definite

cant say only this much was given....

Nice prob, at least its different from the run of the mill stuff.

First consider \int_0^{z^2} f(x) \ dx

With the substitution x=y^2, we have that this integral is equivalent to \int_0^z 2y f(y^2) \ dy

So consider F(t) = \int_0^t 2y f(y^2) \ dy

Remember that F(2) = \int_0^4 f(x) \ dx which is what we are seeking.

From Rolle's Theorem, \exists \ \alpha \in (0,1) such that F'(\alpha) = \frac{F(1) - F(0)}{1-0} = F(1) \ (\because \ F(0) = 0)

Similarly \beta \in (1,2) satisfying

F'(\beta) = \frac{F(2) - F(1)}{2-1} = F(2) - F(1)

Adding we obtain F'(\alpha) + F'(\beta) = F(2)

But from Leibnitz Theorem F'(\alpha) = 2 \alpha f(\alpha^2) and F'(\beta) = 2 \beta f(\beta^2)

Hence

\int_0^4 f(x) dx =F(2) = F'(\alpha) + F'(\beta) = 2 \alpha f(\alpha^2) + 2 \beta f(\beta^2)