Show $f$ is constant zero function on $[a, b]$ if $\int_a^x f(t) dt = \int_x^b f(t) dt \forall x \in [a, b].$

Question

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!

Answer 1

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:

$$\int_x^b f(t) dt=-\int_b^x f(t) dt$$