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.