I don't know if there is a smart way. My dumb way is find the equation of the 3 sides of the triangle, and the integration area is the big triangle formed by y=0, x=3, and y=(4/3)*x, that is, the lower limit of y integral from 0 to upper limit of (4/3)*x, and x integral limits from 0 to 3, then minus the triangle formed by y=0, x=2, and y = (1/2)*x using similar way to find the limits, and then minus the rectangle formed by y=0, x=3, y=1, x=2, of which the limit is obvious, and finally minus the triangle formed by y=1, x=3, and y=((4-1)/(3-2))*(x-2)+1, or y=3*x=1, with the limits for y to be 1 to 3x+1, and the limits for x to be from 2 to 3.

If someone has a smarter way to do it, please share with all of us.

Another way is to find the sum of two triangles, with first one formed by y=(1/2)x, x=2, y=(4/3)x, with y limits from (1/2)x to (4/3)x, and x limits from 0 to 2, plus the second one formed by x=2, y=3x+1, and y=(4/3)x. with y limits from (3x+1) to (4/3)x, and x limits from 2 to 3. It is faster, but not that obvious.

I don't know if there is a smart way. My dumb way is find the equation of the 3 sides of the triangle, and the integration area is the big triangle formed by y=0, x=3, and y=(4/3)*x, that is, the lower limit of y integral from 0 to upper limit of (4/3)*x, and x integral limits from 0 to 3, then minus the triangle formed by y=0, x=2, and y = (1/2)*x using similar way to find the limits, and then minus the rectangle formed by y=0, x=3, y=1, x=2, of which the limit is obvious, and finally minus the triangle formed by y=1, x=3, and y=((4-1)/(3-2))*(x-2)+1, or y=3*x=1, with the limits for y to be 1 to 3x+1, and the limits for x to be from 2 to 3.

If someone has a smarter way to do it, please share with all of us.

Another way is to find the sum of two triangles, with first one formed by y=(1/2)x, x=2, y=(4/3)x, with y limits from (1/2)x to (4/3)x, and x limits from 0 to 2, plus the second one formed by x=2, y=3x+1, and y=(4/3)x. with y limits from (3x+1) to (4/3)x, and x limits from 2 to 3. It is faster, but not that obvious.