Pretty simple question but I have no one to compare with.
We are asked to find the integral $\int \int _A (x^2-2y)dA$ where $A$ is the triangle $\{(1,0),(2,1),(3,4)\}$
What I did:
I found the equation of each side of the triangle, they are
$y=x-1$, $y=2x-2$, $y=3x-5$.
I partitioned the triangle by connecting the dot $(2,1)$ and the dot $(2,2)$. This made it so I have 2 simple regions instead of 1 complex one.
Now I just need to find the integrals $\int \int _{A_1} (x^2-2y)dA_1$ and $\int \int _{A_2} (x^2-2y)dA_2$ and add them, where $A_1=\{(1,0),(2,1),(2,2)\}$ and $A_2=\{(2,1),(2,2),(3,4)\}$
$\int \int _{A_1} (x^2-2y)dA_1 = \int_{1}^{2} \int_{x-1}^{2x-2} (x^2-2y)dydx =\int_{1}^{2} x^2-(2x-2)^2-x^2+(x-1)^2 dx = \int_{1}^{2} -3x^2+2x-3 dx=-7$
$\int \int _{A_2} (x^2-2y)dA_2 = \int_{2}^{3} \int_{3x-5}^{2x-2} (x^2-2y)dydx = \int_{2}^{3} x^2-(2x-2)^2-x^2+(3x-5)^2 = \int_{2}^{3} 5x^2-26x+21dx=-\frac{37}{3}$
$\int \int _{A_1} (x^2-2y)dA_1+\int \int _{A_2} (x^2-2y)dA_2=-7-\frac{37}{3}=-\frac{58}{3}$
Is this correct?
You did not need to partition the triangle. However, your regions of integration are correct, although you have some errors. $\iint_{A_1}(x^2-2y)\mathrm dA_1=\frac{5}{12},$ not $-7$. Specifically, note that $$\begin{aligned}\int_1^2\int_{x-1}^{2x-2}(x^2-2y)dydx&=\int_1^2\left[x^2y-y^2\right]_{x-1}^{2x-2}dx\\&=\int_1^2[x^2(2x-2)-(2x-2)^2-x^2(x-1)+(x-1)^2]dx\\&=\int_1^2[2x^3-2x^2-4x^2+8x-4-x^3+x^2+x^2-2x+1]dx\\&=\int_1^2 (x^3-4x^2+6x-3)dx\\&=\left[\frac{x^4}{4}-\frac{4x^3}{3}+3x^2-3x\right]_{1}^2\\&=4-\frac{32}{3}+12-6-\frac14+\frac43-3+3\\&=10-\frac{28}{3}-\frac14\\&=\frac{120-112-3}{12}\\&=\frac{5}{12}\end{aligned}$$
Also, the second integral is incorrect as well. $\iint_{A_2}(x^2-2y)\mathrm dA_2=\frac{5}{12},$ not $-\frac{37}{3}$. Just go through the integration like I did above and you will find the answer. Be careful! Its easy to get lost in these types of calculations.
Summing both integrals, you get that the total area of this triangle is $2\left(\frac{5}{12}\right)=\frac56$