Absolute Value of Definite Integral

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Theorem

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


Then:

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


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:

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


Hence the result.

$\blacksquare$


Sources