Is there a non-trivial function satisfying $\int_0^1 (f(x)+x)^a \, dx=0$ for $a$ a positive real (or integer)?

Question

Here is the functional equation:

$$\int_0^1 (f(x)+x)^a \, dx=0$$

Is there a function $f(x)$ that satisfies it for any $a>0$ and $f(x)\not\equiv-x$?

What about the case where $a$ is an integer?

Answer 1

Let $S$ be an finite set such that $S \subset [0,1]$, and $M$ an subset with the same ammount of elements of $S$, such that $M \subset \mathbb{R}$.

Now let $\begin{cases} f(x) = -x \text{ if } x\notin S \\ f(x) = g(x) \text{ otherwise} \end{cases}$ Where $g(x)$ is some abritary function $g:S \to M$

Then the integral in question can be computed by splitting it into an finite sum of other integrals, in which $f(x)$ behaves exactly as $f(x)=-x$.

Now, It's fully possible this trickery I did is forbidden by the $f(x) \not\equiv -x$, i don't actually know how restrcitve that notation is. Please do let me know