# How do you find the area under the graph of f(x)=sin(x) on the interval [-pi,pi] ?

Sep 17, 2014

For this question to have to set up a definite integral.

We see that this integral is bounded from $- \pi$ to $\pi$ inclusive.

Also remember that after we integrate we have to subtract the result of the lower bound, $- \pi$ from that of the higher bound, $\pi$.

Be careful with all of those double negatives which resolve to positives.

Lastly, remember from the unit circle that both $\cos \left(- \pi\right)$ and $\cos \left(\pi\right)$ resolve to -1.

${\int}_{- \pi}^{\pi} \sin \left(x\right) \mathrm{dx}$

$= {\left[- \cos \left(x\right)\right]}_{-} {\pi}^{\pi} = - \cos \left(\pi\right) - \left(- \cos \left(- \pi\right)\right)$

$= - \left(- 1\right) - \left(- \left(- 1\right)\right) = 1 - 1 = 0$