$($a$)$ Evaluate the integral below as an infinite series.

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{e}^{-x^2}dx$

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

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

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 $x$ wi all values of $x,$

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

$=1-\frac{x^2}{1!}\frac{x^6}{3!}$ dots

Now we integrate term by term.

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{e}^{-x^2}dx={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(1-\frac{x^2}{1!}\frac{x^6}{3!}$ dots $\frac{(-1{)}^n}{n!})dx$