A question on calculus The question is:

In the accompanying figure the curve represented by $\rm C$ is given by $y= \rm f (x)$ where $\rm f$ is one-to-one and continuous. ${\rm C1}$ and ${\rm C2}$ are respectively $y=2x^2$ and $y=x^2$. At each point $P$ on $\rm {C1}$, assume that the areas $OAP$ and $OPB$ are equal. Here $PA, PB$ are the horizontal and vertical segments. Determine the function $f$.


My progress:
I took any arbitrary point $x &gt; 0$ and calculated the areas and found out that $$\int_{0}^{x}{\rm f}(t)dt=x{\rm f}(x)-(\frac{{\rm f}(x)}{2})^{\frac {3}{2}}.$$Now the problem is that we do not know that $f$ is differentiable or not. If it was, then we could differentiate both sides of the equation(since $f$ continuous is given) w.r.t $x$ to obtain $$f^{'}(x)(x-\frac {3}{4\sqrt {2}}\sqrt {f(x)})=0.$$ Now since $f$ is one-to-one and continuous, we can easily show that $f$ is monotonic and hence $f$ cannot attain a relative extremum anywhere and hence $f^{'}(x)=0$ for no $x$, i.e. for all $x$, $$\sqrt {f(x)}=\frac {4\sqrt {2}x}{3}$$ wherefrom we conclude $f(x)=\frac{32x^{2}}{9}.$  How do we prove that $f$ is differentiable? Also I think we can solve the problem another way round, i.e we first show that $\frac {32x^2}{9}$ indeed is a solution and then that there can be at most one solution of $f$. Can we do it this way, i.e show the uniqueness? Thank you for your help in advance.
(P.S.  It would also be helpful if someone tells me how to draw good diagrams here. I have drawn this in paint and the graphs look horrible.)
Extend the segments $PA$ and $PB$ until they meet the axes at $A'$ and $B\,'$, respectively.



Since arc $OP$ is a graph of $y = 2x^2$ and arc $OB$ is a graph of $y = x^2$, we know that the area enclosed by the figure $OBB\,'$ is half the area enclosed by $OPB\,'$; and the area enclosed by $OPB$ is the difference of those two areas, so it also is half the area of $OPB\,'$.

But we also know that the area of $OPB\,'$, that is, under the parabolic arc $OP$, is one third the area of the rectangle $OB\,'PA'$, and the area of $OPA'$ (between the arc $OP$ and the $y$-axis) is two-thirds of the area of the rectangle.  The area of $OPA$, which is equal to the area of $OPB$, is therefore one quarter of the area of $OPA'$, and $OAA'$ has three-fourths of the area of $OPA'$.

Now, since $f$ is one-to-one, $f^{-1}$ exists on the range of $f$, 
and the arcs $OP$ and $OA$ can be described by the equations 
$x = \sqrt{y/2}$ and $x = f^{-1}(y)$, respectively.  From the preceding facts, 
the areas between those arcs and the $y$-axis satisfy the equation

$$ \frac{3}{4} \int_0^Y \sqrt{\frac{y}{2}} dy = \int_0^Y f^{-1}(y) \, dy $$

where $Y$ is the $y$-coordinate of $P$.

Taking the integral on each side of the equation above as a function of $Y$, by the fundamental theorem of calculus the derivatives of these integrals exist and are equal to  $\sqrt{Y/2}$ and $f^{-1}(Y)$, respectively.  Therefore,

$$ \frac{3}{4} \sqrt{\frac{y}{2}} = f^{-1}(y). $$

If $x = \frac{3}{4} \sqrt{\frac{y}{2}}$, then $y = \frac{32}{9} x^2$, so

$$ x = f^{-1} \left( \frac{32}{9} x^2 \right), $$

and applying $f$ to both sides, we find that

$$  f(x) = \frac{32}{9} x^2. $$
First, show that the function you derived (using stronger assumption) solves the problem. Then show that the problem has unique solution on the class of monotone continuous functions.

I also sketch a proof of uniqueness. Let $f_1$ and $f_2$ be two different monotone continuous functions solving the integral equation. Necessarily $f_1(0)=f_2(0)=0$, hence there exist (using continuity) a points $0\leq t_0&lt;t_1$ such that $f_1(t)=f_2(t)$ for all $t\in [0,t_0]$ and (WLOG) $f_2(t)&lt;f_1(t)$ for $t\in(t_0,t_1]$. Now there exists $t_2&gt;t_1$ such that $f_1(t_1)=f_2(t_2)$. When you sketch this situation, you should realize contradiction.

A regularity result for integral equatlity holds and can be obtained by the Implicit function theorem.