an integral from the 2010 MIT Integration Bee

**Random Variable** · January 28th 2010, 08:48 PM

The following integral was on the qualifying test:

$\int^{1}_{0} \sqrt{1+x \ \sqrt{1+x \ \sqrt{1+x \ \sqrt{...}}}} \ dx$

Any ideas? I'm stumped.

**Drexel28** · January 28th 2010, 08:51 PM

Originally Posted by Random Variable

The following integral was on the qualifying test:

$\int^{1}_{0} \sqrt{1+x \ \sqrt{1+x \ \sqrt{1+x \ \sqrt{...}}}} \ dx$

Any ideas? I'm stumped.

Let $x=\sec^2(z)$ or $x=\sinh^2(z)$ ?

**Random Variable** · January 28th 2010, 10:01 PM

How about the following:

let $y = \sqrt{1+x \ \sqrt{1+x \ \sqrt{1+x \ \sqrt{...}}}}$

then $y^{2} = 1+x\sqrt{1+x \ \sqrt{1+x \ \sqrt{1+x \ \sqrt{...}}}} = 1+xy$

solving for y

$y = \frac{x}{2} \pm \frac{\sqrt{x^{2}+4}}{2}$

then using the positive root since we're integrating over positive values of x

$\int^{1}_{0} \sqrt{1+x \ \sqrt{1+x \ \sqrt{1+x \ \sqrt{...}}}} \ dx \ = \int^{1}_{0} \Big(\frac{x}{2} + \frac{\sqrt{x^{2}+4}}{2} \Big) \ dx$ ?

**Calculus26** · January 28th 2010, 10:07 PM

then you solve for y using quadratic formula ?

y= x/2 + sqrt(x^2 +4)/2 ?

or am I way off base?

**Drexel28** · January 28th 2010, 10:09 PM

Originally Posted by Random Variable

What about the following:

let $y = \sqrt{1+x \ \sqrt{1+x \ \sqrt{1+x \ \sqrt{...}}}}$

then $y^{2} = 1+x\sqrt{1+x \ \sqrt{1+x \ \sqrt{1+x \ \sqrt{...}}}} = 1+xy$ ?

Originally Posted by Calculus26

then you solve for y using quadratic formula ?

y= x/2 + sqrt(x^2 +4)/2 ?

or am I way off base?

No no! That is correct. I didn't realize it was supposed to be an infinite surd

. Yes, that method works...if you can prove that the surd converges

.

P.S. You can choose the $\pm$ according to the fact that $y>0$ for $x\in[0,1]$

**Calculus26** · January 28th 2010, 10:11 PM

Then since y > 0 we take y= x/2 + sqrt(x^2 +4)/2 ?

**Drexel28** · January 28th 2010, 10:15 PM

Originally Posted by Calculus26

Then since y > 0 we take y= x/2 + sqrt(x^2 +4)/2 ?

I would agree with that. the answer is $\frac{1+\sqrt{5}}{4}+\text{arccsch}\left(2\right)$

Math Help - an integral from the 2010 MIT Integration Bee

LinkBack

Thread Tools

Display

an integral from the 2010 MIT Integration Bee

Similar Math Help Forum Discussions

Find the largest integer n, where 2009^n divides (2008^2009^2010 + 2010^2009 ^2008)

Search Tags