First of all, there is an extreme value when x=t, that is to say, when x=t, f'(x) is 0, and f'(x)= 12x? +8ax+b, so 12t? +8at+b=0。

And when x=t, the extreme value is 0, that is, 4t? +4at？ +bt=0

A=-2t b=4t can be expressed by two formulas simultaneously?

So f(x)=4x? -8tx？ +4t？ x=4x(x？ -2tx+t？ )=4x(x-t)？

So f(x)=0 has two solutions, 0 and t.

The figure enclosed by f(x) and x axis is a block from 0 to t.

Then the area t squared /3 can be obtained by integrating f(x) from 0 to t (definite integral).