Multivariable calculus

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

Q: Why does e^(x + y) factor when sin(x + y) does not?

The exponent law e^{a+b} = e^a e^b splits the exponential into one factor per variable. Sine has no such law: \sin(x+y) expands to \sin x\cos y + \cos x \sin y, a sum of two products rather than one product.

Q: What is the value over the rectangle 0 to 2 by 0 to 3?

The same factoring gives \left(e^2 - 1\right)\left(e^3 - 1\right) \approx 121.94. The integrand grows fast, so most of that total comes from the corner near (2,3).

The double integral of e to the x plus y over the unit square equals (e minus 1) squared, about 2.95249. Because e to the x plus y equals e to the x times e to the y, the integrand factors, and the answer is the square of the integral of e to the x from 0 to 1, which is e minus 1.

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

\int_{0}^{1}\int_{0}^{1} e^{x+y}\,dy\,dx = (e-1)^2 \approx 2.95249

Numerically 2.952492, confirmed by quadrature on every build.

The exponent law does the factoring for you

A sum in the exponent is a product of exponentials, which is exactly the shape the separable shortcut wants.

e^{x+y} = e^x e^y

\int_{0}^{1}\int_{0}^{1} e^x e^y\,dy\,dx = \left(\int_{0}^{1} e^x\,dx\right)\left(\int_{0}^{1} e^y\,dy\right) = (e-1)^2

Numerically $e - 1 \approx 1.71828$ , so the answer is about $1.71828^2 \approx 2.95249$ . Both factors are the same integral because the square is symmetric in $x$ and $y$ .

The direct route, for when the law is not obvious

You do not need the exponent law. Integrate in $y$ with $x$ frozen and the constant $e^x$ appears on its own.

\int_{0}^{1} e^{x+y}\,dy = \left[e^{x+y}\right]_{y=0}^{y=1} = e^{x+1} - e^{x} = e^{x}(e-1)

\int_{0}^{1} e^{x}(e-1)\,dx = (e-1)\left[e^x\right]_{0}^{1} = (e-1)^2

The constant $e - 1$ pulls out of the outer integral because it holds no $x$ . Recognising which pieces are constants at each stage is the whole skill in iterated integration.

The mistake: dividing by the exponent

Writing $\int e^{x+y}\,dy = \frac{e^{x+y}}{x+y}$ is wrong. You divide by the derivative of the exponent with respect to the integration variable, not by the exponent itself, and here that derivative is 1.

$\int e^{x+y}\,dy = e^{x+y}$ , since $\frac{\partial}{\partial y}(x+y) = 1$ .
$\int e^{x+3y}\,dy = \frac{e^{x+3y}}{3}$ , since the derivative is 3.
$\int e^{xy}\,dy = \frac{e^{xy}}{x}$ for $x \neq 0$ , since the derivative is $x$ .

Compare this with $\sin(x+y)$ , which does not factor at all. Exponentials turn sums into products and sines do not, so the same looking argument gets very different treatment.

Frequently asked questions

Why does e^(x + y) factor when sin(x + y) does not?

The exponent law $e^{a+b} = e^a e^b$ splits the exponential into one factor per variable. Sine has no such law: $\sin(x+y)$ expands to $\sin x\cos y + \cos x \sin y$ , a sum of two products rather than one product.

What is the value over the rectangle 0 to 2 by 0 to 3?

The same factoring gives $\left(e^2 - 1\right)\left(e^3 - 1\right) \approx 121.94$ . The integrand grows fast, so most of that total comes from the corner near $(2,3)$ .