Find all functions $f(x)$ continuous and non-negative that have the property that the area under the graph is $\frac{1}{3}$ the rectangle with opposite vertices at $(0,0)$ and $(x,f(x))$ and sides parallel to the coordinate axes.
I tried to solve it and I got the following:
$\int_0^x f(x)dx = \frac{1}{3} x f(x) \\ \Rightarrow f(x)=\frac{1}{3} \frac{d}{dx} (xf(x))$
but I could not go further. Any idea?
On
The equation is not correct, it should be $$\int_0^xf(t)dt=\frac{1}{3}xf(x).$$ Using FTC, then, we have $$\begin{split}f(x)&=\frac{1}{3}\frac{d}{dx}(xf(x)) \\&=\frac{1}{3}f(x)+\frac{1}{3}xf'(x)\end{split}$$ so $$\frac{f'(x)}{f(x)}=\frac{2}{x}$$ so $$\ln(f(x))=2\ln(x)+C_1$$ so the general solution is $$f(x)=Cx^2,\,C\in\mathbb{R}.$$
Let $g(x):=xf(x)$. Then
$$\frac{g(x)}x=\frac{g'(x)}3$$ is a separable ODE that integrates as
$$3\log x+c=\log g(x)$$
and finally
$$f(x)=\frac{g(x)}x=cx^2.$$