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**emttim84** Instructions state to find the average value of $\displaystyle f(x,y)$ over the region $\displaystyle R$.

$\displaystyle f(x,y) = e^{x+y}$ with the region R being a triangle with vertices at (0,0), (0,1) and (1,1).

So the base formula for taking the average is:

$\displaystyle Average = \frac{1}{A} \int_R \int f(x,y) dA$

Drawing the graph, I find it is a triangle anchored at the origin. I'm approaching it from the dy position so I have points from $\displaystyle y=0$ to $\displaystyle y=2$ and it's entering the region in question at $\displaystyle x=0$ and exiting the region at $\displaystyle x=y$.

My double integral is as follows:

$\displaystyle Average = \frac{1}{y} \int_0^1 \int_0^y (e^{x+y}) dxdy$