Average value of a double integral over a region

[jonroberts74]

Homework Statement

f(x,y) = $e^{x+y}$ D is the triangle vertices (0,0), (0,1) , (1,0)

Homework Equations

$f(x,y)_{avg}=\frac{\iint_D f(x,y) dA}{\iint_D dA}$

The Attempt at a Solution

$\iint_D dA \Rightarrow \int_{0}^{1}\int_{0}^{-y+1} dxdy = \frac{1}{2}$

$\iint_D f(x,y) dA \Rightarrow \int_{0}^{1}\int_{0}^{-y+1} e^{x+y}dxdy$

$\int_{0}^{1} e - e^y dy = 1$


$f(x,y)_{avg} = \frac{1}{1/2} = 2$

this doesn't seem correct.

 

[jonroberts74]

nvm, was thinking about it incorrectly

 

[HallsofIvy]

Homework Statement

f(x,y) = $e^{x+y}$ D is the triangle vertices (0,0), (0,1) , (1,0)

Homework Equations

$f(x,y)_{avg}=\frac{\iint_D f(x,y) dA}{\iint_D dA}$

The Attempt at a Solution

$\iint_D dA \Rightarrow \int_{0}^{1}\int_{0}^{-y+1} dxdy = \frac{1}{2}$

Or: the area of a triangle is (1/2)(height)(base)= (1/2)(1)(1)= 1/2.

$\iint_D f(x,y) dA \Rightarrow \int_{0}^{1}\int_{0}^{-y+1} e^{x+y}dxdy$

And this you can write as
$$\int_0^1 e^x\left(\int_0^{1- x}e^y dy\right)dx$$
$$= \int_0^1 e^x\left[e^y\right]_0^{1- x}dx= \int_0^1 e^x\left[e^{1- x}- 1\right]dx$$
$$=\int_0^1 e- e^x dx= \left[ex- e^x\right]_0^1= (e- e)- (0- 1)= 1$$

$\int_{0}^{1} e - e^y dy = 1$


$f(x,y)_{avg} = \frac{1}{1/2} = 2$

this doesn't seem correct.

Yes, it is correct.