Question

Find the average value of $f\left(x\right)=\left(x^3-3x^2+2x\right)\text{on}\left[0,\frac{1}{2}\right]$.

Answer 1

$f\left(x\right)=x^3-3x^2+2x$

at $x=0$

$f\left(x\right)=0$

at $x=\frac{1}{2}$

$\therefore f\left(x\right)={\left(\frac{1}{2}\right)}^3-3{\left(\frac{1}{2}\right)}^2+2\times \frac{1}{2}$

$=\frac{1}{8}-\frac{3}{4}+1=\frac{1-6+8}{8}$

$=\frac{3}{8}$

$\therefore$ average value of $f\left(x\right)$ between $[0,\frac{1}{2}]$

$f\left(x\right)=\frac{f\left(0\right)+f\left(\frac{1}{2}\right)}{2}=\frac{\frac{3}{8}}{2}$

$=\frac{3}{16}$