# Example 1

Use Simpson's Rule to approximate the integral of f(*x*) = *x*^{3} on the interval [1, 2].

1/6 (f(1) + 4 f(1.5) + f(2))(2 - 1) = 3.75

The actual value of the integral is 3.75. This is exact because the error for
Simpson's rule is O(h^{4}) and we are integrating a cubic function.

# Example 2

Use Simpson's Rule to approximate the integral of
f(*x*) = *e*^{-0.1 x}
on the interval [2, 5].

1/6 (f(2) + 4 f(3.5) + f(5))(5 - 2) = 2.122006886

The actual value of the integral is 2.122000934 .

# Example 3

Suppose we can only evaluate a function at the integers (for example, when we
are periodically sampling a signal).
Use three applications of Simpson's Rule and two applications of Simpson's 3/8 Rule
to approximate the integral of response with a decaying transient
f(*x*) = cos(x) + *x* *e*^{−x} on the
interval [0, 6].

Thus, we calculate:

and

The actual integral is -7 exp(-6) + sin(6) + 1 = 0.7032332366, and therefore, three applications of Simpson's rule with intervals of width 2 appears to be more accurate than two applications of Simpson's 3/8th rule with intervals of width 3.

Copyright ©2005 by Douglas Wilhelm Harder. All rights reserved.