Multivariable calculus

Double Integral of e^(y^2) Over a Triangle

Q: Is there really no antiderivative for e^(y^2)?

Not among elementary functions. Liouville's theorem settles it: no finite combination of powers, exponentials, logarithms and trigonometric functions differentiates to e^{y^2}. The related \int e^{-y^2}\,dy is written with the error function for the same reason.

Q: How do I spot an order-reversal problem?

The inner integral has no closed form while the integrand depends on only one variable. Reversing puts that variable on the outside, and the inner step contributes a factor of the other variable, which is usually the missing piece a substitution needs. The classic three are e^{y^2}, \sin\!\left(y^2\right) and \frac{\sin y}{y}.

The double integral of e^(y^2) over the triangle with 0 <= x <= y and 0 <= y <= 1 equals (e - 1)/2, about 0.859141. Integrating in x first is the only workable order: it produces y e^(y^2), whose antiderivative is e^(y^2)/2. The other order has no elementary inner integral.

f(x,y) = e^{y^2}

\int_0^1 \int_0^{y} e^{y^2}\,dx\,dy = \frac{e-1}{2}

Numerically 0.859141, confirmed by quadrature on every build.

Reverse the order, because e^(y^2) has no elementary antiderivative

Written as a type 1 region the triangle is $0 \le x \le 1$ with $x \le y \le 1$ , giving $\int_0^1\int_x^1 e^{y^2}\,dy\,dx$ . The inner integral is a dead end: $e^{y^2}$ has no antiderivative among the elementary functions, which is exactly why this problem is set.

Describe the same triangle as type 2 instead. For each $y$ in $[0,1]$ the horizontal slice runs from $x = 0$ to $x = y$ .

\int_0^1 \int_0^{y} e^{y^2}\,dx\,dy = \int_0^1 e^{y^2}\left[x\right]_{x=0}^{x=y}\,dy = \int_0^1 y\,e^{y^2}\,dy

The inner step manufactured a factor of $y$ , and that factor is precisely what the substitution $u = y^2$ , $du = 2y\,dy$ requires.

\int_0^1 y\,e^{y^2}\,dy = \left[\frac{e^{y^2}}{2}\right]_0^1 = \frac{e-1}{2}

The mistake: swapping limits without redrawing the region

The mechanical swap turns $\int_0^1\int_x^1 e^{y^2}\,dy\,dx$ into $\int_x^1\int_0^1 e^{y^2}\,dx\,dy$ , which is meaningless since outer limits can never contain a variable. The panicked version, $\int_0^1\int_0^1$ , integrates over the whole unit square and gives a different number.

Reversing the order means re-describing the region, and a sketch is the only reliable route. This triangle has vertices $(0,0)$ , $(0,1)$ and $(1,1)$ . Read by columns it is $x \le y \le 1$ ; read by rows it is $0 \le x \le y$ .

Outer limits are always numbers.
Inner limits are the two boundaries you meet crossing the region in the inner variable's direction.
Check a corner: at $y = 1$ the row runs from $x = 0$ to $x = 1$ , the full top edge of the triangle.

A second slip is claiming $\frac{e^{y^2}}{2y}$ as an antiderivative of $e^{y^2}$ . Differentiating it returns $e^{y^2} - \frac{e^{y^2}}{2y^2}$ , which is not the integrand. Chain rule factors cannot be divided out after the fact.

Why (e-1)/2 is the right size

On this triangle $e^{y^2}$ runs from $1$ up to $e \approx 2.718$ , and the triangle has area $\frac12$ . So the answer had to land between $\frac12$ and $\frac{e}{2} \approx 1.359$ .

It lands at $0.859141$ . Dividing by the area gives the average value $e - 1 \approx 1.718$ , comfortably between the minimum $1$ and the maximum $e$ , as it must be for a continuous function on a connected region.

Frequently asked questions

Is there really no antiderivative for e^(y^2)?

Not among elementary functions. Liouville's theorem settles it: no finite combination of powers, exponentials, logarithms and trigonometric functions differentiates to $e^{y^2}$ . The related $\int e^{-y^2}\,dy$ is written with the error function for the same reason.

How do I spot an order-reversal problem?

The inner integral has no closed form while the integrand depends on only one variable. Reversing puts that variable on the outside, and the inner step contributes a factor of the other variable, which is usually the missing piece a substitution needs. The classic three are $e^{y^2}$ , $\sin\!\left(y^2\right)$ and $\frac{\sin y}{y}$ .