1. ## Area between curves

Find the area between $$f(x)=sin^5(x)$$ and $$g(x)=sin^3(x)$$ both with domain $$[0,\pi]$$.

On $$[0,\pi]$$ the point of intersection is 0 and $$\frac{\pi}{2}$$. So,

$\[\int_{0}^{\frac{\pi}{2}}[sin^3(x)-sin^5(x)]dx$

$=\int_{0}^{\frac{\pi}{2}}[(sin^2x)sin(x)-(sin^2x)^2sin(x)]\ dx$

$=\int_{0}^{\frac{\pi}{2}}[(1-cos^2x)sin(x)-(1-cos^2x)^2sin(x)]\ dx$

$u=cos(x); \ du=-sin(x)$

$\int_{0}^{1}[(1-u^2)-(1-u^2)^2]\ du$

$= u - \frac{1}{3}u^3 - u + \frac{2}{3}u^3-\frac{1}{5}u^5\biggr|_{0}^{1}$

$= 1 - \frac{1}{3} - 1 +\frac{2}{3} - \frac{1}{5} = \frac{2}{15}$

However, the answer in the back of the book says the answer is $\frac{4}{15}$. Did I do anything wrong?

2. Yes. You failed to observe that your point of intersection represents also a point of symmetry. When you changed the limits from $[0,\pi]$ to $[0,\pi/2]$, you should have multiplied the entire expression by 2.