$$J=\int_0^1\sqrt{1-x^4}$$
$$K=\int_0^1\sqrt{1+x^4}$$
$$L=\int_0^1\sqrt{1-x^8}$$
A) J

Question

$$J=\int_0^1\sqrt{1-x^4}$$
$$K=\int_0^1\sqrt{1+x^4}$$
$$L=\int_0^1\sqrt{1-x^8}$$
A) J<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.

theopenstudyowl · Answer

#walframa

loser66 · Answer

Thanks, but no.

meowlover17 · Answer

lol @theopenstudyowl

theopenstudyowl · Answer

Meow...lol

meowlover17 · Answer

=^-^=

loser66 · Answer

Please, dont' mess my post

meowlover17 · Answer

@welshfella

loser66 · Answer

@ganeshie8  any shortest way to find the answer without graphing or calculating?

nincompoop · Answer

visualise it

loser66 · Answer

graphing?

ganeshie8 · Answer

Ahh graphing looks neat!

nincompoop · Answer

you don't want to "graph" it, so visualize

ganeshie8 · Answer

ohk i misread you don't want to graph/evaluate...

loser66 · Answer

@nincompoop  what is difference between graph and visualize?

ganeshie8 · Answer

visualize = graphing in ur head

nincompoop · Answer

GRAPH WOULD INVOLVE YOU ACTUALLY GRAPHING IT, VISUALIZE THAT INVOLVES JUST MENTAL

anonymous · Answer

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 · Answer

THERE YOU GO!

ganeshie8 · Answer

For $x\in (0,1)$ do we have $x^m \lt x^n$ if $m \gt n$  ?

ganeshie8 · Answer

then apply concentrationalizing logic

loser66 · Answer

Got it. Thanks you all. :)

anonymous · Answer

You're welcome :)

ganeshie8 · Answer

looks the question boils down to simply putting the given integrands in ascending order

loser66 · Answer

When graphing, J >1, and others <1, hence A is a correct answer. But I would like to shorten the time solving it. :)

ganeshie8 · Answer

$$x^4 \gt x^8 \implies  -x^4 \lt -x^8 \implies 1-x^4\lt 1-x^8 \implies \sqrt{ 1-x^4}\lt \sqrt{ 1-x^8}$$

loser66 · Answer

yyyyyyyyyyyyyyyes!! that 's good.

loser66 · Answer

just from 0 to 1, that's true, right?

ganeshie8 · Answer

oh forgot to put that, yes all that is valid only for $x\in (0,1)$

loser66 · Answer

Thanks a lot.

ganeshie8 · Answer

np sometimes it is hard to see $x^4 \lt x$ for $x\in (0,1)$

ganeshie8 · Answer

if we let $x = 0.9$, we get $x^4 = 0.9*0.9*0.9*x$

which is clearly less than $x$

ganeshie8 · Answer

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 · Answer

yeah, and we know that 1+ whatever >1, hence K>1 , others <1, good interpretation. :)

ganeshie8 · Answer

yeah that was easy :)