How to show $f(x)$ is $0$ in following problem?

Question

Let $f$ be a continuous function defined on $[a, b]$. Assume that there exist constants $α$ and $β$ with $(α ≠ β)$ such that
$$\alpha\int_a^x f(u)du + β\int_x^bf(u)du = 0 $$ for all $x$ belonging to $[a,b]$. Show that $f(x) = 0 $ for all $x$ belonging to $[a,b]$.

My attempt: if we take $x = a$, then we get $\int_a^bf(x)dx = 0$. However this does not imply $f(x)$ is $0$.

Answer 1

Hint:\begin{align}\frac{d}{dx}\left(\alpha\int_a^xf(u)du+\beta\int_x^bf(u)du\right)=(\alpha -\beta)f(x)\end{align}

Answer 2

Note that \begin{align*} \alpha \int_a^x f + \beta \int_x^b f = \beta\int_a^b f + (\alpha - \beta)\int_a^x f. \end{align*} In particular, setting $x= 0$ shows that $\int_a^bf = 0$. Thus $\int_a^x f$ for all $x\in[a, b]$. Since $f$ is continuous, it must vanish everywhere.