Show $f$ is constant zero function on $[a, b]$ if $\int_a^x f(t) dt = \int_x^b f(t) dt \forall x \in [a, b].$ I've started by trying to use the Fundamental Theorem of Calculus and have that
$$\int_a^x f(t) dt - \int_x^b f(t) dt = 0$$
$$= F(x) - F(a) - (F(b) - F(x))$$
$$0 = 2F(x) - F(a) - F(b)$$
but am not sure how to proceed to show that $f$ is the constant zero function.
I saw this very similar question:   Suppose $f:[0,1] \Rightarrow \mathbb{R}$ is continuous and $\int_0^x f(x)dx = \int_x^1 f(x)dx$. Prove that $f(x) = 0$ for all $x$, but am not sure how to form the "one line proof" using the Fundamental Theorem of Calculus mentioned in the comments.
Thanks for any tips!

Just differentiate both sides to get:
$$f(x)=-f(x)$$
for all $x$, then $f(x)=0$ for all $x$.
Recall that the fundamental theorem of calculus tells you that for $f(x)$ continuous:
$$\frac{d}{dx}\int_a^x f(t) dt=f(x)$$
Then note that:
$$\int_x^b f(t) dt=-\int_b^x f(t) dt$$
