Question

Find the average value of f(x) = 7x^2(x^3+1)6 over the interval [0, 2].

Answer 1

The average value is

`\displaystyle\frac1{2-0}\int_0^27x^2(x^3+1)^6\,\mathrm dx`

Let
`u=x^3+1`, so that
`\mathrm du=3x^2\,\mathrm dx`:

`\displaystyle\frac12\int_1^9\frac73u^6\,\mathrm du=\frac16u^7\bigg|_1^9=\frac{9^7-1}6=\frac{2,391,484}3`