Q. 34 - Calculus

Question

ind the Maclaurin series for $e^{-x^2}$ , and use it to approximate ${\int }_0^1{e}^{-x^2}dx$ to within 0.001 of its value. How many terms would you need to approximate the integral to within ${10}^{-6}$ of its value? 

Step-by-Step Solution

Answer

The Maclaurin series can be given as

$e^{-x^2}=1-x^2+\frac{x^4}{2}-\frac{x^6}{3!}+\frac{x^8}{4!}-...$

On integrating we get,

${\int }_0^1{e}^{-x^2}dx=1-\frac{1}{3}+\frac{1}{10}-\frac{1}{42}+\frac{1}{216}-...$

And we require first 6 terms to get corrected values upto 0.0001

Step 1: Given information

We are given the power series of $e^x$

Step 2: Find the power series of e - x 2

We know the Maclaurin series of $e^x$ which can be given as

$$\begin{gathered} e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\frac{x^4}{4!}+... \\ substitute x=-x^2 \\ {e}^{-x^2}=1-x^2+\frac{x^4}{2}-\frac{x^6}{3!}+\frac{x^8}{4!}-... \\ Now integrating \\ {\int }_0^1{e}^{-x^2}dx={\int }_0^{1}1-x^2+\frac{x^4}{2}-\frac{x^6}{3!}+\frac{x^8}{4!}-...dx \\ {\int }_0^1{e}^{-x^2}dx={{[x-\frac{x^3}{3}+\frac{x^5}{10}-\frac{x^7}{42}+\frac{x^9}{216}-......]}^1}_0 \\ {\int }_0^1{e}^{-x^2}dx=1-\frac{1}{3}+\frac{1}{10}-\frac{1}{42}+\frac{1}{216}-... \end{gathered}$$

We reuqire first 6 terms to get values corrected upto 0.0001 of its value