what is the form of $f(x)$ 
Given that area of OPB and OPA are same, could any one help me to find the the $f(x)$

The area of OPB:
$$\int_0^t (2x^2 - x^2) \ dx = \int_0^t x^2 \ dx = \left[\frac{1}{3} x^3\right]_0^t = \frac{1}{3} t^3.$$
The area of OPA:
$$\int_0^s (f(x) - 2x^2) \ dx + \int_s^t (2t^2 - 2x^2) \ dx = \int_0^s f(x) \ dx + 2 (t - s) t^2 - \int_0^t 2x^2 \ dx \\ = \int_0^s f(x) \ dx + 2 t^3 - 2 t^2 s - \frac{1}{3} t^3 = \int_0^s f(x) \ dx + \frac{4}{3} t^3 - 2 t^2 s.$$
Since they should be equal, we get the condition
$$\int_0^s f(x) \ dx + t^3 - 2 t^2 s = 0 \quad \Longleftrightarrow \quad \int_0^s f(x) \ dx = t^2(2s - t).$$
Furthermore we have the conditions $f(0) = 0$, $f(s) = 2t^2$ and $f$ is increasing at least on $(0, s)$. Several functions will satisfy all these conditions.
