---
tag: 0H60H1Q0SX
kind: statement
role: proposition
name: Changing the order of integration
aliases: [reversing-order-of-integration]
msc: [26B15]
uses: [0R0T1VWH5X, 0P4WX7Y502, 0TXCQ9CST1]
sources:
  - url: https://en.wikipedia.org/wiki/Order_of_integration_(calculus)
    locator: Problem statement
  - url: https://en.wikipedia.org/wiki/Multiple_integral
    locator: Double integral over a normal domain
same_as:
  wikidata: Q3825182
  wikipedia: Order of integration (calculus)
---

## Statement

Let $D \subseteq \mathbb{R}^2$ be a region that has two descriptions,
$$D = \{(x, y) : a \le x \le b, \ \varphi_1(x) \le y \le \varphi_2(x)\} = \{(x, y) : c \le y \le d, \ \psi_1(y) \le x \le \psi_2(y)\},$$
where $\varphi_1 \le \varphi_2$ are continuous on $[a, b]$ and $\psi_1 \le \psi_2$ are continuous on $[c, d]$.
Let $f \colon D \to \mathbb{R}$ be continuous.
Then
$$\int_a^b \int_{\varphi_1(x)}^{\varphi_2(x)} f(x, y) \, dy \, dx = \int_c^d \int_{\psi_1(y)}^{\psi_2(y)} f(x, y) \, dx \, dy,$$
and both equal the double integral of $f$ over $D$.

## Remarks

The proof applies [[Fubini's theorem|0R0T1VWH5X]] on a rectangle that contains $D$ to $f$ times the indicator function of $D$.
The main work is to describe $D$ again in the other order. One way is to sketch the region.
Example: $\int_0^1 \int_x^1 e^{y^2} \, dy \, dx$ has no elementary inner antiderivative. The region is $0 \le x \le y \le 1$, so the integral equals $\int_0^1 \int_0^y e^{y^2} \, dx \, dy = \int_0^1 y e^{y^2} \, dy = (e - 1)/2$.
For unbounded regions or discontinuous integrands, a change of order needs extra hypotheses.

## Explanations

### L1 human

An iterated integral over a region can sometimes be hard in one order and easy in the other.
Changing the order means describing the same region by slicing it in the other direction, and then integrating in the new order.