Reversal of Limits of Definite Integral

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Theorem

Let $\closedint a b$ be a closed real interval.

Let $f: \closedint a b \to \R$ be an integrable real function.

Let:

$\ds \int_a^b \map f x \rd x$

be the definite integral of $f$ over $\closedint a b$.


Let $a \le b$.

Then:

$\ds \int_a^b \map f x \rd x = -\int_b^a \map f x \rd x$


Proof

\(\ds \int_a^b \map f x \rd x + \int_b^a \map f x \rd x\) \(=\) \(\ds \int_a^a \map f x \rd x\) Sum of Integrals on Adjacent Intervals for Integrable Functions
\(\ds \) \(=\) \(\ds 0\) Definite Integral on Zero Interval
\(\ds \leadsto \ \ \) \(\ds \int_a^b \map f x \rd x\) \(=\) \(\ds -\int_b^a \map f x \rd x\)

Sources