# 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:

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

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 | ||||||||||

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

## Sources

- 1968: Murray R. Spiegel:
*Mathematical Handbook of Formulas and Tables*... (previous) ... (next): $\S 15$: General Formulas involving Definite Integrals: $15.10$ - 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 13.2$ - 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 13.16$