Question

How do you use a Power Series to estimate the integral `int_0^0.01e^(x^2)dx` ?

Answer 1

Since

`e^x=1+x+x^2/{2!}+cdots`,

`e^{x^2}=1+x^2+x^4/{2!}+cdots`

`int_0^{0.01}e^{x^2}dx=int_0^{0.01} (1+x^2+x^4/{2!}+cdots)dx`

`=[x+x^3/3+x^5/{10}+cdots]_0^{0.01}`

`=0.01+(0.01)^3/3+(0.01)^5/10+cdots`

`approx 0.01`

I hope that this was helpful.