Triangle Inequality for Integrals/Real

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

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

Proof
From Negative of Absolute Value, we have $$\forall a \in \left[{a \,. \, . \, b}\right]: - \left|{f \left({t}\right)}\right| \le f \left({t}\right) \le \left|{f \left({t}\right)}\right|$$.

Thus from Relative Sizes of Definite Integrals, $$- \int_a^b \left|{f \left({t}\right)}\right| dt \le \int_a^b f \left({t}\right) dt \le \int_a^b \left|{f \left({t}\right)}\right| dt$$.

Hence the result.