Chapter 7.8, Problem 68ES

A convergent improper integral over an infinite interval can be approximated by first replacing the infinite limit(s) of integration by finite limit(s), then using a numerical integration technique, such as Simpson’s rule, to approximate the integral with finite limit(s). This technique is illustrated in these exercises.

Suppose that the integral in Exercise 66 is approximated by first writing it as

${\displaystyle {\int }_0^{+\infty }{e}^{-x^2}dx={\displaystyle {\int }_0^k{e}^{-x^2}}dx+{\displaystyle {\int }_k^{+\infty }{e}^{-x^2}dx}}$

then dropping the second term, and then applying Simpson’s rule to the integral

${\displaystyle {\int }_0^k{e}^{-x^2}dx}$

The resulting approximation has two sources of error; the error Simpson’s rule and the error

$E={\displaystyle {\int }_k^{+\infty }{e}^{-x^2}}dx$

that result from discarding the second term. We call E the truncation error.

(a) Approximate the integral in Exercise 66 by applying Simpson’s rule with $n=10$ subdivisions to the integral

${\displaystyle {\int }_0^3{e}^{-x^2}dx}$

Round your answer to four decimal places and compare it to $\frac{1}{2}\sqrt{\pi }$ rounded to four decimal places.

(b) Use the result that you obtained in Exercise 52 and the fact that $e^{-x^2}\le \frac{1}{3}x{e}^{-x^2}\text{for }x\ge 3$ to show that the truncation error for the approximation in part (a) satisfies $0<E<2.1\times {10}^{-5}.$