Exponential double integral

**emttim84** · March 11th 2008, 12:07 PM

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

$f(x,y) = e^{x+y}$ 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:

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

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 $y=0$ to $y=2$ and it's entering the region in question at $x=0$ and exiting the region at $x=y$ .

My double integral is as follows:

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

Here's where I run into trouble because I'm not sure how to use a U sub to my advantage on this problem. If I do $u = x+y$ so $du = 1 + y dx$ , well I don't have a Y in the function, but obviously I do have a 1, so how would I approach this problem? Would I do $\frac{1}{y}du=1dx$ so I have:

$Average = \frac{1}{y^2} \int_0^1 (e^{y} + e^{2y}) dy$

Or would I do something different?

**Krizalid** · March 11th 2008, 12:28 PM

You set up correctly the double integral.

By makin' $u=x+y$ for the inner integral, we have $du=dx,$ we just treat $y$ as a constant, since we're takin' the derivative with respect to $x.$ Now, find the integration limits and evaluate the rest.

What does mean $A$ ? Area?

**Opalg** · March 11th 2008, 12:33 PM

Originally Posted by emttim84

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

$f(x,y) = e^{x+y}$ 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:

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

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 $y=0$ to $y=2$ and it's entering the region in question at $x=0$ and exiting the region at $x=y$ .

My double integral is as follows:

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

What's the 1/y doing outside the integral? The quantity A is supposed to be the area of the region over which you are taking the average. Since this is a triangle, with base and height both equal to 1, its area is 1/2, so 1/A should be 2.

Originally Posted by emttim84

Here's where I run into trouble because I'm not sure how to use a U sub to my advantage on this problem. If I do $u = x+y$ so $du = 1 + y dx$ , well I don't have a Y in the function, but obviously I do have a 1, so how would I approach this problem?

You don't need to make a substitution. Think of the integral as $2\int_0^1\!\! \int_0^y e^{x+y}\, dxdy = 2\int_0^1\! \left(\int_0^y e^xe^y\, dx\right)dy$ . As far as the inner integral is concerned, y is a constant, so you can integrate it as $2\int_0^1 \!\!\left(e^y\bigl[ e^x\bigr]_0^y\right)dy = 2\int_0^1\!\!(e^{2y}-e^y)dy$ . The rest I'll leave to you.

**emttim84** · March 11th 2008, 02:18 PM

Wow, yeah I feel kinda embarrassed after missing the fact that $x+y$ with Y as the constant is a lot easier than what I must have been thinking, $xy$ .

So with the area being 2, here's what I have:

$Average = 2 \int_0^1 (e^{2y} - e^y) dy$

Integrating both terms, I have:

$2 \frac{1}{2} [e^{2y} - e^y]$ so the constants in front cancel out.

Evaluating them from 0 to 1 I have:

$Average = [(e^2 - e^1) - (e^0 - e^0)] = e^2 - e^1$

Final answer: $Average = e^2 - e^1$

Look good to you guys?

**Krizalid** · March 11th 2008, 02:24 PM

No, the value of the double integral is $(e-1)^2.$ Check your computations when evaluating the last integral.

**emttim84** · March 11th 2008, 02:40 PM

Ok, integrating them without making algebra mistakes, I have:

$Average = 2 \int_0^1 (e^{2y} - e^y) dy = 2 (\frac{1}{2}e^{2y} - e^y)$ evaluated from 0 to 1 which simplifies to:

$Average = e^{2y} - 2e^y$ evaluated from 0 to 1 which results in:

$Average = [(e^{2(1)} - 2e^1) - (e^{2(0)} - 2e^0)]$

Simplifes to:

$Average = e^2 - 2e^1 - (-1) = e^2 - 2e^1 + 1$

And $e^2 - 2e^1 + 1 = (e-1)^2$

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