Does there exist $u(x)\neq 0$ such that integral is equal to $0$? Let $\theta \in \mathbb R$. Does there a function $u(x)\neq 0$, not depending on $\theta$, such that
$$\int\limits_{0}^{\theta} u(x)(\theta-x) dx = 0 ?$$

I spend much time to try guessing $u(x)$, like $u(x)=x$, $u(x)=x^2$, $u(x)=x^3$, but the integral is not equal $0$.

So does there exist $u(x)\neq 0$ such that integral is equal to $0$?
Is there a way to find $u(x)$ without guessing?
If $\  \displaystyle\int_0^\theta u(x)(\theta-x)xdx =0\ $ for all values of $\ \theta\ $ and $\ u\ $ does not depend on theta, then
\begin{align}
0&amp;=\frac{d}{d\theta} \int_0^\theta u(x)(\theta-x)dx\\
&amp;=u(\theta)(\theta-\theta) + \int_0^\theta u(x)dx\\
&amp;= \int_0^\theta u(x)dx\ .
\end{align}
Now differentiating the equation $\  \displaystyle\int_0^\theta u(x)dx=0\ $ with respect to $\ \theta\ $ gives
$
u(\theta)=0\ 
$.  If $\ u\ $ is continuous, this must be true for all values of $\ \theta\ $. If $\ u\ $ is not required to be continuous, it must still be true with the possible exception of for values of $\ \theta\ $ lying within some set of measure zero.
As $\theta-x$ is a straight line $\int_0^{\theta}\theta-xdx = \frac{\theta^2}{2}$.Now there exists a $c$ in $(0,\theta) $ such that $\int_0^{c}\theta-xdx = \int_c^{\theta}\theta-xdx $ so, now define $u(x) = 1$ for x in $[0,c]$ and $u(x)=-1$ in  $(c,\theta]$ and we have value of  that integral $0$.