Intermediate Value Theorem for Integrals

Suppose $f$ is a continuous function on $[a,b]$. Then $\exists x \in (a,b)$:

$$f(x)=\frac{1}{b-a}\int_a^b f$$

Proof

Suppose $f_{\text{max}} = M = f(x_0)$ and $f_\text{min} = m=f(y_0)$.

When $M=m$: $f$ is a constant function. Proof is trivial.

Otherwise:

$$m(b-a) \le \int_a^b f \le M(b-a)$$

Then there exists $x \in (x_0, y_0)$. $$