Equivalent form of a double integral. I am looking at the second question of this problem set:

The iterated integral $\int_0^1 \int_{y/2}^1 e^{x^2} dx \, dy$ can be expressed as
(a) $\int_0^1 \int_0^{2x} e^{x^2} dy \, dx$
(b) $\int_0^1 \int_0^{2y} e^{x^2} dy \, dx$
(c) $\int_{y/2}^1 \int_0^1 e^{x^2} dy \, dx$
(d) $\int_0^1 \int_0^{2y} e^{x^2} dx \, dy$
(e) $\int_0^1 \int_y^1 e^{x^2} dy \, dx + \int_0^1 \int_0^y e^{x^2} dx \, dy$

The first thing I tried was to change the order of integration which gave me
$$
\int_0^1 \int_{y/2}^1 e^{x^2} dx \, dy =
\int_0^{1/2}\int_0^{2x}e^{x^2} dy \, dx + \int_{1/2}^1\int_0^1 e^{x^2} dy \, dx = 
e^{1/4}-1+\int_{1/2}^1e^{x^2} dx
$$
Since this didn't look like any of the available options, I tried to exclude them:


*

*I excluded (b),(c) and (e) as these are (non-constant) functions of $y$, in contrast to the original integral.

*(a) is wrong because it integrates the same positive function $f(x,y)=e^{x^2}$ over a proper superset of the original area of integration (with the difference of the two areas having positive measure).

*(d) is greater than the original: It integrates the same positive function over a different area, say $B$, than the original area, call it $A$. $B \setminus A$ contains the triangle $T$ defined by the points $(1,\frac{1}{2}),(1,1),(2,1)$ which is congruent to $A \setminus B$. The values of $e^{x^2}$ are greater at the points of $T$ compared to the points of $A \setminus B$. Alternatively, using wolframalpha: original $ \approx 1.2$, (d) $\approx 3.05$
So, I have managed to exclude every option. What am I missing?
 A: Note that the original region of integration $R$ is the trapezoid with vertices $(0,0), (1,0), (1,1), (1/2,1)$.
(a) is incorrect because the region of integration is a strict superset of $R$, containing all points in $R$ yet also including the triangle $2 > 2x > y > 1$.
(b) is notationally incorrect because the inner integral is evaluated with respect to $y$ but the upper limit of integration is a function of $y$.
(c) is also notationally incorrect since the outermost limit of integration is not a fixed interval and thus cannot evaluate to a number.
(d) is incorrect because the region of integration is a triangle with vertices $(0,0), (2,1), (0,1)$.  It is easy to see--even without computation--that the integral over the region formed by the union of the two triangular regions $1 > y > 2x > 0$ and $1/2 < x/2 < y < 1$ is much, much greater than that of the triangular region $0 < y < x/2 < 1/2$, since the integrand is constant with respect to $y$, and very rapidly increasing as a function of $x$.
(e) is notationally incorrect for the same reason that (b) is incorrect.
