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Average value of a double integral over a region

  1. Jul 28, 2014 #1
    1. The problem statement, all variables and given/known data
    f(x,y) = ##e^{x+y}## D is the triangle vertices (0,0), (0,1) , (1,0)



    2. Relevant equations

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



    3. 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.
     
  2. jcsd
  3. Jul 28, 2014 #2
    nvm, was thinking about it incorrectly
     
  4. Jul 29, 2014 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

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

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


    Yes, it is correct.
     
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