$\iint_D (x^2+y)\mathrm dx\mathrm dy$ by definition

Question

Calculate $$\iint_D (x^2+y)\mathrm dx\mathrm dy$$ where $D = \{(x,y)\mid -2 \le x \le 4,\ 5x-1 \le y \le 5x+3\}$ by definition.

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Plotting the set $D$, we notice that it is a parallelogram. I tried to divide it into equal parallelograms, i.e. take $x_i = -2 + \frac{6i}{n}$ and $y_j = 5x - 1+ \frac{4j}{m}$. The definition requires to calculate the sum $$\sigma = \sum_{i=1}^n\sum_{j=1}^m f(\xi_i, \eta_j)\mu(D_{ij})$$ I was thinking about choosing $\xi_i = x_i$ and $\eta_j = y_j(x_i)$. However, this seems a bit strange. Also, finding the area of $D_{ij}$ doesn't seem to be convenient.

Any help is appreciated.

Answer 1

Your setup is fine. Notation-wise, your $\eta_j$ must be $\eta_{i,j}$ since it depends on both $i$ and $j$.

Having fixed that, you could directly compute the area of each sub-parallelogram, which would be the safer bet since it would ensure that they actually are all the same. Or you could take for granted that they are the same and just divide the total area by $mn$. Here the area is $6 \cdot 4=24$ so the area of each sub-parallelogram is $24/mn$.

Assuming that your setup is correct, that leaves you to look at

$$\frac{24}{mn} \sum_{i=1}^n \sum_{j=1}^m (-2+6i/n)^2 + 5(-2+6i/n)-1+4j/m$$

which is pretty straightforward, you just need the leading order asymptotics of $\sum_{i=1}^n i$ and $\sum_{i=1}^n i^2$.

Answer 2

You can integrate directly using Fubini's Theorem: $$\int_D f(x,y) dx dy=\int_{D_x}\bigg(\int_{D_y}f(x,y)dy\bigg)dx=\int_{D_y}\bigg(\int_{D_x}f(x,y)dx\bigg)dy$$

Where $D_x$ and $D_y$ represent the constraints on $x$ and $y$.

In this case: $$\iint_D (3-2x)(4y+5)dxdy=\int_{-2}^4\bigg(\int_{5x-1}^{5x+3}(3-2x)(4y+5)dy\bigg)dx\\=\int_{-2}^4(3-2x)\bigg(\int_{5x-1}^{5x+3}(4y+5)dy\bigg)dx$$

which is straightforward to evaluate by treating $x$ as a constant when integrating $dy$, and vice versa.