1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integral calculus involving Fubini

  1. May 2, 2012 #1
    Integral calculus involving Fubini's Theorem

    1. The problem statement, all variables and given/known data
    [tex]f(x,y) = x + y, if: x^2 ≤ y ≤ 2x^2 [/tex]
    [tex]f(x,y) = 0, otherwise[/tex]
    Evaluate [tex]\iint_\textrm{I}f[/tex] where I = [0,1] x [0,1]


    2. Relevant equations

    For a Jordan domain K in ℝ^n, let h: K → ℝ and g: K → ℝ be continuous bounded functions with the property that
    [tex]h(x) ≤ g(x)[/tex] for all points x in K.
    Define
    D = {(x,y) in ℝ^(n+1): x in K, h(x) ≤ y ≤ g(x)}.
    Suppose that the function f: D → ℝ is continuous and bounded. Then
    [tex]\int_\textrm{D}f = \int_\textrm{K}\int_{h(x)}^{g(x)}f(x,y)dydx [/tex]
    3. The attempt at a solution
    This was my answer
    [tex]\iint_\textrm{I}f = \int_{0}^{1}\int_{x}^{2x^2}(x + y)dydx[/tex]
    [tex] = \int_{0}^{1}\left[xy + \frac{1}{2}y^2\right]_{x}^{2x^2}dx[/tex] etc
    until I eventually came to 1/5 as my answer.

    However the grader wrote that this [tex]\int_{0}^{1}\int_{x}^{2x^2}(x + y)dydx[/tex] is the wrong integral. Can anyone help me figure out why?
     
    Last edited: May 2, 2012
  2. jcsd
  3. May 2, 2012 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Sure. If x=1 and y=2x^2, the upper limit of your integral, gives you the point (x,y)=(1,2). That's not in I. Did you draw a graph of the integration region and your boundary curves? I'd suggest you try that.
     
  4. May 2, 2012 #3
    I did
    ZecdCl.jpg

    Okay, I would guess that it's
    [tex]\iint_\textrm{I}f = \int_{0}^{\frac{1}{\sqrt{2}}}\int_{x^2}^{2x^2}(x + y)dydx[/tex]

    Am I warm?
     
    Last edited: May 3, 2012
  5. May 2, 2012 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Getting better. From x=0 to x=1/sqrt(2) your upper bound for y is 2*x^2. From x=1/sqrt(2) to x=1 it's just 1. You should probably think about splitting the integral up into parts if you are doing it that way. And shouldn't the lower bound be x^2?
     
    Last edited: May 2, 2012
  6. May 3, 2012 #5
    I just realized I've been writing h(x) = x instead of x^2 as the lower bound. That's not the main reason it's wrong though, yes?
     
  7. May 3, 2012 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It's one. The other is that the upper bound is wrong between x=1/sqrt(2) and x=1.
     
  8. May 3, 2012 #7
    Okay my next guess is this:
    [tex]\iint_\textrm{I}f = \int_{0}^{\frac{1}{\sqrt{2}}}\int_{x^2}^{2x^2}(x + y)dydx + \int_{\frac{1}{\sqrt{2}}}^{1}\int_{x^2}^{1}(x + y)dydx[/tex]
     
  9. May 3, 2012 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    That looks ok to me. Does it work? It's getting kind of late here, so I might be a little fuzzy.
     
  10. May 3, 2012 #9
    I've got this:
    [tex]\frac{45\sqrt{2} - 32}{80\sqrt{2}}[/tex]
    which is about 0.5. It feels right I guess.

    Thank you for your help.
     
  11. May 3, 2012 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I get something different. Might want to check that. You could also integrate dx first. Then you wouldn't need to split the integral up and it should give you the same answer.
     
  12. May 3, 2012 #11
    Okay I went over it again and found a mistake so the new answer is:
    [tex]\frac{21\sqrt{2} - 18}{40\sqrt{2}}[/tex]

    I also tried your suggestion of integrating dx first and set this up:

    [tex]\int_{0}^{1}\int_{\sqrt{0.5y}}^{\sqrt{y}}(x+y) dx dy[/tex]
    which led to a different answer:
    [tex]\frac{21\sqrt{2} - 16}{40\sqrt{2}}[/tex]

    I guess I made a mistake somewhere that I can't spot.
    Either of those match what you got?
     
  13. May 3, 2012 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The second one looks good.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook