Question

Prove that (beware: integrals)...

Answer 1

\[\pi \le \int\limits_{0}^{\pi} \sqrt{1+\sin^4 (x)} dx \le \pi \sqrt{2}\]

Answer 2

can't be that hard can it?

Answer 3

the lower bound comes from the fact that the minimum value of \(1+\sin^4(x)\) is 1

Answer 4

ok

Answer 5

i bet you can do the upper one same way

Answer 6

what is the max of \[1+\sin^4(x)\]?

Answer 7

2

Answer 8

so max of the square root is?

Answer 9

oohhhh sqrt{2}

Answer 10

lol light goes on...

Answer 11

is that all we have to state? I have a feeling it's waaayyy too easy lol

Answer 12

you gotta admit, scared as you was about "proof" this was pretty easy

Answer 13

lol, that was so straight forward.... dang xD

Answer 14

well you might mention that the lower bound is the min of the function times the length of the path \(\pi\) in this case and so on, to make it look nice

Answer 15

awesome, sounds great :D

Answer 16

thanks :)

Answer 17

yw