**Introduction to Simpson's Rule**

Simpson’s Rule takes our philosophy another step: If rectangles are good, and trapezoids better, then why not approximate by curves? In Simpson’s Rule, we approximate by *parabolas*.

We have a continuous function *f* on the interval [ *a, b* ] and we have a partition *P* = { *x* _{0} , *x* _{1} ,..., *x _{k}* } of our partition as usual. It is convenient in this technique to assume that we have an even number of intervals in the partition.

Now each rectangle, over each segment of the partition, is capped off by an arc of a parabola. Figure 8.46 shows just one such rectangle. In fact, for each pair of intervals [ *x* _{2 j −2} , *x* _{2 j −1} ], [ *x* _{2 j −1} , *x* _{2 j} ], we consider the unique parabola passing through the endpoints

Note that a parabola *y* = *Ax*^{2} + *Bx* + *C* has three undetermined coefficients, so three points as in (*) will determine *A, B, C* and pin down the parabola.

In fact (pictorially) the difference between the parabola and the graph of *f* is so small that the error is almost indiscernible. This should therefore give rise to a startling accurate approximation, and it does.

Summing up the areas under all the approximating parabolas (we shall not perform the calculations) gives the following approximation to the integral:

If it is known that the fourth derivative *f* ^{(iv)} ( *x* ) satisfies | *f* ^{(iv)} ( *x* )| ≤ *M* on [ *a, b* ], then the error resulting from Simpson’s method does not exceed

**Examples**

**Example 1**

Use Simpson’s Rule to calculate to two decimal places of accuracy.

**Solution 1**

If we set *f* ( *x* ) = *e* ^{− x2} then it is easy to calculate that

*f* ^{(iv)} ( *x* ) = *e* ^{− x2} · [12 − 72 *x*^{2} + 32 *x*^{4} ].

Thus | *f* ( *x* )| ≤ 12 = *M* .

In order to achieve the desired degree of accuracy, we require that

Simple manipulation yields

This condition is satisfied when *k* = 2.

Thus our job is easy. We take the partition *P* = {0, 1/2, 1}. The sum arising from Simpson’s Rule is then

Comparing with the “exact value” 0.746824 for the integral that we noted in Example 8.29, we find that we have achieved two decimal places of accuracy.

It is interesting to note that if we had chosen a partition with *k* = 6, as we did in Example 8.29, then Simpson’s Rule would have guaranteed an accuracy of

or nearly four decimal places of accuracy.

**Example 2**

Estimate the integral

using Simpson’s Rule with a partition having four intervals. What degree of accuracy does this represent?

**Solution 2**

Of course this example is parallel to Example 8.30, and you should compare the two examples. Our function is *f* ( *x* ) = 1/(1 + *x* ^{2} ) and our partition is *P* = {0, 1/4, 2/4, 3/4, 1}. The sum from Simpson’s Rule is

Comparing with Example 8.30, we see that this answer is accurate to four decimal places. We invite the reader to do the necessary calculation with the Simpson’s Rule error to term to confirm that we could have predicted this degree of accuracy.

**You Try It** : Estimate the integral

using both the Trapezoid Rule and Simpson’s Rule with a partition having six points. Use the error term estimate to state what the accuracy prediction of each of your calculations is. If the software or is available to you, check the answers you have obtained against those provided by these computer algebra systems.

Find practice problems and solutions for these concepts at: Applications of the Integral Practice Test.

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