Question

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

Answer 1

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.)

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I hope that this was helpful