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Loser66
 one year ago
\[J=\int_0^1\sqrt{1x^4}\]
\[K=\int_0^1\sqrt{1+x^4}\]
\[L=\int_0^1\sqrt{1x^8}\]
A) J<L<1<K
B) J<L<K<1
C) L<J<1<K
D) L<J<K<1
E) L<1<J<K
please, help.
Loser66
 one year ago
\[J=\int_0^1\sqrt{1x^4}\] \[K=\int_0^1\sqrt{1+x^4}\] \[L=\int_0^1\sqrt{1x^8}\] A) J<L<1<K B) J<L<K<1 C) L<J<1<K D) L<J<K<1 E) L<1<J<K please, help.

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MeowLover17
 one year ago
Best ResponseYou've already chosen the best response.1lol @theopenstudyowl

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Please, dont' mess my post

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 any shortest way to find the answer without graphing or calculating?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Ahh graphing looks neat!

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.0you don't want to "graph" it, so visualize

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3ohk i misread you don't want to graph/evaluate...

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@nincompoop what is difference between graph and visualize?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3visualize = graphing in ur head

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.0GRAPH WOULD INVOLVE YOU ACTUALLY GRAPHING IT, VISUALIZE THAT INVOLVES JUST MENTAL

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wouldn't \(\sqrt{1x^{8}}\) have steep climbs and descents than \(\sqrt{1x^{4}}\) ? I would expect because of the increased steepness in \(\sqrt{1x^{8}}\) that it's integral would be of greater value. \(\sqrt{1+x^{4}}\) Would never tend towards the xaxis. It should be very similar to \(\sqrt{1x^{4}}\) But increasing as x increases. So trying to logic it out, I would say J < L < K

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3For \(x\in (0,1)\) do we have \(x^m \lt x^n\) if \(m \gt n\) ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3then apply concentrationalizing logic

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Got it. Thanks you all. :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3looks the question boils down to simply putting the given integrands in ascending order

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0When graphing, J >1, and others <1, hence A is a correct answer. But I would like to shorten the time solving it. :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\[x^4 \gt x^8 \implies x^4 \lt x^8 \implies 1x^4\lt 1x^8 \implies \sqrt{ 1x^4}\lt \sqrt{ 1x^8}\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0yyyyyyyyyyyyyyyes!! that 's good.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0just from 0 to 1, that's true, right?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3oh forgot to put that, yes all that is valid only for \(x\in (0,1)\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3np sometimes it is hard to see \(x^4 \lt x\) for \(x\in (0,1)\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3if we let \(x = 0.9\), we get \(x^4 = 0.9*0.9*0.9*x\) which is clearly less than \(x\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3it 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
 one year ago
Best ResponseYou've already chosen the best response.0yeah, and we know that 1+ whatever >1, hence K>1 , others <1, good interpretation. :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3yeah that was easy :)
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