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## Homework Statement

Evaluate ##\int\int_{R} (x+2)(y+1) \; dx \; dy## where ##R## is the pentagon with vertices ##(\pm 1,0)##, ##(\pm 2,1)## and ##(0,2)##.

## Homework Equations

## The Attempt at a Solution

After drawing ##R## I split ##R## into two sections ##R_1## (left half) and ##R_2## (right half) so we now calculate $$\int\int_{R_1} (x+2)(y+1) \; dx \; dy+\int\int_{R_2} (x+2)(y+1) \; dx \; dy.$$

From ##R_1## we have ##y = 2+\frac{x}{2} \Longleftrightarrow x = 2y-4##.and ##y = -1+x \Longleftrightarrow x = y+1##. From ##R_2## we have ##y = 2-\frac{x}{2} \Longleftrightarrow x = 4-2y## and ##y = -1-x \Longleftrightarrow x = -1-y##.

Does this inner integral become

$$\int_{y+1}^{2y-4} (x+2)(y+1) \; dx+\int_{-1-y}^{4-2y} (x+2)(y+1) \; dx.$$

Then what does the outer integral become (clearly the bounds of integration are scalars but what?)