Triangle Inequality for Integrals/Real

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

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

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


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

Thus from Relative Sizes of Definite Integrals:


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

Hence the result.