anonymous
  • anonymous
Can someone walk me through how to switch the order of integration on some double integrals?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\text{Original integral:}~\int_0^1\int_{y/2}^{1/2}e^{-x^2}dxdy\]
anonymous
  • anonymous
I tried changing the bounds, and I got \[\int_0^{1/2}\int_0^{2x}f(x,y)dydx\]Am I on the right track so far?
ganeshie8
  • ganeshie8
Looks good!

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anonymous
  • anonymous
Im kind of confused on what to do next though.
ganeshie8
  • ganeshie8
next integrate
ganeshie8
  • ganeshie8
start with the inner integral : \[\large \int_0^{1/2} \color{blue}{\int_0^{2x}e^{-x^2}dy}~dx\]
anonymous
  • anonymous
OH! I forgot that I can actually integrate it now. So, it becomes: \[\int_0^{1/2}2xe^{-x^2}dx=-(e^{-x^2})|_0^{1/2}=1-e^{-1/4}\]Is that right?
ganeshie8
  • ganeshie8
Perfect! you may use wolfram to double check http://www.wolframalpha.com/input/?i=%5Cint_0%5E1%5Cint_%7By%2F2%7D%5E%7B1%2F2%7De%5E%7B-x%5E2%7Ddxdy
anonymous
  • anonymous
Alright, thanks! I thought I had to do something to the integrand, and I didn't recognize that with the change to dydx, I could actually integrate.
ganeshie8
  • ganeshie8
Thats it! changing order of integration gave us that factor 2x which was useful in u-substitution
anonymous
  • anonymous
Thanks. :)
ganeshie8
  • ganeshie8
np:)

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