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 $f(x)$
 A: 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.
A: I will think of the dashed line as not being a boundary, more like an awkward hint, maybe to break up the integral, or maybe to integrate with respect to $y$. We integrate with respect to $y$.  But breaking up is better, no fractional exponents.      
It is easy to show that the area of $OPB$ is $(1/3)t^3$.
Let us guess that the answer is $y=2k^2x^2$.  Then $x=\frac{1}{k\sqrt{2}}y^{1/2}$.
The area of $OPA$ is
$$\int_0^{2t^2} \left(\frac{1}{\sqrt{2}}y^{1/2}-\frac{1}{k\sqrt{2}}y^{1/2}\right)dy.$$
Calculate. We get $(1-1/k)(4/3)t^3$. This should be $(1/3)t^3$, so $1-1/k=1/4$, $k=4/3$, and therefore the equation is $y=(32/9)x^2$.
Now we can appeal to the geometrically obvious uniqueness. Or else we can let the inverse function of $f$ be $g$, set up the integral, and differentiate under the integral sign (Fundamental Theorem of Calculus). Same result.
