anonymous
  • anonymous
double integral question.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
can someone please help me solve this? http://gyazo.com/d88e00524337c49ec6734dcb2b9ad5b1
amistre64
  • amistre64
its nice they give you a box to work in
amistre64
  • amistre64
let y+1 = some constant, say k \[\int_{0}^{2}\frac{1}{(x+k)^3}dx\] \[\int_{0}^{2}(x+k)^{-3}dx\]

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amistre64
  • amistre64
whats your solution to this part?
anonymous
  • anonymous
|dw:1370890838800:dw|
experimentX
  • experimentX
http://www.wolframalpha.com/input/?i=Integrate%5BIntegrate%5B1%2F%28x%2By%2B1%29%5E3%2C+%7Bx%2C+0%2C+2%7D%5D%2C%7By%2C+0%2C+1%7D%5D
anonymous
  • anonymous
@amistre64 solution \[-\frac{ (x + k) ^ 2 }{ 2} + C\]
anonymous
  • anonymous
i forgot the (-) on the 2 up there
amistre64
  • amistre64
that would be the indefinite integral yes; lets apply the limits of 0 to 2, recall that \[\int_{a}^{b}f(x)~dx=F(b)-F(a)\]
amistre64
  • amistre64
\[\left(-\frac{ (2 + k) ^{-2} }{ 2}\right)-\left(-\frac{ (0 + k) ^{-2} }{ 2} \right)\] \[-\frac{ (2 + k) ^{-2} }{ 2}+\frac{ ( k) ^{-2} }{ 2} \] \[\frac12(k^{-2}-(2 + k) ^{-2})~:~k=y+1 \] \[\frac12((y+1)^{-2}-(y+3) ^{-2})\] integrate that from 0 to 1
amistre64
  • amistre64
\[\frac12\left(\int_{0}^{1}(y+1)^{-2}dy-\int_{0}^{1}(y+3)^{-2}dy\right)\]
anonymous
  • anonymous
from your 2nd step, am i not to replace k = 2? and k = 0? @amistre64
amistre64
  • amistre64
no, k is just some constant with respect to x your integrating "x" from 0 to 2 .... not "k" from 0 to 2
anonymous
  • anonymous
why not with respect to "y"?
amistre64
  • amistre64
because your order of integration is dxdy dx first, then dy
amistre64
  • amistre64
\[\int_{0}^{1}\left(\int_{0}^{2}\frac{1}{(x+y+1)^3}~dx\right)dy\] the inside stuff is dx .. anything that is not xed is constant, therefore y+1 = k \[...\left(\int_{0}^{2}\frac{1}{(x+k)^3}~dx\right)...\]
amistre64
  • amistre64
\[...\left(\int_{0}^{2}\frac{1}{(x+k)^3}~dx\right)...\] \[...\left(-\frac{1}{2(2+k)^2}+\frac{1}{2(0+k)^2}\right)...\] replace k with y+1 \[...\left(-\frac{1}{2(2+y+1)^2}+\frac{1}{2(y+1)^2}\right)...\] \[\frac12\int_{0}^{1}\left(\frac{1}{(y+1)^2}-\frac{1}{(y+3)^2}\right)dy\]
anonymous
  • anonymous
hmmm....i got 1/24 as final answer....
amistre64
  • amistre64
you most likely added things incorrectly
amistre64
  • amistre64
\[\frac12\int_{0}^{1}\left(\frac{1}{(y+1)^2}-\frac{1}{(y+3)^2}\right)dy\] \[\frac12\left(\frac{1}{-(y+1)}-\frac{1}{-(y+3)}\right)\] \[\frac12\left(\frac{1}{-(1+1)}-\frac{1}{-(1+3)}\right)-\frac12\left(\frac{1}{-(0+1)}-\frac{1}{-(0+3)}\right)\] \[\frac12\left(\frac14-\frac12\right)-\frac12\left(\frac13-1\right)\] \[\frac12\left(\frac14-\frac12-\frac13+1\right)\] \[\frac12\left(\frac{3-6-4+12}{12}\right)\] \[\frac12\left(\frac{5}{12}\right)\]
anonymous
  • anonymous
ouch!...i am redoing this thing!
amistre64
  • amistre64
:) the algebra/arithmetic is usually the culprit
anonymous
  • anonymous
exactly. i did it a second time and got 11/24. i'm doing it again!!
anonymous
  • anonymous
@amistre64 got it finally. my friend also got 5/24 :). thank you very much!!
amistre64
  • amistre64
:) good job

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