The question is:
Find the area of the region between y=sin(x) and y=x
for 0 ≤ x ≤ pi/2.
It's going to be a definite integral with bounds from 0 to pi/2, right? But what is the equation and how do I get it?
Thank you!
$\displaystyle x$ is greater than $\displaystyle \sin(x)$ for all $\displaystyle x>0$, so you want to find the area under $\displaystyle y=x$ and subtract the area under $\displaystyle y=\sin(x)$. The value you want is:
$\displaystyle \int_0^{\pi/2}x\,dx-\int_0^{\pi/2}\sin x\,dx$
This can be rewritten as a single integral: $\displaystyle \int_0^{\pi/2}(x-\sin x)\,dx$