# Area between curves

#### BrownianMan

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?

#### TKHunny

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.

• BrownianMan and AllanCuz