Integral between Limits is Independent of Direction

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Theorem

Let $f$ be a real function which is integrable on the interval $\openint a b$.

Then:

$\displaystyle \int_a^b \map f x \rd x = \displaystyle \int_a^b \map f {a + b - x} \rd x$


Proof

Let $z = a + b - x$.

Then:

$\dfrac {\d z} {\d x} = -1$

and:

$x = a \implies z = a + b - a = b$
$x = b \implies z = a + b - b = a$


So:

\(\displaystyle \int_a^b \map f {a + b - x} \rd x\) \(=\) \(\displaystyle \int_b^a \map f z \paren {-1} \rd z\) Integration by Substitution
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b \map f z \rd z\) Reversal of Limits of Definite Integral
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b \map f x \rd x\) renaming variables

$\blacksquare$