Question

How to use the Fundamental Theorem of Calculus to evaluate ?

Answer 1

Split the integral into two pieces.

Explanation:

`int_-pi^pi f(x) dx = int_-pi^0 f(x) dx + int_0^pi f(x) dx`

`= int_-pi^0 6x dx + int_0^pi -9sin(x) dx`

`= 3x^2]_-pi^0 + 9 cosx]_0^pi`

`= [3(0)^2-3(-pi)^2] + [9cos(pi)-9cos(0)]`

`= -3pi^2-18`

Answer 2

`int_(-pi)^pif(x)dx=3pi^2-9`

Explanation:

The Fundamental Theorem of Calculus states that `int_a^bf(x)dx=[g(x)]_a^b=g(b)-g(a)`, where `g(x)` is the antiderivative of `f(x)`, ie `g'(x)=f(x)`.

Since the given function `f(x)` is a compound function with different definitions each side of `0`, we have to split the integral into `2` parts and select the limits of integration accordingly, that is,

`int_(-pi)^pif(x)dx=int_(-pi)^0 6xdx+int_0^pi-9sinxdx`

`=6[x^2/2]_(-pi)^0+[9cosx]_0^pi`

`=6(0-pi^2/2)+9(cospi-cos0)`

`=3pi^2-9`

`=20.6088`