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Math Help - Exponential double integral

  1. #1
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    Exponential double integral

    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?
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  2. #2
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    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?
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  3. #3
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    Quote Originally Posted by emttim84 View Post
    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.

    Quote Originally Posted by emttim84 View Post
    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.
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  4. #4
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    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?
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  5. #5
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    No, the value of the double integral is (e-1)^2. Check your computations when evaluating the last integral.
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  6. #6
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    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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