How can one find the value of this integral?
$$\int_M \frac{20x^3}{1+y^5}d(x,y)$$
with $M := \{(x,y)^T \in \mathbb{R}^2: 0 \leq x \leq 1, x \leq y \leq 1 \}$
So we integrate over the rectangle M and I think that we can rewrite this as
$$\int_x^1 \int_0^1 \frac{20x^3}{1+y^5}$$
But how do we continue? What about the x?
On
First draw a picture. The region involved is the boundary and interior of the triangle with vertices (0,0), (0,1) and (1,1). Shade in the region and look at it carefully. Then you integrate $$ \frac{20x^3}{1+y^5}$$ with respect to x, remembering that $y$ is being held constant, from $x=0$ to $x=y$, obtaining $$\frac{5y^4}{1+y^5}$$. You must integate this expression with respect to $y$ from $y=0$ to $y=1$, which is extremely easy, because the numerator is the derivative of the denominator. A sensible question is "Why did we integrate first with respect to $x$ and then with respect to $y$ ? Could we not have integrated first with respect to $y$ and then with respect to $x$ ?" We could have done the question that way, eventually obtaining the same answer, but it would have been MUCH harder. One of the techniques for success in mathematics is to be intelligently lazy.
The set $M$ is a triangle, not a rectangle. The idea is to compute$$\int_0^1\int_x^1\frac{20x^3}{1+y^5}\,\mathrm dy\,\mathrm dx,$$but this is hard to do directly. Instead, compute\begin{align}\int_0^1\int_0^y\frac{20x^3}{1+y^5}\,\mathrm dx\,\mathrm dy&=\int_0^1\frac{5y^4}{1+y^5}\,\mathrm dy\\&=\left[\log(1+y^5)\right]_{y=0}^{y=1}\\&=\log2.\end{align}