In this case, `b-a=1`, so to find the average value, we simply need to evaluate:

`int_0^1x(3x^2-1)^3dx`

Observing the graph, we notice that it has value near 0 for most values of *x* between 0 and 1. We **estimate** the value of the average to be somewhere around *y* = 0.5.

`int_0^1x(3x^2-1)^3dx`

Put `u=3x^2-1`, then `du = 6x dx`

So `(du)/6=x\ dx`

`int_0^1x(3x^2-1)^3dx=1/6int_(x=0)^(x=1)u^3du`

`=1/6[u^4/4]_(x=0)^(x=1)`

`=1/24[(3x^2-1)^4]_0^1`

`=1/24[(3(1)^2-1)^4-(3(0)^2-1)^4]`

`=1/24[16-1]`

`=15/24`

`=5/8`

So the average value required is `0.625` units. This is consistent with our earlier estimate.