Triangle Inequality for Integrals/Real

Theorem
Let $f$ be a real function which is continuous on the closed interval $\left[{a. . b}\right]$.

Then:
 * $\displaystyle \left|{\int_a^b f \left({t}\right) dt}\right| \le \int_a^b \left|{f \left({t}\right)}\right| dt$

Proof
From Negative of Absolute Value, we have for all $a \in \left[{a. . b}\right]$:


 * $\displaystyle - \left|{f \left({t}\right)}\right| \le f \left({t}\right) \le \left|{f \left({t}\right)}\right|$

Thus from Relative Sizes of Definite Integrals:


 * $\displaystyle- \int_a^b \left|{f \left({t}\right)}\right| dt \le \int_a^b f \left({t}\right) dt \le \int_a^b \left|{f \left({t}\right)}\right| dt$

Hence the result.