<s>[INST] Area under curves , two given curves and finding function of 3rd curve from relationship between their area . Let $C_1,C_2$ be the graphs of $y=x^{2} , y=2x, 0<x<1$ respectively. Let $C_3$ be the graph of an unknown function $y=f(x), 0<c<1$ and $f(0)=0$ for a point $P$ on $C_1$ let the line pass through $P$, parallel to the axes, meet $C_2$ and $C_3$ at $Q$ and $R$ respectively . If every position of $P ($$on $ $C_1)$ the areas of the shaded region $OPQ $ and $ORP$ are equal then find $f(x)$ ? 
diagram : 

 [/INST] Let the x-coordinate of $P$ be $x$. Then the points are: $P(x,x^2), Q(\frac{1}{2}x^2,x^2),R(x,f(x))$. Let $S$ be the projection of point $P$ onto the x-axis. and $T$ the projection of points $P,Q$ onto the y-axis.
Then 
$$\begin{align}
Area(OPQ)&=Area(OSPT)-Area(OSP)-Area(OQT)\\[1em]
         &=x\cdot x^2-\frac{1}{3}x^3-\frac{1}{2}(\frac{1}{2}x^2)(x^2) \\[1em]
         &=\frac{2}{3}x^3-\frac{1}{4}x^4 \\[1em]
         &=Area(ORP)
\end{align}$$
So $$\int_0^X{x^2-f(x)\,dx}=\frac{2}{3}X^3-\frac{1}{4}X^4$$
By the fundamental theorem of calculus, we then have
$$X^2-f(X)=\frac{d}{dX}\left[\frac{2}{3}X^3-\frac{1}{4}X^4\right]=2X^2-X^3$$
so
$$\boxed{f(x)=x^3-x^2}$$
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