Multivariable calculus

Double Integral of x + y Over the Unit Triangle

Q: Can I do this integral in the order dx dy?

Yes, and by symmetry the setup is the mirror image: \int_0^1 \int_0^{1-y} (x+y)\,dx\,dy, which again gives \frac13. Neither order is harder here, because the region and the integrand are both symmetric in x and y.

The double integral of x + y over the triangle bounded by x = 0, y = 0 and x + y = 1 equals 1/3. The inner integral in y collapses to (1 - x^2)/2, and integrating that from 0 to 1 gives 1/3. By symmetry, x and y each contribute 1/6.

f(x,y) = x + y

\int_0^1 \int_0^{1-x} (x+y)\,dy\,dx = \frac{1}{3}

Numerically 0.333333, confirmed by quadrature on every build.

Set up the slices, then push through the algebra

The slanted edge is $x + y = 1$ . Solving for the top of a vertical slice gives $y = 1 - x$ , so for $x$ in $[0,1]$ the slice runs from $y = 0$ up to $y = 1-x$ .

\int_0^{1-x} (x+y)\,dy = \left[xy + \frac{y^2}{2}\right]_{y=0}^{y=1-x} = x(1-x) + \frac{(1-x)^2}{2}

Expand before integrating, because the $x$ terms cancel in a way that is easy to miss.

x - x^2 + \frac{1 - 2x + x^2}{2} = \frac{1 - x^2}{2}

\int_0^1 \frac{1-x^2}{2}\,dx = \frac{1}{2}\left(1 - \frac{1}{3}\right) = \frac{1}{3}

The mistake: reading (1-x)^2 as 1 - x^2

Almost every wrong answer on this problem comes from the squared binomial. $(1-x)^2 = 1 - 2x + x^2$ , and the middle term is not optional. Dropping it leaves the inner value $x - \frac{3}{2}x^2 + \frac12$ and a final answer of $\frac12$ instead of $\frac13$ .

You cannot catch this at the endpoints, since both versions happen to agree at $x = 0$ and $x = 1$ . Test the middle instead. At $x = \frac12$ the slice is $0 \le y \le \frac12$ , so the inner integral is computable by hand.

\int_0^{1/2}\left(\frac{1}{2} + y\right)\,dy = \frac{1}{4} + \frac{1}{8} = \frac{3}{8}

The correct formula gives $\frac{1 - 1/4}{2} = \frac38$ . The faulty one gives $\frac58$ . One interior test point separates them.

Expand $(1-x)^2$ in full, every time.
Test the inner result at an interior value, not only at the ends.
The inner result must vanish at $x = 1$ , where the slice has zero length.

A symmetry check worth knowing

Swapping $x$ and $y$ maps this triangle onto itself, so $\iint_T x\,dA$ and $\iint_T y\,dA$ have to be equal. Each is $\frac16$ , and they must add to the answer.

\iint_T x\,dA = \int_0^1 x(1-x)\,dx = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}, \qquad \frac{1}{6} + \frac{1}{6} = \frac{1}{3}

The triangle has area $\frac12$ , so the average value of $x+y$ on it is $\frac{1/3}{1/2} = \frac23$ . That agrees with the centroid, which sits at $\left(\frac13, \frac13\right)$ , and $\frac13 + \frac13 = \frac23$ .

Frequently asked questions

Why is the average not 1/2, the midpoint of the values 0 and 1?

Because those values are not spread evenly over the triangle. The level set $x + y = c$ is a segment of length $c\sqrt{2}$ inside the region, so larger values of $x+y$ occupy more area. Weighting by that length gives $\frac{\int_0^1 c^2\,dc}{\int_0^1 c\,dc} = \frac{2}{3}$ .

Can I do this integral in the order dx dy?

Yes, and by symmetry the setup is the mirror image: $\int_0^1 \int_0^{1-y} (x+y)\,dx\,dy$ , which again gives $\frac13$ . Neither order is harder here, because the region and the integrand are both symmetric in $x$ and $y$ .