Find the area of the region bounded by the graph of function $f(x) = \sin{x}$ and the $x$ axis on the interval $\left[-\frac{\pi}{3},\frac{2\pi}{3}\right]$.
My thought process is to just to integrate $\sin{x}$ which is –$\cos{x}+c $ and then plug the values on the interval starting with $-\frac{\pi}{3}$ which gives $-\frac{1}{2}$ and then $\frac{2\pi}{3}$ which give $\frac{1}{2}$. Using the form F(a)-F(b) should give 2. Except I end up getting $-\frac{1}{2}-\frac{1}{2} = -1 $ What am I doing wrong? I think I have a major mis understanding with how problems like this work.
Best Answer
$\sin x$ is not positive on $[-\pi/3,0]$, which means its contribution to the raw integral is negative. But an area can only be positive.
You need to compute the integral over $[-\pi/3,0]$ and $[0,2\pi/3]$ separately, then negate the first component – to account for the negativity of $\sin$ there – before adding up.