Triangle Inequality for Integrals/Real
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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): $\S 13.24$