Double integral question

  • Thread starter kieranl
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  • #1
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Homework Statement



Sketch the region of integration and then evaluate the double integral:


Homework Equations



[tex]\int\int[/tex]x2exydA over the region R= {(x,y), y<=x<=1, 0<=y<=1}

The Attempt at a Solution



I have managed to do half of the problem and integrate it respect to x but then have no idea how to finish the problem.

I mangaged to get upto [tex]\int[/tex]ey(2/y2-2/y3-1/y)+ey2(y-2/y+2/y3) dy

can someone please tell me how to finish it or if i have done something wrong so far??

cheers
 
Last edited:

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi kieranl! Welcome to PF! :smile:

(have a ≤ :wink:)
[tex]\int\int[/tex]x2exydA over the region R= {(x,y), y<=x<=1, 0<=y<=1}

I mangaged to get upto [tex]\int[/tex]ey(2/y2-2/y3-1/y)+ey2(y-2/y+2/y3) dy

uhh? :confused:

start again, and integrate over dy first! :smile:
 
  • #3
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but if i integrated by dy first wouldnt i end up with an answer containing y's and not numbers??
 
  • #4
tiny-tim
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but if i integrated by dy first wouldnt i end up with an answer containing y's and not numbers??

If you integrate with respect to dy, then you eliminate y, and only have x's.
 
  • #5
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but i have to integrate over y<=x<=1 for dx so the final answer would then contain y's?
 
  • #6
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Draw a picture of the region R. Then use the picture to write R in the form

R={(x,y):a<x<b, f(x)<y<g(x)}.
 
  • #7
tiny-tim
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Hi kieranl! :smile:

(have an integral: ∫ and a ≤ :wink:)
but i have to integrate over y<=x<=1 for dx so the final answer would then contain y's?

No, you have to integrate over 0 ≤ y ≤ x ≤ 1 …

whether you do it over x first or over y first is up to you. :wink:
 

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