Change the order of integration and evaluate the double integral of e^(y^3) with x from 0 to 2 and y from the square root of 2x up to 2.

Step-by-step solution

1. 1

   See why the given order is a dead end. The inner integral asks for $\int e^{y^3}\,dy$, and $e^{y^3}$ has no elementary antiderivative. Swapping the order is not a stylistic choice here - it is the only way to finish, because integrating in $x$ first produces an extra factor of $y^2$, which is exactly what the substitution $u=y^3$ needs.

2. 2

   Describe the region from the limits. The limits say $0\le x\le 2$ and $\sqrt{2x}\le y\le 2$. The lower boundary $y=\sqrt{2x}$ is the right half of the parabola $y^2=2x$, i.e. $x=\dfrac{y^2}{2}$. Note that at $x=2$ the lower limit already equals the upper limit $y=2$, so the region closes there.

3. 3

   Re-slice horizontally. For a fixed $y$ between $0$ and $2$, the region runs from the left edge $x=0$ across to the parabola $x=\dfrac{y^2}{2}$. So

   $R=\left\{(x,y):0\le y\le 2,\ 0\le x\le \tfrac{y^2}{2}\right\},\qquad \int_0^2\!\!\int_0^{y^2/2} e^{y^3}\,dx\,dy.$

4. 4

   Do the inner integral. $e^{y^3}$ is constant with respect to $x$, so it just picks up the width of the slice:

   $\int_0^{y^2/2}e^{y^3}\,dx=\frac{y^2}{2}\,e^{y^3}.$

5. 5

   Substitute $u=y^3$. Then $du=3y^2\,dy$, so $y^2\,dy=\tfrac{du}{3}$, and the limits $y=0,2$ become $u=0,8$:

   $\int_0^2\frac{y^2}{2}e^{y^3}\,dy=\frac{1}{6}\int_0^8 e^{u}\,du=\frac{1}{6}\left[e^u\right]_0^8=\frac{e^8-1}{6}.$

6. 6

   Sanity-check the size. $\dfrac{e^8-1}{6}\approx 496.66$, and a direct numerical evaluation of the original iterated integral gives $496.66$ as well, so the swap preserved the region correctly.