Double Integrals over General Regions — Question 6

Question 6 – Solution

Strategy. Draw the triangle $0\le y\le x\le 1$; reversing the order creates an inner interval whose length supplies the factor needed for substitution.

Step 1: Region The displayed bounds say $0\le y\le 1$ and $y\le x\le 1$. Equivalently, $$D=\{(x,y):0\le x\le 1,\ 0\le y\le x\}.$$

Step 2: Why reverse In the given order, the inner antiderivative $\int e^{x^2}dx$ is not elementary. With $dy$ first, $e^{x^2}$ is constant in the inner variable.

Step 3: Reversed evaluation $$\begin{align*} I&=\int_0^1\int_0^x e^{x^2}\,dy\,dx =\int_0^1xe^{x^2}\,dx\\ &=\frac 12\int_0^1e^u\,du =\boxed{\frac{e-1}{2}},\qquad u=x^2. \end{align*}$$ The integrand and region are positive, so the positive result is consistent. Differentiating $e^{x^2}/2$ returns $xe^{x^2}$.