Multivariable calculus

Double Integral of exp(x + y) Over the Unit Square

Q: Is (e-1)^2 the same as e^2 - 1?

No, and the difference is large. (e-1)^2 = e^2 - 2e + 1 \approx 2.9525, while e^2 - 1 \approx 6.389. Squaring a difference is not the same as subtracting from a square.

Q: What changes if the square becomes 0 to 2 in each variable?

The same split applies, with each single integral becoming \int_{0}^{2} e^{t}\,dt = e^2 - 1. The double integral is then (e^2-1)^2 \approx 40.82.

The double integral of e raised to the power x plus y over the unit square, with x and y each from 0 to 1, equals (e - 1) squared, about 2.9525. The integrand factors into a function of x times a function of y, so the answer is the product of two copies of the integral of e to the x, each equal to e - 1.

f(x,y) = e^{x+y}

(e-1)^2

Numerically 2.952492, confirmed by quadrature on every build.

Split the exponent into a product, then use Fubini

The exponential law $e^{x+y} = e^x e^y$ turns a function that looks entangled into a clean product. That is the whole method here.

\int_{0}^{1}\!\!\int_{0}^{1} e^{x+y}\,dy\,dx = \int_{0}^{1} e^{x}\left(\int_{0}^{1} e^{y}\,dy\right)\,dx

The inner integral is a number, $e - 1$ , so it slides straight out of the outer integral.

= (e-1)\int_{0}^{1} e^{x}\,dx = (e-1)(e-1) = (e-1)^2 \approx 2.9525

Expanded, $(e-1)^2 = e^2 - 2e + 1$ . Both forms are correct; the factored one shows where the answer came from.

The mistake: turning e^(x+y) into e^x + e^y

This is the single most common slip on this integral. Exponentials convert addition in the exponent into multiplication, never into addition of the outputs.

e^{x+y} = e^{x}e^{y} \quad \text{and} \quad e^{x+y} \neq e^{x} + e^{y}

The wrong version is easy to catch numerically. Integrating $e^x + e^y$ over the unit square gives $2(e-1) \approx 3.4366$ , which overshoots the true $2.9525$ by more than $16\%$ .

A one-second check: at $x = y = 0$ the true integrand is $e^0 = 1$ , while $e^x + e^y$ would be $2$ . If the integrand is already wrong at a corner, no amount of careful integrating will save the answer.

Reading the number as a volume

The integrand is positive everywhere, so $2.9525$ is the volume of the solid between the unit square and the surface $z = e^{x+y}$ .

That number should look reasonable. On the unit square the integrand ranges from $e^{0} = 1$ at the origin to $e^{2} \approx 7.389$ at the far corner, and the base has area $1$ , so the volume has to sit between $1$ and $7.389$ .

Lower bound: $1 \cdot \text{area} = 1$ , using the smallest value of the integrand.
Upper bound: $e^{2} \cdot \text{area} \approx 7.389$ , using the largest.
The true value $2.9525$ sits comfortably inside, which is the cheapest sanity check available on any double integral.

Frequently asked questions

Is (e-1)^2 the same as e^2 - 1?

No, and the difference is large. $(e-1)^2 = e^2 - 2e + 1 \approx 2.9525$ , while $e^2 - 1 \approx 6.389$ . Squaring a difference is not the same as subtracting from a square.

What changes if the square becomes 0 to 2 in each variable?

The same split applies, with each single integral becoming $\int_{0}^{2} e^{t}\,dt = e^2 - 1$ . The double integral is then $(e^2-1)^2 \approx 40.82$ .