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
\(\ds \leadsto \ \ \) \(\ds \int_a^b \map f x \rd x\) \(=\) \(\ds -\int_b^a \map f x \rd x\)


Due to the organization of pages at $\mathsf{Pr} \infty \mathsf{fWiki}$, this argument is circular.
In particular: Careful with the above. Down at the bottom of Sum of Integrals on Adjacent Intervals for Integrable Functions use is made of this result. There has been a lot of activity on this area of mathematics, and several people have contributed, possibly without having completely followed through all the chains of implications.
My bad. Either way this has to be proven independently since the case $a = b$ was not considered in that theorem.
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Sources

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