Cauchy-Bunyakovsky-Schwarz Inequality/Definite Integrals

Theorem
Let $f$ and $g$ be real functions which are continuous on the closed interval $\closedint a b$.

Then:
 * $\ds \paren {\int_a^b \map f t \, \map g t \rd t}^2 \le \int_a^b \paren {\map f t}^2 \rd t \int_a^b \paren {\map g t}^2 \rd t$

Proof
where:

The quadratic equation $A x^2 + 2 B x + C$ is non-negative for all $x$.

It follows (using the same reasoning as in Cauchy's Inequality) that the discriminant $\paren {2 B}^2 - 4 A C$ of this polynomial must be non-positive.

Thus:
 * $B^2 \le A C$

and hence the result.