# Absolute Value of Definite Integral

## Theorem

Let $f$ be a real function which is continuous on the closed interval $\closedint a b$.

Then:

$\displaystyle \size {\int_a^b \map f t \rd t} \le \int_a^b \size {\map f t} \rd t$

## Proof

From Negative of Absolute Value, we have for all $a \in \closedint a b$:

$-\size {\map f t} \le \map f t \le \size {\map f t}$

Thus from Relative Sizes of Definite Integrals:

$\displaystyle -\int_a^b \size {\map f t} \rd t \le \int_a^b \map f t \rd t \le \int_a^b \size {\map f t} \rd t$

Hence the result.

$\blacksquare$