Loser66
  • Loser66
\[J=\int_0^1\sqrt{1-x^4}\] \[K=\int_0^1\sqrt{1+x^4}\] \[L=\int_0^1\sqrt{1-x^8}\] A) J
Mathematics
katieb
  • katieb
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theopenstudyowl
  • theopenstudyowl
#walframa
Loser66
  • Loser66
Thanks, but no.
MeowLover17
  • MeowLover17

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theopenstudyowl
  • theopenstudyowl
Meow...lol
MeowLover17
  • MeowLover17
=^-^=
Loser66
  • Loser66
Please, dont' mess my post
MeowLover17
  • MeowLover17
Loser66
  • Loser66
@ganeshie8 any shortest way to find the answer without graphing or calculating?
nincompoop
  • nincompoop
visualise it
Loser66
  • Loser66
graphing?
ganeshie8
  • ganeshie8
Ahh graphing looks neat!
nincompoop
  • nincompoop
you don't want to "graph" it, so visualize
ganeshie8
  • ganeshie8
ohk i misread you don't want to graph/evaluate...
Loser66
  • Loser66
@nincompoop what is difference between graph and visualize?
ganeshie8
  • ganeshie8
visualize = graphing in ur head
nincompoop
  • nincompoop
GRAPH WOULD INVOLVE YOU ACTUALLY GRAPHING IT, VISUALIZE THAT INVOLVES JUST MENTAL
anonymous
  • anonymous
Wouldn't \(\sqrt{1-x^{8}}\) have steep climbs and descents than \(\sqrt{1-x^{4}}\) ? I would expect because of the increased steepness in \(\sqrt{1-x^{8}}\) that it's integral would be of greater value. \(\sqrt{1+x^{4}}\) Would never tend towards the x-axis. It should be very similar to \(\sqrt{1-x^{4}}\) But increasing as x increases. So trying to logic it out, I would say J < L < K
nincompoop
  • nincompoop
THERE YOU GO!
ganeshie8
  • ganeshie8
For \(x\in (0,1)\) do we have \(x^m \lt x^n\) if \(m \gt n\) ?
ganeshie8
  • ganeshie8
then apply concentrationalizing logic
Loser66
  • Loser66
Got it. Thanks you all. :)
anonymous
  • anonymous
You're welcome :)
ganeshie8
  • ganeshie8
looks the question boils down to simply putting the given integrands in ascending order
Loser66
  • Loser66
When graphing, J >1, and others <1, hence A is a correct answer. But I would like to shorten the time solving it. :)
ganeshie8
  • ganeshie8
\[x^4 \gt x^8 \implies -x^4 \lt -x^8 \implies 1-x^4\lt 1-x^8 \implies \sqrt{ 1-x^4}\lt \sqrt{ 1-x^8}\]
Loser66
  • Loser66
yyyyyyyyyyyyyyyes!! that 's good.
Loser66
  • Loser66
just from 0 to 1, that's true, right?
ganeshie8
  • ganeshie8
oh forgot to put that, yes all that is valid only for \(x\in (0,1)\)
Loser66
  • Loser66
Thanks a lot.
ganeshie8
  • ganeshie8
np sometimes it is hard to see \(x^4 \lt x\) for \(x\in (0,1)\)
ganeshie8
  • ganeshie8
if we let \(x = 0.9\), we get \(x^4 = 0.9*0.9*0.9*x\) which is clearly less than \(x\)
ganeshie8
  • ganeshie8
it is less than \(x\) because multiplying anything by a proper fraction always produces a smaller number (0.9 = 9/10 is a proper fraction)
Loser66
  • Loser66
yeah, and we know that 1+ whatever >1, hence K>1 , others <1, good interpretation. :)
ganeshie8
  • ganeshie8
yeah that was easy :)

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