Triangle Inequality for Integrals/Real

Theorem

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

Then:

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

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:

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

Hence the result.

$\blacksquare$

Also see

• Triangle Inequality for Integrals, of which this is a special case

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 13.24$