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

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Answer 1 · 2014-09-21T21:52:42

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.