Question

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

Answer 1

Since

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

we have

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

Now, consider the integral in question.

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

by integrating term by term,

`=sum_{n=0}^infty[{(-1)^nx^{n+3/2}}/{(2n+1)!(n+3/2)}]_0^{0.01}`

`={(0.01)^{3/2}}/{1!cdot3/2}-{(0.01)^{5/2}}/{3!cdot 5/2}+{(0.01)^{7/2}}/{5!cdot 7/2}-cdots`

`={2(0.1)^3}/{3cdot 1!}-{2(0.1)^5}/{5cdot 3!}+{2(0.1)^7}/{7cdot 5!}-cdots`

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