## Loser66 one year ago $J=\int_0^1\sqrt{1-x^4}$ $K=\int_0^1\sqrt{1+x^4}$ $L=\int_0^1\sqrt{1-x^8}$ 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.

1. theopenstudyowl

#walframa

2. Loser66

Thanks, but no.

3. anonymous

lol @theopenstudyowl

4. theopenstudyowl

Meow...lol

5. anonymous

=^-^=

6. Loser66

7. anonymous

@welshfella

8. Loser66

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

9. nincompoop

visualise it

10. Loser66

graphing?

11. ganeshie8

Ahh graphing looks neat!

12. nincompoop

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

13. ganeshie8

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

14. Loser66

@nincompoop what is difference between graph and visualize?

15. ganeshie8

visualize = graphing in ur head

16. nincompoop

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

17. 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

18. nincompoop

THERE YOU GO!

19. ganeshie8

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

20. ganeshie8

then apply concentrationalizing logic

21. Loser66

Got it. Thanks you all. :)

22. anonymous

You're welcome :)

23. ganeshie8

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

24. Loser66

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

25. ganeshie8

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

26. Loser66

yyyyyyyyyyyyyyyes!! that 's good.

27. Loser66

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

28. ganeshie8

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

29. Loser66

Thanks a lot.

30. ganeshie8

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

31. ganeshie8

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

32. 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)

33. Loser66

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

34. ganeshie8

yeah that was easy :)