() We could also choose, in which case we use method=midpoint. ApproimateInt(f(), =..Pi, partition=8, method=midpoint, output=plot);

5 An Approimation of the Integral of f() = ep(sin(*)) on the Interval [, Pi] Using a Midpoint Riemann Sum Area: Partitions: 8 f() The problem, though, is that there is no way to determine how accurate the estimates are, if the Left-Hand Rule or the Right-Hand Rule is used. The material in Section 7.7 deals with approimating integrals, and also providing bounds on the error. For instance, to evaluate to 6 decimal places, all that is needed is to choose an big enough, where determining what "big enough" is, is not too difficult. The two main approimation techniques are the Trapezoid Rule and Simpson's Rule. The Trapezoid Rule approimates using trapezoids, and the resulting formula is: To see where these trapezoids come from, use method=trapezoid. ApproimateInt(f(), =..Pi, partition=8, method=trapezoid,

6 output=plot); An Approimation of the Integral of f() = ep(sin(*)) on the Interval [, Pi] Using the Trapezoid Rule Area: Partitions: 8 f() Simpson's Rule requires that be even, and does the following: It takes the points,, and, and finds the parabola that all three of these points lie on. Then it approimates the area under the curve and between and by the area under the curve and between and. It then continues with the points,, and and does the same thing, etc. The formula which is obtained from this is: To use Simpson's Rule, we use the option method=simpson, and to see the parabolas which approimate can be seen by also including output=plot. ApproimateInt(f(), =..Pi, partition=8, method=simpson, output=plot);

7 An Approimation of the Integral of f() = ep(sin(*)) on the Interval [, Pi] Using Simpson's Rule Area: Partitions: 8 f() Now, I will go around the room and give you a value of. Substitute this value for N in the epression below. I will ask for the approimations that you get. ApproimateInt(f(), =..Pi, partition=n, method=simpson, output=plot); The approimations seem to converge, and we could probably guess the actual value, but our guess would only be a guess. There are actual error bounds for the Midpoint Rule, the Trapezoid Rule, and Simpson's Rule: If for all between and, then the errors in approimation by using the Trapezoid Rule and the Midpoint Rule are: and If for all between and, then the error in approimation by using

8 Simpson's Rule is: We will use these error bounds to approimate our integral to 6 places by using the Trapezoid Rule and Simpson's Rule, by finding bounds on the second and fourth derivatives of, then choosing big enough so thatthe error is at most, which will guarantee that we have 6 decimal places accurate. I will show you how to do this below. First, we will graph the second derivative of and see where its second derivative is the largest in absolute value. plot ((D@@)(f)(), =..Pi); 4 From the graph, it appears that for all between and, so we can let Since we want to have an error of at most, we want to have

9 . We can use Maple to find for us: solve (.8*(Pi-)^3/(4 * n^) = ^(-6), n); (5) This means we need many trapezoids:. to be at least Now we use the Trapezoid Rule with this ApproimateInt(f(), =..Pi, partition=3736, method=trapezoid, output=plot); An Approimation of the Integral of f() = ep(sin(*)) on the Interval [, Pi] Using the Trapezoid Rule Area: Partitions: 3736 f() So our integral is approimately Now we use Simpson's Rule. First, we need to find how large the fourth derivative of gets in absolute value. Again, we will plot the function: plot ((D@@4)(f)(), =..Pi);

10 5 5 It appears that we can let or so. Now we want to find out when solve ((Pi-)^5 / n^4 = ^(-6), n); evalf (%); (6) (7) using Simpson's Rule, n must be even; so we will let n=34 and get the approimation: ApproimateInt(f(), =..Pi, partition=34, method=simpson, output=plot);

11 An Approimation of the Integral of f() = ep(sin(*)) on the Interval [, Pi] Using Simpson's Rule Area: Partitions: 34 f() which matches what we got earlier.