$($b$)$ Evaluate the integral below correct to within an error of $0\mathrm{.}001.$

${\displaystyle \underset{0}{\overset{1}{}}}{e}^{-x^2}$

Solution

$($a$)$ First we find the Maclaurin series for $f(x)=e^{-x^2}.$ Although it's possible to use the direct method, let's find it simply by replacing $x$ with $-x^2$ in the series for $e^x$ given in Table $1$ in the book. Thus for all values of $x,$

$e^{-x^2}={\displaystyle \underset{n=0}{\overset{}{}}}\frac{(-x^2{)}^n}{n!}={\displaystyle \underset{n=0}{\overset{}{}}}(-1{)}^n\frac{x^{2n}}{n!}$

$=1-\frac{x^2}{1!}+\frac{x^6}{2!}+$dots

Now we integrate term by term.

This series converges for all $x$ because the original series for $e^{-x^2}$ converges for all $x.$

$($b$)$ The fundamental Theorem of Calculus gives

${\displaystyle \underset{0}{\overset{1}{}}}{e}^{-x^2}dx=[x-,+\frac{x^5}{5*2!}-\frac{x^7}{7*3!}+$dots${]}_0^1$

$=-\frac{1}{3}+,-\frac{1}{42}+\frac{1}{216}-$dots

~~$-\frac{1}{3}+,-\frac{1}{42}+\frac{1}{216}$~~

The Alternating Series Estimation Theorem shows that the error involved in this approximation is less than $0\mathrm{.}001.($b$)$ Evaluate the integral below correct to within an error of $0\mathrm{.}001.$

${\displaystyle \underset{0}{\overset{1}{}}}{e}^{-x^2}$

Solution

$($a$)$ First we find the Maclaurin series for $f(x)=e^{-x^2}.$ Although it's possible to use the direct method, let's find it simply by replacing $x$ with $-x^2$ in the series for $e^x$ given in Table $1$ in the book. Thus for all values of $x,$

$e^{-x^2}={\displaystyle \underset{n=0}{\overset{}{}}}\frac{(-x^2{)}^n}{n!}={\displaystyle \underset{n=0}{\overset{}{}}}(-1{)}^n\frac{x^{2n}}{n!}$

$=1-\frac{x^2}{1!}+\frac{x^6}{2!}+$dots

Now we integrate term by term.

This series converges for all $x$ because the original series for $e^{-x^2}$ converges for all $x.$

$($b$)$ The fundamental Theorem of Calculus gives

${\displaystyle \underset{0}{\overset{1}{}}}{e}^{-x^2}dx=[x-,+\frac{x^5}{5*2!}-\frac{x^7}{7*3!}+$dots${]}_0^1$

$=-\frac{1}{3}+,-\frac{1}{42}+\frac{1}{216}-$dots

~~$-\frac{1}{3}+,-\frac{1}{42}+\frac{1}{216}$~~

The Alternating Series Estimation Theorem shows that the error involved in this approximation is less than $0\mathrm{.}001.$