Question

as nin types out his humongous reply, try this integral\[\int \limits_0^2\sqrt{x+ \sqrt{x+\sqrt{x+\cdots}}}dx\]

Answer 1

if we call this expression \(y\) then \(y^2 - y = x\) so probably a substitution like \(t^2 - t = x\) should work? i'm not sure

Answer 2

I'm getting \(\dfrac{19}{6}\)

Answer 3

oh great, how? i'm just not good at integrals in general.

Answer 4

I have used your method and solved \(y\)

Answer 5

quadratic equation?

Answer 6

And dx with respect to y would be?

Answer 7

Yes

\[\sqrt{x+ \sqrt{x+\sqrt{x+\cdots}}}=y \\~\\\implies x+y = y^2\\~\\\implies y = \frac{1}{2}(1+\sqrt{4x+1}) \]

Answer 8

simple... i'll post more

Answer 9

\(1- \sqrt{4x+1}\)
can never be positive or 0 ?
why wasn't it considered....

Answer 10

i know y must be positive..

Answer 11

when x = 0, the integral evaluates to 0.

for x > 0, the value of integrand is clearly positive.
\(1-\sqrt{4x+1}\)is never positive in the given range. so yeah we simply reject this root

Answer 12

gotcha :)

Answer 13

OH MY GOD I JUST REALISED I DERIVED AN EULER SUB ABOVE. LOL

Answer 14

"if we call this expression \(y\) then \(y^2 - y = x\) so probably a substitution like \(t^2 - t = x\) should work? i'm not sure"

that is an euler's substitution