Question

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

Answer 1

Since

`cosx=sum_{n=0}^infty(-1)^n{x^{2n}}/{(2n)!}`,

we have

`cos(sqrt{x})=sum_{n=0}^infty{(-1)^n(sqrt{x})^{2n}}/{(2n)!}=sum_{n=0}^infty{(-1)^nx^{n}}/{(2n)!}`

Now, consider the integral in question.

`int_0^{0.01}cos(sqrt{x})dx=int_0^{0.01}sum_{n=0}^infty{(-1)^nx^{n}}/{(2n+1)!}dx`

by integrating term by term,

`=sum_{n=0}^infty[{(-1)^nx^{n+1}}/{(2n)!(n+1)}]_0^{0.01}`

`={(0.01)^1}/{0!cdot1}-(0.01)^2/{2!cdot2}+{(0.01)^3}/{4!cdot3}-cdots`

By adding a few terms of the above series, we can approximate the value of the original definite integral.