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

1 Answer
Oct 21, 2014

Since

#cosx=1-x^2/{2!}+x^4/{4!}-cdots#,

#cos(x^2)=1-x^4/{2!}+x^8/{4!}-cdots#.

So,

#int_0^{0.01} cos(x^2)dx#

#=int_0^{0.01}(1-x^4/{2!}+x^8/{4!}-cdots)dx#

#=[x-x^5/{5cdot2!}+x^9/{9cdot4!}-cdots]_0^{0.01}#

#=0.01-(0.05)^5/{5cdot2!}+(0.01)^9/{9cdot4!}-cdots#

#approx 0.01#

(Note that from the second term on, terms are negligible.)


I hope that this was helpful