anonymous
  • anonymous
find all values x* described in the mean value theorem for integrals. f(x)=s/x^2 [1,3]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
|dw:1358283986640:dw|
anonymous
  • anonymous
@zepdrix
amoodarya
  • amoodarya
|dw:1358284021076:dw|

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amoodarya
  • amoodarya
|dw:1358284117061:dw|
anonymous
  • anonymous
i really appreciate your help, but i'm having a little trouble reading your work.. :'(
amoodarya
  • amoodarya
|dw:1358284164559:dw|
anonymous
  • anonymous
i'still not comprehending... it's really hard to understand your work.. :"(
amoodarya
  • amoodarya
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anonymous
  • anonymous
thank you.. but anyone else know how ot do this differently? or clearly? :D
amoodarya
  • amoodarya
which part of that is complicated for you ?
anonymous
  • anonymous
the formula you used, which is different from what i have on my review sheet, and i still can't read someof your work :'(
anonymous
  • anonymous
ah.., let me try this myself :) i think i now get what you did!
amoodarya
  • amoodarya
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anonymous
  • anonymous
ok, i dont.get.this. where did the "S" come from?
amoodarya
  • amoodarya
s in figure is area of f(x)
anonymous
  • anonymous
could you try doing it without the "s"?
amoodarya
  • amoodarya
yeah without s i.e. f(x)= 1/x^2 ?
anonymous
  • anonymous
yes :)
amoodarya
  • amoodarya
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amoodarya
  • amoodarya
look if ok go on
anonymous
  • anonymous
oh... ok. im gonna have lf again. thank you:)
anonymous
  • anonymous
i meant try myself again.
amoodarya
  • amoodarya
do you know the mean value of integrals theory?
anonymous
  • anonymous
ok, so here is what i did. \[\frac{ 1 }{ 3-1 }\int\limits_{1}^{3}\frac{ 1 }{ x ^{x} }dx = \frac{ 1 }{ 2 }\int\limits_{1}^{3}x ^{-2}dx ^{}\] \[\frac{ 1 }{ 2 }(-x ^{-1})\]from 1 to 3, \[\frac{ 1 }{ 2 }(-3^{-1}+1^{-1}) = \frac{ 1 }{ 2 }(\frac{ 1 }{ -3 }+\frac{ 1 }{ 1 }) = \frac{ 1 }{ 2 }(\frac{ 2 }{ 3 })=\frac{ 1 }{ 3 }\]
anonymous
  • anonymous
and i dont know what to do next. but i do know that i have found(x*) so far.

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