Area between curves

Oct 2009
114
2
Find the area between \(\displaystyle \[f(x)=sin^5(x)\]\) and \(\displaystyle \[g(x)=sin^3(x)\]\) both with domain \(\displaystyle \[[0,\pi]\]\).

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

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

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

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

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

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

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

\(\displaystyle = 1 - \frac{1}{3} - 1 +\frac{2}{3} - \frac{1}{5} = \frac{2}{15}\)


However, the answer in the back of the book says the answer is \(\displaystyle \frac{4}{15}\). Did I do anything wrong?
 
Aug 2007
3,171
860
USA
Yes. You failed to observe that your point of intersection represents also a point of symmetry. When you changed the limits from \(\displaystyle [0,\pi]\) to \(\displaystyle [0,\pi/2]\), you should have multiplied the entire expression by 2.