Results 1 to 6 of 6

Thread: Exponential double integral

  1. #1
    Junior Member
    Joined
    Nov 2007
    Posts
    53

    Exponential double integral

    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:

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

    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 $\displaystyle u = x+y$ so $\displaystyle du = 1 + y dx$, 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 $\displaystyle \frac{1}{y}du=1dx$ so I have:

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

    Or would I do something different?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,656
    Thanks
    14
    You set up correctly the double integral.

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

    What does mean $\displaystyle A$? Area?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Quote Originally Posted by emttim84 View Post
    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:

    $\displaystyle Average = \frac{1}{y} \int_0^1 \int_0^y (e^{x+y}) dxdy$
    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 $\displaystyle u = x+y$ so $\displaystyle du = 1 + y dx$, 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 $\displaystyle 2\int_0^1\!\! \int_0^y e^{x+y}\, dxdy = 2\int_0^1\! \left(\int_0^y e^xe^y\, dx\right)dy$. As far as the inner integral is concerned, y is a constant, so you can integrate it as $\displaystyle 2\int_0^1 \!\!\left(e^y\bigl[ e^x\bigr]_0^y\right)dy = 2\int_0^1\!\!(e^{2y}-e^y)dy$. The rest I'll leave to you.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Nov 2007
    Posts
    53
    Wow, yeah I feel kinda embarrassed after missing the fact that $\displaystyle x+y$ with Y as the constant is a lot easier than what I must have been thinking, $\displaystyle xy$.

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

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

    Integrating both terms, I have:

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

    Evaluating them from 0 to 1 I have:

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

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

    Look good to you guys?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,656
    Thanks
    14
    No, the value of the double integral is $\displaystyle (e-1)^2.$ Check your computations when evaluating the last integral.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Nov 2007
    Posts
    53
    Ok, integrating them without making algebra mistakes, I have:

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

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

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

    Simplifes to:

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

    And $\displaystyle e^2 - 2e^1 + 1 = (e-1)^2$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Jun 2nd 2010, 02:25 AM
  2. double exponential decay and tau calculation
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: Mar 11th 2010, 06:27 PM
  3. Marginal Distribution - Double exponential
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: Dec 9th 2009, 07:08 AM
  4. Help with double integral of exponential
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Dec 5th 2008, 09:51 PM
  5. Replies: 6
    Last Post: May 18th 2008, 06:37 AM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum