Multivariable calculus

Double Integral of (x + y)^2 Over the Unit Square

Q: Why is there no factor of one half from the substitution?

Because u = x + y has \frac{\partial u}{\partial y} = 1 while x is held fixed, so du = dy. With (x + 2y)^2 the derivative would be 2, and the antiderivative in y would be \frac{(x+2y)^3}{6}.

Q: What is the value over the square from -1 to 1 in both variables?

It is \frac{8}{3}. Expanding, \iint x^2\,dA = \frac{4}{3} and \iint y^2\,dA = \frac{4}{3}, while the cross term 2\iint xy\,dA vanishes because x is odd on [-1,1].

The double integral of (x + y) squared over the unit square, with x and y each running from 0 to 1, equals 7/6, about 1.1667. Integrating in y first gives one third of (x + 1) cubed minus x cubed, and integrating that from 0 to 1 gives 7/6.

f(x,y) = (x + y)^2

\int_{0}^{1}\int_{0}^{1} (x+y)^2\,dy\,dx = \frac{7}{6}

Numerically 1.166667, confirmed by quadrature on every build.

The inner integral is a shift, so no extra factor appears

Hold $x$ fixed and substitute $u = x + y$ . Because $x$ is frozen, $\frac{\partial u}{\partial y} = 1$ and $du = dy$ exactly, with no constant to divide by.

\int_{0}^{1} (x+y)^2\,dy = \left[\frac{(x+y)^3}{3}\right]_{y=0}^{y=1} = \frac{(x+1)^3 - x^3}{3}

Both endpoints matter. The $x^3$ comes from the lower limit $y = 0$ , and dropping it is the single most common slip on this problem.

\int_{0}^{1} \frac{(x+1)^3 - x^3}{3}\,dx = \frac{1}{3}\left[\frac{(x+1)^4}{4} - \frac{x^4}{4}\right]_{0}^{1} = \frac{1}{3}\left(\frac{16 - 1}{4} - \frac{1 - 0}{4}\right) = \frac{7}{6}

Expanding first is the same work in a different order

Since $(x+y)^2 = x^2 + 2xy + y^2$ , one substitution becomes three monomials, and each of the three factors over the square.

\iint_R x^2\,dA + 2\iint_R xy\,dA + \iint_R y^2\,dA = \frac{1}{3} + 2\cdot\frac{1}{4} + \frac{1}{3} = \frac{7}{6}

Agreement between the two routes is the cheapest check available on your algebra, because past the setup they share no steps at all.

The mistake: squaring the integral instead of integrating the square

Computing $\iint_R (x+y)\,dA = 1$ and then squaring gives 1, not $\frac{7}{6}$ . Averaging and then squaring is never larger than squaring and then averaging, and it is strictly smaller unless the function is constant. The gap $\frac{7}{6} - 1 = \frac{1}{6}$ is exactly the variance of $x + y$ on the square.

$\int (x+y)^2\,dy = \frac{(x+y)^3}{3}$ needs $\frac{\partial}{\partial y}(x+y) = 1$ . With $(x+2y)^2$ the antiderivative is $\frac{(x+2y)^3}{6}$ instead.
Evaluate at both limits: $\frac{(x+1)^3}{3}$ alone is missing the $-\frac{x^3}{3}$ from $y = 0$ , and it gives $\frac{5}{4}$ rather than $\frac{7}{6}$ .
Expanding is never wrong here, only slower, so use it when the substitution feels shaky.

Sanity check the size: on the unit square $(x+y)^2$ runs from 0 to 4, and it exceeds 1 on exactly half the square, the half above the diagonal $x + y = 1$ . So an average just above 1 is right, and anything near 4 is not.

Frequently asked questions

Why is there no factor of one half from the substitution?

Because $u = x + y$ has $\frac{\partial u}{\partial y} = 1$ while $x$ is held fixed, so $du = dy$ . With $(x + 2y)^2$ the derivative would be 2, and the antiderivative in $y$ would be $\frac{(x+2y)^3}{6}$ .

What is the value over the square from -1 to 1 in both variables?

It is $\frac{8}{3}$ . Expanding, $\iint x^2\,dA = \frac{4}{3}$ and $\iint y^2\,dA = \frac{4}{3}$ , while the cross term $2\iint xy\,dA$ vanishes because $x$ is odd on $[-1,1]$ .