Multivariable calculus

Double Integral of x exp(xy) Over the Unit Square

Q: Why does the inner integral come out to e^x - 1 with no extra constant?

Because \frac{\partial}{\partial y}e^{xy} = x\,e^{xy}, the integrand is already the exact y derivative of e^{xy}. Evaluating at y=1 and y=0 gives e^{x} - e^{0} = e^{x}-1.

Q: Is e - 2 really positive?

Yes. e \approx 2.71828, so e - 2 \approx 0.7183. The integrand is non-negative on the square, so a negative answer would have signalled an arithmetic slip.

The double integral of x times e to the power xy over the unit square, with x and y each from 0 to 1, equals e - 2, about 0.7183. Integrating in y first is the key: the inner integral collapses to e to the x, minus 1, with no integration by parts needed.

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

e-2

Numerically 0.718282, confirmed by quadrature on every build.

Choose the order that makes the inner integral trivial

Hold $x$ fixed and integrate in $y$ . Then $x$ is a constant, and the leading $x$ is exactly the constant that $e^{xy}$ needs when you undo the chain rule.

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

The $x$ out front vanished into the antiderivative rather than surviving as a factor, which is what makes this order so cheap. Now finish in $x$ .

\int_{0}^{1}\left(e^{x} - 1\right)\,dx = \Big[e^{x} - x\Big]_{0}^{1} = (e - 1) - 1 = e - 2 \approx 0.7183

What the other order costs you

Both orders are legal, since the integrand is continuous on the square. They are not equally pleasant. Integrating in $x$ first means holding $y$ fixed and facing $\int_{0}^{1} x\,e^{xy}\,dx$ , which needs integration by parts.

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

It looks singular at $y = 0$ , but the singularity is removable: the limit there is $\frac{1}{2}$ , which is just $\int_{0}^{1} x\,dx$ . To finish, you have to recognise the expression as a quotient-rule derivative.

\frac{d}{dy}\left(\frac{e^{y}-1}{y}\right) = \frac{e^{y}}{y} - \frac{e^{y}-1}{y^{2}} \quad \Longrightarrow \quad \int_{0}^{1}\left(\frac{e^{y}}{y} - \frac{e^{y}-1}{y^{2}}\right)\,dy = \left[\frac{e^{y}-1}{y}\right]_{0}^{1} = (e-1) - 1 = e - 2

So the other order does close, and on the same $e - 2$ . It costs integration by parts, a removable singularity to argue about, and a derivative you have to read backwards. The first order cost one line.

The habit worth building: before integrating anything, glance at both inner integrals and pick the one that needs no technique at all.

The mistake: treating the leading x as an inert constant

A frequent error is to pull $x$ out of the inner integral and then also treat $e^{xy}$ as if the $x$ inside it were not there, writing $x \int_{0}^{1} e^{y}\,dy = x(e-1)$ .

That drops the $x$ from the exponent, which changes the function entirely. Test it inside the square at $x = \tfrac{1}{2}$ , $y = 1$ : the real integrand is $\tfrac{1}{2}e^{1/2} \approx 0.824$ , the corrupted one is $\tfrac{1}{2}e \approx 1.359$ .

Pulling a factor out is legal, forgetting a variable inside a composition is not. When you factor $x$ out of $x\,e^{xy}$ , the $e^{xy}$ that remains still depends on $x$ , so it cannot be integrated as $e^{y}$ .

Frequently asked questions

Why does the inner integral come out to e^x - 1 with no extra constant?

Because $\frac{\partial}{\partial y}e^{xy} = x\,e^{xy}$ , the integrand is already the exact $y$ derivative of $e^{xy}$ . Evaluating at $y=1$ and $y=0$ gives $e^{x} - e^{0} = e^{x}-1$ .

Is e - 2 really positive?

Yes. $e \approx 2.71828$ , so $e - 2 \approx 0.7183$ . The integrand is non-negative on the square, so a negative answer would have signalled an arithmetic slip.