Chapter 5.10, Problem 60E

To determine

To estimate the value of ${\displaystyle \underset{0}{\overset{{}^{\infty }}{\int }}{e}^{-x^2}dx}$ by splitting the expression

  ${\displaystyle \underset{0}{\overset{{}^{\infty }}{\int }}{e}^{-x^2}dx}={\displaystyle \underset{0}{\overset{{}^4}{\int }}{e}^{-x^2}dx}+{\displaystyle \underset{4}{\overset{{}^{\infty }}{\int }}{e}^{-x^2}dx}$ and also, find the value of ${\displaystyle \underset{4}{\overset{{}^{\infty }}{\int }}{e}^{-4x}dx}$

Explanation of Solution

Given information: ${\displaystyle \underset{0}{\overset{{}^{\infty }}{\int }}{e}^{-x^2}dx}={\displaystyle \underset{0}{\overset{{}^4}{\int }}{e}^{-x^2}dx}+{\displaystyle \underset{4}{\overset{{}^{\infty }}{\int }}{e}^{-x^2}dx}$

The value of first integral is evaluated by using simpson’s rule by taking $n=8$ .

Concept used:

The expression ${\displaystyle \underset{0}{\overset{{}^{\infty }}{\int }}{e}^{-x^2}dx}=\frac{\sqrt{\pi }}{2}=0.886226$

Formula used:

1. If $f(x)$ is continuous on $[a,b]$ , then by simpson’s rule:

     ${\displaystyle \underset{a}{\overset{{}^b}{\int }}y(x)dx}=\frac{h}{3}\left[(y_o+y_n)+4(y_1+y_3+\mathrm{....}+y_{n-1})+2(y_2+y_4+\mathrm{....}+y_{n-2})\right];h=\frac{b-a}{n}$

   Proof:

   Here, $a=0,b=4\text{and}n=8$

   The value of $h$ is $0.5$ .

   The values of $y(x)$ for different values of $x$ is given in the following table as:

   $x$	0	0.5	1	1.5	2	2.5	3	3.5	4
   $y(x)=e^{-x^2}$	1	0.7788	0.3678	0.1053	0.0183	0.001930	0.00012	0.0000047	0.00000011
   	$y_o$	$y_1$	$y_2$	$y_3$	$y_4$	$y_5$	$y_6$	$y_7$	$y_8$

   Using Simpson’s rule, the integral becomes as:

     ${\displaystyle \underset{0}{\overset{{}^4}{\int }}{e}^{-x^2}dx}=\frac{0.5}{3}\left[\begin{array}{l}(1+0.00000011)+4(0.7788+0.1053+0.00193+0.0000047)+\\ 2(0.3678+0.0183+0.00012)\end{array}\right]$

     ${\displaystyle \underset{0}{\overset{{}^4}{\int }}{e}^{-x^2}dx}=0.166\left[1.00000011+3.5441388+0.77244\right]=0.8825520991$

   Which is the required value of first integral.

   According to the given statement:

     ${\displaystyle \underset{0}{\overset{{}^{\infty }}{\int }}{e}^{-x^2}dx}={\displaystyle \underset{0}{\overset{{}^4}{\int }}{e}^{-x^2}dx}+{\displaystyle \underset{4}{\overset{{}^{\infty }}{\int }}{e}^{-x^2}dx}$

     $\Rightarrow 0.886226=0.8825520991+{\displaystyle \underset{4}{\overset{{}^{\infty }}{\int }}{e}^{-x^2}dx}$

     $\Rightarrow {\displaystyle \underset{4}{\overset{{}^{\infty }}{\int }}{e}^{-x^2}dx}=0.886226-0.8825520991=0.0036$

   which is the required vale of the second integral.

     ${\displaystyle \underset{4}{\overset{{}^{\infty }}{\int }}{e}^{-4x}dx}={\left[\frac{e^{-4x}}{-4}\right]}_4^{\infty }=0+\frac{e^{-16}}{4}=\text{0}\text{.0000000288}$