If $\int_a^b f(x) \ \mathrm{d}x = \int_a^b g(x) \ \mathrm{d}x$ then $\exists x \in [a,b]$ with $f(x) = g(x).$ I am trying to prove the following: 

Take $f, g:[a,b] \to \mathbb{R}$ such that $f$ and $g$ are continuous. If
$$\int_a^b f(x) \ \mathrm{d}x = \int_a^b g(x) \ \mathrm{d}x,$$

then there exists some $c \in [a,b]$ such that $f(c) = g(c).$ 

Here's my current proof. I'd welcome any feedback regarding correctness and clarity. 

Current Proof

Assume that there exists no such $c$. There are then three possibilities. First, it is possible that $f(x) &gt; g(x)$ $\forall$ $x \in [a,b]$. However, this cannot be, since then we would have
$$\int_a^b f(x) \ \mathrm{d}x &gt; \int_a^b g(x) \ \mathrm{d}.x$$

Similarly, we cannot have $g(x) &gt; f(x)$ $\forall$ $x \in [a,b]$, since then
$$\int_a^b f(x) \ \mathrm{d}x &lt; \int_a^b g(x) \ \mathrm{d}x.$$
Thus, there exists some $x \in [a,b]$ such that $f(x) &gt; g(x)$ and some $y \neq x$ such that $g(y) &gt; f(y).$ Assume without loss of generality that $x &lt; y.$ Consider a new function $h:[a,b] \to \mathbb{R}$ defined by
$$h(x) = f(x) - g(x).$$
Clearly, $h$ is continuous, as the difference of two continuous functions. From the above, we have that $h(x) &gt; 0$ and $h(y) &lt; 0.$ Apply the Intermediate Value Theorem to $h$ on the interval $(x,y).$ Thus, there exists some $c \in (x, y)$ such that $f(c) = 0$. Since $(x, y) \subset [a, b]$, we have found an element of $[a,b]$ such that $h(c) = 0 \implies f(x) = g(x).$
While your proof is correct, there are ways you could simplify it.  For example, as noted in the comments, you could apply Rolle's theorem to $\int_a^x f(x') - g(x') dx'$ since you know it will be zero at the endpoints $a$ and $b$.
It can be proved:
    ∫{a}^{b}f(x)dx=∫{a}^{b}g(x)dx
    and
    f(x)≥g(x), for all x∈[a,b],
    then there is c∈[a,b] such that f(c)=g(c).