EX 4.10 expected value of EX 4.9

Expected value in EX 4.9 amount of gravel.

The pdf of weekly gravel sales $X$ was

$$f\left(x\right)=\{\begin{array}{cc}\frac{3}{2}\left(1-x^2\right)& 0\le x\le 1\\ 0& \text{ otherwise }\end{array}$$ $$\begin{array}{rl}E\left(X\right)& ={\int }_{-\infty }^{\infty }x\cdot f\left(x\right)dx\\ & ={\int }_0^{1}x\cdot \frac{3}{2}\left(1-x^2\right)dx\\ & =\frac{3}{2}{\int }_0^1\left(x-x^3\right)dx\\ & =\frac{3}{2}\left(\frac{x^2}{2}-\frac{x^4}{4}\right){|}_{x=0}^{x=1}\\ & =\frac{3}{8}\end{array}$$