proof that $\int_{a}^{x} = \int_{x}^{b}$ I want to show that if $f$ is a continuous function on the interval $[a,b]$ then there must exist some $x \in [a,b]$ such that: $$\int_{a}^{x} = \int_{x}^{b}$$
Intuitively this seems very easy and I can see why it is true, its just the structure of the proof that I'm confused about.
Is it sufficient to say that since $f$ is continuous, $\int_{a}^{x}$ exists $\forall x \in [a,b]$ and in particular we can find some $x_{0}$ such that: $$\int_{a}^{x_{0}} = \frac{1}{2} \int_{a}^{b}$$ then since: $$\int_{a}^{x_{0}}+\int_{x_{0}}^{b} = \int_{a}^{b} $$ we get that: $$\int_{x_{0}}^{b} = \int_{a}^{b} - \int_{a}^{x_{0}} $$
$$\int_{x_{0}}^{b} = \int_{a}^{b} - \frac{1}{2} \int_{a}^{b} $$
so: $$\int_{x_{0}}^{b} = \frac{1}{2} \int_{a}^{b} $$
and we get that: 
$$\int_{a}^{x_{0}} = \int_{x_{0}}^{b}$$
Is this rigorous enough? should I try a different method maybe using partitions and upper/lower sums? is there a sort of, mean value theorem equivalent for integrals with area instead of the derivative?
Thanks guys.

We know that there exists a continuous function on the interval $x \in [a, b]$, $F(x)$, such that:
$$
\int_a^bf(t)dt = F(a) - F(b) \\
\int_a^xf(t)dt = F(x) - F(a) \\
\int_x^bf(t)dt = F(b) - F(x) \\
$$
Set the last two equations to be true:
$$
F(x) - F(a) = F(b) - F(x) \longrightarrow
F(x) = \frac{F(a)+ F(b)}{2}
$$
Using the Intermediate Value Theorem, such an $x$ must exist on the interval $x \in [a, b]$ since it can be shown that the the average of $F(a)$ and $F(b)$ is certainly between the two values of $F(a)$ and $F(b)$.
