Multivariable calculus

Double Integral of exp(-x-y) Over a Triangle

Q: Would integrating in the other order be harder?

No, and by symmetry it gives the same expression. With y outer from 0 to 1 and x inner from 0 to 1-y, the inner integral is e^{-y} - e^{-1} and the answer is again 1 - 2/e.

Q: How large is 1 - 2/e compared with the area of the region?

The triangle has area 0.5 and the integral is about 0.2642, so the average value of e^{-x-y} on the triangle is roughly 0.528. That sits sensibly between e^{-1} \approx 0.368 and 1.

The double integral of e to the power minus x minus y over the triangle with vertices at the origin, (1,0) and (0,1) equals 1 - 2/e, about 0.2642. The inner integral in y runs from 0 to 1 - x and gives e to the minus x, minus 1/e, which then integrates to the answer.

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

1 - \dfrac{2}{e}

Numerically 0.264241, confirmed by quadrature on every build.

Reading the triangle into bounds

The region is bounded by $x = 0$ , $y = 0$ and the line $x + y = 1$ . Solving that line for $y$ gives the ceiling $y = 1 - x$ , and $x$ itself ranges over $[0,1]$ .

\iint_{D} e^{-x-y}\,dA = \int_{0}^{1}\!\!\int_{0}^{1-x} e^{-x-y}\,dy\,dx

Split the exponential as $e^{-x}e^{-y}$ so the constant factor comes out of the inner integral.

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

The outer integral, and why the constant term appears

The inner answer simplified because $e^{-x} \cdot e^{x-1} = e^{-1}$ . That constant is the value of the integrand along the hypotenuse: every vertical slice stops on the line $x + y = 1$ , and $e^{-x-y}$ equals $e^{-1}$ everywhere on it, whichever slice you are on.

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

Numerically $1 - 2/e \approx 0.2642$ . The integrand is positive throughout, and the triangle has area $1/2$ , so a value between $0$ and $1/2$ is exactly what to expect.

The mistake: integrating over the square instead of the triangle

Because $e^{-x-y}$ splits so cleanly, it is tempting to use the product shortcut and integrate each variable from $0$ to $1$ independently.

\left(\int_{0}^{1} e^{-x}\,dx\right)\left(\int_{0}^{1} e^{-y}\,dy\right) = \left(1 - e^{-1}\right)^{2} \approx 0.3996

That is the integral over the whole unit square, which has twice the area of the triangle and gives $0.3996$ instead of $0.2642$ . The product shortcut requires constant bounds, and $y = 1-x$ is not constant.

The difference of about $0.1353$ is the contribution of the upper triangle where $x + y > 1$ , which the correct setup excludes.

Frequently asked questions

Would integrating in the other order be harder?

No, and by symmetry it gives the same expression. With $y$ outer from $0$ to $1$ and $x$ inner from $0$ to $1-y$ , the inner integral is $e^{-y} - e^{-1}$ and the answer is again $1 - 2/e$ .

How large is 1 - 2/e compared with the area of the region?

The triangle has area $0.5$ and the integral is about $0.2642$ , so the average value of $e^{-x-y}$ on the triangle is roughly $0.528$ . That sits sensibly between $e^{-1} \approx 0.368$ and $1$ .