Evaluation of $\int_0^2\int_{y/ 2}^1 ye^{-x^3} dxdy$

Question

Evaluate $$\int_0^2\int_{y/2}^1 ye^{-x^3} dxdy$$

The integral cannot be evaluated by elementary methods, since $e^{-x^3}$ has no elementary antiderivative. However it can be interpreted as $\int\int_T ye^{-x^3}dA $ where $T$ is the triangle with vertices $(0,0),(1,0),(1,2)$.
Writing this double integral as an iterated integral in the other order leads to an easy calculation: $$\int_0^1\int_{0}^{2x} ye^{-x^3} dydx=\cdots={2\over 3}(1-e^{-1}) $$

How do we know $$\int_0^2\int_{y/2}^1 ye^{-x^3} dxdy$$ can be interpreted as $\iint_T ye^{-x^3}dA $ ? Where did that $T$ come from ?

Answer 1

$$T = \{ (x,y): 0 \le y \le 2, \frac{y}{2} \le x \le 1\}$$

Sketch out the picture and you can see the triangle.

Notice that we can describe $T$ also as

$$T = \{ (x,y): 0 \le x \le 1, 0 \le y \le 2x\}$$

Answer 2

The limits of integration for a$$\int_0^2\int_{y\over 2}^1 ye^{-x^3} dxdy$$ are $0\le y\le 2$ and for each $y$, $ y/2\le x\le 1.$

The graph of this region is a triangle bounded by the $x$-axis, the line $ y=2x$, and the vertical line $x=1$.

When you describe this triangle in the other order you will find the new limits as in $$ \int_0^1\int_{0}^{2x} ye^{-x^3} dydx=\cdots={2\over 3}(1-e^{-1})$$