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

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Answer 1 · 2014-09-17T22:39:47

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

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

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

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

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

$\int_{- π}^{π} sin (x) d x$ $int_(-pi)^pi sin(x) dx$

$= {[- cos (x)]}_{- π}^{π} = - cos (π) - (- cos (- π))$ $=[-cos(x)]_-pi^pi=-cos(pi)-(-cos(-pi))$

$= - (- 1) - (- (- 1)) = 1 - 1 = 0$ $=-(-1)-(-(-1))=1-1=0$

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