# Integral between Limits is Independent of Direction

## 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

$\blacksquare$