Analysis of Integrals Suppose that $f : [a,b] \rightarrow R$ is continuous. Suppose that$\int_a^x = \int_x^b f $ for all $x ∈ [a,b]$. Show that $f(x)=0$ for all $x$ in $[a,b]$
I have the following:
$$\int_a^xf- \int_x^bf =0$$
$$\implies\int_a^bf=F(b)-F(a)=0$$
$$\implies F(b)=F(a)$$.
I am unsure if this is the right track, and if it is how to come to the conclusion.
 A: We have
$$\int_a^xf- \int_x^bf =0$$
and by Chasle relation
$$\int_a^xf+ \int_x^bf =\int_a^bf=\int_a^af=0$$
so
$$\int_a^xf =0$$
so differentiate and we have $f(x)=0$.
A: Take the derivative on both sides of 
$$
\int_a^x f(x)\; dx = \int_x^bf(x)\; dx
$$
and get
$$\begin{align}
\frac{d}{dx} \int_a^x f(x)\; dx &= \frac{d}{dx}(-\int_b^xf(x)\; dx) \quad\implies\\
f(x) &= - f(x).
\end{align}
$$
This means that 
$$
f(x) = 0
$$
for all $x\in (a,b)$. Now $f$ is continuous on $[a,b]$ so you also have $f(a) = f(b) = 0$.
A: Suppose $c\in(a,b)$ with $f(c)=\alpha>0$. 
By continuity, there is an $h>0$ with $c+h<b$ so that $f(x)>\alpha/2$ for all $x\in[c,c+h]$.
We then have
$$
\int_a^c f(x)\, dx<\int_a^{c+h} f(x)\,dx
$$
and
$$
\int_{c+h}^b f(x)\, dx<\int_c^{b} f(x)\,dx.
$$
If 
$$\int_a^{c+h} f(x)\,dx = \int_{c+h}^b f(x)\, dx,$$
we would have
$$
\int_a^c f(x)\, dx\ne\int_c^{b} f(x)\,dx.
$$
It follows if $f$ satisfies your hypotheses, then $f$ takes no positive value in $(a,b)$; and thus by continuity, no positive value in $[a,b]$.
Similarly, one can show $f$ takes no negative value in $[a,b]$.
