Question

Find the average of f(x)=x^3-x over [0,2]

Answer 1

9514 1404 393

Answer:

  1

Step-by-step explanation:

The average value of the function is the integral over the interval, divided by the width of the interval.

  $\displaystyle\overline{y}=\frac{1}{2-0}\int_0^2{f(x)}\,dx=\left.\frac{1}{2}\left(\frac{x^4}{4}-\frac{x^2}{2}\right)\right|_0^2=\frac{1}{2}\left(\frac{2^4}{4}-\frac{2^2}{2}\right)=\frac{4-2}{2}\\\\\boxed{\overline{y}=1}$