How to find the original function from a definite integral. I have that $\int_{0}^{x} f(x) \,dx = 2x,$ and I would like to find $f(x)$.
I am not even sure how to begin.
I would appreciate any help!
 A: Note that the derivative of $\int_0^x g(t)dt$ is $g(x)$ whenever $g$ is continuous (See Fundamental theorem of calculus as mentioned by David). To get $g(\tilde x)$ you can calculate $\frac{d}{dx}\left.\int_0^x g(t)dt\right|_{x=\tilde x}$ ... 
A: You could say: 
find f(x) for  $\int_{0}^{a} f(x) \,dx = 2\ a$
By inspection f(x) = 2 is one solution.
$ Anyone $ among an infinite set of functions that removes area from first half and deposits on the second half satisfies it. It is up to your imagination or familiarity with function forms to suggest a solution.
One simple example where center point $(x,1)$ is chosen constant:
$$ f(x) = 2 + A \, \sin \frac {\pi \, x}{\lambda} $$ where $ A, \lambda $ are constants. $\lambda$ is wave length of sine-wave. Other functions which are "odd" with respect to this point are possible.
If you choose segments in ratio 1:3 , area average height should be 3:1, and so on.
A: By using the Leibniz rule then $f$ defined by
\begin{align}
\int_{ax}^{bx} f(t) \, dt = c \, x,
\end{align}
where $a, b, c$ are constants, can be found in the following way:
\begin{align}
\frac{d}{dx} \, \int_{ax}^{bx} f(t) \, dt &= \frac{d}{dx} \, (c \, x) = c \\
\int_{ax}^{bx} \frac{d}{dx}(f(t)) \, dt + f(bx) \, \frac{d(bx)}{dx} - f(ax) \, \frac{d(ax)}{dx} &= c \\
b \, f(bx) - a \, f(ax) &= c \tag{1}
\end{align}
One form of $f(t)$ of (1) is given by
\begin{align}
f(t) = \frac{c}{b-a}.
\end{align}
For the case of $a=0, b=1, c=2$ this leads to $f(t) = 2$. 
