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Fubini computes $\int_{0}^{1} \int_{- 1}^{1} x \exp (x y) d x d y$ by reversing the order

Example

The continuous integrand $x \exp (x y)$ satisfies $\int_{0}^{1} \int_{- 1}^{1} x \exp (x y) d x d y = e - e^{- 1} - 2.$ Reversing the order avoids an awkward antiderivative in $x$ .

Facts & Assumptions

Given: The displayed integral over $[- 1, 1] \times [0, 1]$ .

[L1]

A continuous function on a rectangle may be integrated in either repeated order (A continuous function on a closed rectangle has repeated Riemann integrals in every coordinate order, all equal to its multiple integral).

[L2]

The derivative of the exponential is the exponential (The exponential function is smooth and $(\exp)^{'} = \exp$ ), and the second fundamental theorem evaluates integrals of derivatives (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ ).

[L3]

Exponential values are positive and $\exp (- 1) = \exp (1)^{- 1}$ (The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ ).

Verification

technique · computation

1.1

By continuity and [L1], reverse the order and integrate in $y$ first. Since $\partial_{y} \exp (x y) = x \exp (x y)$ , [L2] gives $\int_{0}^{1} x \exp (x y) d y = e^{x} - 1$ , including $x = 0$ .

L1L2

2.1

A second application of [L2] gives $\int_{- 1}^{1} (e^{x} - 1) d x = (e - e^{- 1}) - 2.$

L2step 1.1

3.1

Using [L3] for the negative endpoint yields the stated value $e - e^{- 1} - 2$ .

L3step 2.1∎

Depends on

Used by

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Sources

J. Lebl, Basic Analysis II, Exercise 10.2.1 (standard reference, not scraped)

Fubini computes ∫01∫−11xexp⁡(xy) dx dy∫01​∫−11​xexp(xy)dxdy by reversing the order