Question

How do you use the Maclaurin series for sin(8x^2) to evaluate the integral a = 0 b = 0.65?

Answer 1

`int_0^0.65 sin(8x^2) dx ~~ 0.3136`

Explanation:

`sin(t) = sum_(n=0)^oo (-1)^n/((2n+1)!) t^(2n+1)`

Put `t = 8x^2` to get:

`sin(8x^2) = sum_(n=0)^oo ((-1)^n8^(2n+1))/((2n+1)!) x^(4n+2)`

Then:

`int_0^c sin(8x^2) dx = int_0^c sum_(n=0)^oo ((-1)^n8^(2n+1))/((2n+1)!) x^(4n+2) dx`

`color(white)(int_0^c sin(8x^2) dx) = [ sum_(n=0)^oo ((-1)^n8^(2n+1))/((4n+3)(2n+1)!) x^(4n+3) ]_0^c`

`color(white)(int_0^c sin(8x^2) dx) = sum_(n=0)^oo ((-1)^n8^(2n+1))/((4n+3)(2n+1)!) c^(4n+3)`

`color(white)(int_0^c sin(8x^2) dx) = 8/3 c^3 - 256/21 c^7 + 4096/165 c^11 - 131072/4725 c^15 + 1048576/53865 c^19 - 33554432/3586275 c^23 + O(c^25)`

Evaluated for `c = 0.65` gives us `0.31359 + O(c^25)`

A better approximation for the integral is `0.313618`.