Reversal of Limits of Definite Integral

Theorem

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

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

Let:

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

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

Let $a \le b$.

Then:

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

Proof

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