Double integral $$ \int\int_R x^3 dA $$
The region is bounded by the triangle (0, 0), (7, 1), (1, 7)
Here is my thinking, I need to set up my bounds such that
$$\int_0^7\int_{\frac{1}{7}y}^{-y+8} x^3 dxdy$$
If this is correct I find the area by taking the integral with respect to x , evaluate it and take the second integral with respect to y and evaluate it.
Does this seem correct?
 A: For direct integration to work, you need the following integral:
$$
\int_0^7\int^{\min(7x, 8-x)}_{\frac{x}{7}} x^3 dydx
$$
since for each $x$, the vertical line segment inside the region goes from $\left(x,\frac{x}{7}\right)$ up to whichever edge of the triangle it hits first.
Consider my rather crude picture of your region below:

You need to do the integral over each of the two regions (yellow and blue) separately (that is what the $\min$ in the integral bound amounts to). Therefore you get
$$
\int_0^1\int_{\frac x7}^{7x} x^3 dydx + \int_1^7\int_{\frac x7}^{8-x} x^3dydx
$$
You can, of course, do the integral in the direction $dxdy$, in which case it will, by the same argument, be
$$
\int_0^1\int_{\frac y7}^{7y} x^3 dxdy + \int_1^7\int_{\frac y7}^{8-y} x^3dxdy
$$
A: Alternative:
$$\int\int_{R}x^{3}dA=\int\int_{S}x^{3}dA-\int\int_{T}x^{3}dA$$ where
$S$ is the triangle determined by $\left(0,0\right),\left(1,7\right),\left(8,0\right)$
and $T$ is the triangle determined by $\left(0,0\right),\left(7,1\right),\left(8,0\right)$.
$$\int\int_{S}x^{3}dA-\int\int_{T}x^{3}dA=\frac{1}{2}\times8\times7-\left[\int_{0}^{7}\int_{0}^{\frac{1}{7}x}x^{3}dydx+\int_{7}^{8}\int_{0}^{8-x}x^{3}dydx\right]$$$$=28-\left[\int_{0}^{7}\frac{1}{7}x^{4}dx+\int_{7}^{8}x^{3}\left(8-x\right)dx\right]$$
