Multivariable calculus

Double Integral of x e^(xy) Over [0,1] by [0,2]

Q: How do I know to integrate in y first?

Look at the loose factor. Here it is x, and \frac{\partial}{\partial y}\left(e^{xy}\right) = x e^{xy}, so x is exactly what the y substitution consumes. Integrating in x first leaves x unmatched and forces integration by parts.

Q: What happens at x = 0, where the antiderivative has a 1/x in it?

Nothing goes wrong. The factor x in front cancels the \frac{1}{x} before the limits are applied, leaving e^{xy}. At x = 0 the inner integral is e^{0} - e^{0} = 0, which matches an integrand of 0 along that edge.

The double integral of x times e to the xy over the rectangle x from 0 to 1 and y from 0 to 2 equals (e squared minus 3) divided by 2, about 2.19453. Integrate in y first: the factor x is exactly the derivative the substitution needs, so the inner integral is e to the 2x minus 1.

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

\int_{0}^{1}\int_{0}^{2} x e^{xy}\,dy\,dx = \frac{e^2 - 3}{2} \approx 2.19453

Numerically 2.194528, confirmed by quadrature on every build.

Pick the order that avoids integration by parts

With $x$ frozen, $\frac{\partial}{\partial y} e^{xy} = x e^{xy}$ . The integrand is already an exact derivative in $y$ , so the inner integral needs no work at all.

\int_{0}^{2} x e^{xy}\,dy = \left[e^{xy}\right]_{y=0}^{y=2} = e^{2x} - 1

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

Numerically that is about 2.19453. The lower limit contributes the $-\frac{1}{2}$ from $\frac{e^0}{2}$ , which is easy to lose.

What the other order costs

Integrating in $x$ first means $\int x e^{xy}\,dx$ with $y$ frozen, and now the factor $x$ is not the derivative of the exponent. That needs integration by parts.

\int_{0}^{1} x e^{xy}\,dx = \frac{e^{y}}{y} - \frac{e^{y} - 1}{y^{2}}

That function is fine on $(0,2]$ and has a removable discontinuity at $y = 0$ , where the limit is $\frac{1}{2}$ . Integrating it from 0 to 2 does return 2.19453, so the order is a choice about labour rather than about correctness.

The rule of thumb: look for the variable whose differential the extra factor already matches. Here $x\,dy$ pairs with $d\left(e^{xy}\right)$ , so integrate in $y$ first.

The mistake: dividing by the wrong variable

The antiderivative of $e^{xy}$ with respect to $y$ is $\frac{e^{xy}}{x}$ : you divide by $\frac{\partial}{\partial y}(xy) = x$ . Dividing by $y$ instead produces $\frac{e^{xy}}{y}$ , whose $y$ derivative is not the integrand.

Integrating in $y$ , divide by $\frac{\partial}{\partial y}(xy) = x$ .
Integrating in $x$ , divide by $\frac{\partial}{\partial x}(xy) = y$ .
Here the leading $x$ cancels that $\frac{1}{x}$ exactly, which is what makes this order clean.

No division by zero sneaks in at $x = 0$ . The cancellation happens before the division, and the inner result $e^{2x} - 1$ is simply 0 there, matching an integrand that is 0 along that whole edge.

Frequently asked questions

How do I know to integrate in y first?

Look at the loose factor. Here it is $x$ , and $\frac{\partial}{\partial y}\left(e^{xy}\right) = x e^{xy}$ , so $x$ is exactly what the $y$ substitution consumes. Integrating in $x$ first leaves $x$ unmatched and forces integration by parts.

What happens at x = 0, where the antiderivative has a 1/x in it?

Nothing goes wrong. The factor $x$ in front cancels the $\frac{1}{x}$ before the limits are applied, leaving $e^{xy}$ . At $x = 0$ the inner integral is $e^{0} - e^{0} = 0$ , which matches an integrand of 0 along that edge.