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Iterated integrals

  1. Mar 6, 2012 #1
    1. The problem statement, all variables and given/known data

    Calculate the given iterated integrals ∫02 dy ∫0yy2 * exy dx


    My attempt:

    20dy[exy*y]y0

    = ∫20 ey*y*y - ex*0*0

    = ∫20ey2*y dx

    = [ey^2]*y]20 = 2e4

    Is this correct?
     
    Last edited: Mar 6, 2012
  2. jcsd
  3. Mar 6, 2012 #2

    Mark44

    Staff: Mentor

    I think you skipped a step going from the above to the line below. Written in a more usual form, your integral would be
    [tex]\int_{y = 0}^2~\int_{x = 0}^y y^2 e^{xy}~dx~dy[/tex]

    (You can see what I did in LaTeX by clicking the integral above.)
    Show how you get from the step above to the step where you've carried out the inner integration.


    No - see above. I get (1/2)e4 - 5/2
     
  4. Mar 6, 2012 #3
    I dont understand what I have done wrong at all
     
  5. Mar 6, 2012 #4

    Mark44

    Staff: Mentor

    The error is here:
    [tex]\left . ye^{xy}\right |_{x = 0}^y[/tex]

    You need to replace x by 0, not y.
     
  6. Mar 6, 2012 #5
    Im still confused I get:

    [ey^2*y-y]20 =

    (e^2^2 - 2) - (e^0^2 - 0) =

    2e^4 - 2 ?

    Im lost
     
  7. Mar 6, 2012 #6

    Mark44

    Staff: Mentor

    You're skipping steps. You have the outer integrand right, but you have made a mistake when you integrated this integral.
    [tex]\int_0^2 (ye^{y^2} - y)dy[/tex]

    Split this into two integrals and carry out the two integrations.
     
    Last edited: Mar 6, 2012
  8. Mar 6, 2012 #7
    Oh I have totally lost my comprehensive view now.....
     
  9. Mar 6, 2012 #8

    Mark44

    Staff: Mentor

    I don't know what you mean by this.
     
  10. Mar 6, 2012 #9
    How do I get from:

    = ∫(y = 0 to 2) (ye^(y^2) - y) dy

    = (1/2)e^(y^2) - (1/2)y^2 {for y = 0 to 2}

    ?
     
  11. Mar 6, 2012 #10
    y0 ye^(y^2) - y dy

    to

    = (1/2)e^(y^2) - (1/2)y^2


    {for y = 0 to 2}
     
  12. Mar 6, 2012 #11

    Mark44

    Staff: Mentor

    Split the integral into two.
    Use substitution to do the first integral.
     
  13. Mar 6, 2012 #12
    This didnt get me further from start, but thanks anyway.
     
  14. Mar 6, 2012 #13

    Mark44

    Staff: Mentor

    When you evaluate the above at 2 and 0, what do you get?

    Sorry, I misunderstood what you were asking before.
     
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