Given a continuous function $f(x)$ defined on a range $x\in[a,b]$, such that the function has equal values in points symmetric to $x = {a+b\over 2}$, prove that: $$ \int_a^bf(x)dx = 2\int_{a}^{a+b\over 2}f(x)dx $$
I've started by using the fundamental theorem of calculus, namely: $$ \int_a^bf(x)dx = \int_a^{a+b\over 2}f(x)dx + \int_{a+b\over 2}^bf(x)dx = \\ F\left({a+b\over 2}\right) - F(a) +F(b) - F\left({a+b\over 2}\right) $$
On the other hand: $$ 2\int_{a}^{a+b\over 2}f(x)dx = 2\left(F\left({a+b\over 2}\right)- F(a)\right) \\ = F\left({a+b\over 2}\right) - F(a) + F\left({a+b\over 2}\right) - F(a) $$
Since the function is symmetric with respect to $x_0$ then: $$ F(a) = F(b) \tag 1 $$ Thus: $$ \begin{align} I &= F\left({a+b\over 2}\right) - F(a) + F\left({a+b\over 2}\right) - F(b) \\ &= F\left({a+b\over 2}\right) - F(a) - \left(F(b) - F\left({a+b\over 2}\right)\right)\\ &=\int_a^{a+b\over 2}f(x)dx - \int_{a+b\over 2}^bf(x)dx \tag2 \end{align} $$
Which is obviously nonsense. I believe the error is in $(1)$
So in case, the sign in $(2)$ is plus as opposed to minus then: $$ \int_a^{a+b\over 2}f(x)dx + \int_{a+b\over 2}^bf(x)dx = \int_a^bf(x)dx $$
Is the idea to prove the statement "correct in general"? Where did I make a mistake?
In the seconf integral, do the next C.V. $t=a+b-x$: so that $dx=-dt$ and for $x=b$ $t=a$ and for $x=\frac{a+b}{2}$ $t=\frac{a+b}{2}$. So
$$\int_{\frac{a+b}{2}}^bf(x)dx = -\int_{\frac{a+b}{2}}^{a}f(a+b-t)dt = \int_a^{\frac{a+b}{2}}f\left(\frac{a+b}{2}+\underbrace{\left(\frac{a+b}{2}-t\right)}_{>0}\right)dt = \int_a^{\frac{a+b}{2}}f\left(\frac{a+b}{2}-\left(\frac{a+b}{2}-t\right)\right)dt =\int_a^{\frac{a+b}{2}}f(t)dt $$