What is $f$, where $x \int_x^{x+1} f(t)\,dt = \int_0^x f(t)\,dt$? Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous and bounded function such that$$x \int_x^{x+1} f(t)\,dt = \int_0^x f(t)\,dt$$for any $x \in \mathbb{R}$. What sort of function is $f$, what's it like?
 A: You have, for all real $x$, that
$$\begin{equation}\begin{aligned}
\int_{0}^{x+1}f(t)dt & = \int_{0}^{x}f(t)dt + \int_{x}^{x+1}f(t)dt \\
& = x\int_{x}^{x+1}f(t)dt + \int_{x}^{x+1}f(t)dt \\
& = (x+1)\int_{x}^{x+1}f(t)dt
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Also, directly from the original relation, you have
$$\int_{0}^{x+1}f(t)dt = (x+1)\int_{x+1}^{x+2}f(t)dt \tag{2}\label{eq2A}$$
Comparing the $2$ equations shows that, apart from $x = -1$,
$$\int_{x}^{x+1}f(t)dt = \int_{x+1}^{x+2}f(t)dt \tag{3}\label{eq3A}$$
By continuity of $f$, I believe you can state \eqref{eq3A} also holds for $x = -1$. For this to be true for all real $x$ means that, for $f$ being bounded, $f$ must be a periodic function with a period of $1$ (see Update #$2$ below for a detailed discussion of this).
Next, note you have from the original relation that
$$\begin{equation}\begin{aligned}
x\int_{x}^{x+1}f(t)dt & = \int_{0}^{x}f(t)dt \\
x\left(\int_{0}^{x+1}f(t)dt - \int_{0}^{x}f(t)dt\right) & = \int_{0}^{x}f(t)dt \\
x\int_{0}^{x+1}f(t)dt & = (x+1)\int_{0}^{x}f(t)dt
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Now, as suggested by the question comment, differentiate both sides of \eqref{eq4A} and use the Fundamental Theorem of Calculus to get
$$\begin{equation}\begin{aligned}
\int_{0}^{x+1}f(t)dt + xf(x+1) & = \int_{0}^{x}f(t)dt + (x+1)f(x) \\
\int_{x}^{x+1}f(t)dt & = (x+1)f(x) - xf(x+1)
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
Since it was shown earlier that $f$ is periodic with a period of $1$, you have that $f(x + 1) = f(x)$, so \eqref{eq5A} becomes
$$\int_{x}^{x+1}f(t)dt = f(x) \tag{6}\label{eq6A}$$
Now, multiplying both sides $x$ and using the original relation, you get
$$xf(x) = \int_{0}^{x}f(t)dt \tag{7}\label{eq7A}$$
Next, if $f$ is differentiable, then differentiating \eqref{eq7A} gives
$$xf'(x) + f(x) = f(x) \implies xf'(x) = 0 \implies f'(x) = 0 \tag{8}\label{eq8A}$$
This leads to $f$ being a constant function, as stated in this answer.
Update: As suggested in the answer, in \eqref{eq7A}, since $x$ is differentiable and the right side is differentiable, then dividing by $x$ means that $f(x) = \frac{1}{x}\int_{0}^{x}f(t)dt$ is differentiable at all points except possibly $0$, but due to the periodicity of $f$, it's also differentiable there.
Update #$2$: In \eqref{eq3A}, change $x$ to $x + \epsilon$ for some $\epsilon \gt 0$ to get
$$\int_{x + \epsilon}^{x + \epsilon + 1}f(t)dt = \int_{x + \epsilon + 1}^{x + \epsilon + 2}f(t)dt \tag{9}\label{eq9A}$$
Subtract \eqref{eq3A} from this to get
$$\begin{equation}\begin{aligned}
\int_{x + 1}^{x + \epsilon + 1}f(t)dt - \int_{x}^{x + \epsilon}f(t)dt & = \int_{x + 2}^{x + \epsilon + 2}f(t)dt - \int_{x + 1}^{x + \epsilon + 1}f(t)dt \\
\int_{x + 2}^{x + \epsilon + 2}f(t)dt & = 2\int_{x + 1}^{x + \epsilon + 1}f(t)dt - \int_{x}^{x + \epsilon}f(t)dt
\end{aligned}\end{equation}\tag{10}\label{eq10A}$$
Using a concept like that of a Riemann sum, but with just one interval with the width going to $0$, along with continuity of $f$, as $\epsilon \to 0$ in \eqref{eq10A}, you get
$$f(x + 2) = 2f(x + 1) - f(x) \tag{11}\label{eq11A}$$
Consider if there's any $x = x_0$ where
$$f(x_0 + 1) = f(x_0) + \delta, \; \delta \neq 0 \tag{12}\label{eq12A}$$
If so, then from \eqref{eq11A} you get
$$\begin{equation}\begin{aligned}
f(x_0 + 2) & = 2(f(x_0) + \delta) - f(x) \\
& = f(x_0) + 2\delta
\end{aligned}\end{equation}\tag{13}\label{eq13A}$$
You can fairly easily use induction to show $f(x_0 + n) = f(x_0) + n\delta$ for all natural $n$. However, this means $f(x)$ is unbounded either above (if $\delta \gt 0$) or below (if $\delta \lt 0$). Since $f(x)$ is bounded, this means $\delta = 0$ in \eqref{eq12A}. This shows $f(x)$ is periodic with a period of $1$.
A: Let us define a function $F:\Bbb R \rightarrow \Bbb R$ sending any $x\neq 0$ to $\frac 1 x \int_0^xf(t)dt$ and sending $0$ to $f(0)$.
Then we have:

*

*The function $xF$ is a $\mathcal C^1$ function, with derivative $(xF)' = f$.
This is tautological for $x \neq 0$, and is clear for $x = 0$.


*The function $F$ is continuous.
It is trivial outside $x = 0$, and at $x = 0$ it's also an easy consequence of the continuity of $f$.


*The function $F$ is periodic: $F(x + 1) = F(x)$.
It is just a rewriting of the original condition on $f$ for $x \neq 0, -1$, and then follows from 2 for these two cases.


*The function $F$ is a $\mathcal C^1$ function.
For $x \neq 0$ this is clear from 1, and for $x = 0$ it follows from 3.

By 1, we have $xF' + F = f$. By 2 and 3, the function $F$ is bounded.
Since the function $f$ is also bounded, we conclude that the function $xF'$ is bounded.
However, this is only possible when $F' = 0$.
Therefore $F$ is a constant function and so is $f$.
A: Let $f(x)=k$ where $k$ is a constant real number then
$$x \int_x^{x+1} f(t)\,dt = xk$$
and also $$ \int_0^x f(t)\,dt=xk$$
which satisfies the desired condition.
I did not find any other  function which satisfies your conditions.
A: Let
$$ F(x)=\int_0^xf(t)dt. $$
Then one has
$$ x(F(x+1)-F(x))=F(x) $$
or
$$ xF(x+1)=(x+1)F(x) $$
or
$$ \frac{F(x+1)}{x+1}=\frac{F(x)}{x} $$
which holds for any $x\neq0,-1$. So
$$ \frac{F(x)}{x}\equiv C$$
or
$$F(x)=Cx$$
which implies that $f(x)\equiv C$.
