Question

Integrate:
\[\int_0^2\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}~dx\]

Answer 1

Source: http://math.mit.edu/~sswatson/pdfs/qualifying_round_2014.pdf

Answer 2

sorry - i've no idea how to do this one

Answer 3

No worries, it's intended to be challenging ;)

Answer 4

That was fun =)

Answer 5

\[\int_0^2\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}~dx = \int_0^2 \frac{1+\sqrt{1+4x}}{2}~dx\]

Answer 6

let \[\sqrt{x+\sqrt{x+\sqrt{x...}}}=y\]
\[\sqrt{x+y}=y\]
\[x+y=y^2,y^2-y-x=0,y=\frac{ 1\pm \sqrt{1+4x} }{ 2 }\]

Answer 7

as y is positive
hence \[y=\frac{ 1+\sqrt{1+4x} }{ 2 }\]

Answer 8

tha'ts really clever

Answer 9

Nice work guys!

Answer 10

a brilliant substitution!

Answer 11

That was a piece of beautiful music!!

Answer 12

:) cwrw you might be interested in ramanujan nested radicals, look it up in google when free

Answer 13

@cwrw238

Answer 14

Here's a related one, \[\Large \int\limits_0^1 \cos(\cos(\cos(...)))dx\]

Answer 15

@ganeshie8 thanks I will

Answer 16

http://mathworld.wolfram.com/DottieNumber.html

Answer 17

http://www.wolframalpha.com/input/?i=%28sqrt%28gelfond%27s+constant%29%29%5Ei

Answer 18

I expected a challenging one , but this was simple

Answer 19

Maybe because i have solved nested radicals like these too many times