Melody

Oh dear, these two regions are not the same area.  What a pity.

I guess I better keep thinking!

You cannot find the log of a negative number so I am goinf for the positive values.( this is not great mathematical reasoning here:

I'd believe Heureka if I were you   LOL

\(\displaystyle\int_0^1 x+\frac{1}{x+\frac{1}{x+\frac{1}{x+....}}}dx\\ =\displaystyle\int_1^{\frac{1+\sqrt5}{2}} y(1+y^{-2})dy\\ =\displaystyle\int_1^{\frac{1+\sqrt5}{2}} y+y^{-1}dy\\ =\left[ \frac{y^2}{2}+lny \right ]_1^{\frac{1+\sqrt5}{2}}\\ =\left[ ({\frac{1+\sqrt5}{2})^2}\div2+ln(\frac{1+\sqrt5}{2}) -\frac{1}{2}\right ]\\ =\left[ {\frac{(1+\sqrt5)^2}{8}}+ln(\frac{1+\sqrt5}{2}) -\frac{4}{8}\right ]\\ =\left[ \frac{6+2\sqrt5-4}{8}+ln(\frac{1+\sqrt5}{2}) \right ]\\ =\left[ \frac{2+2\sqrt5}{8}+ln(\frac{1+\sqrt5}{2}) \right ]\\ = \frac{1+\sqrt5}{4}+ln(\frac{1+\sqrt5}{2}) \\ \)

Interesting :)

Let

\(y=x+\frac{1}{x+\frac{1}{x+\frac{1}{x+....}}}\\ so\\ y=x+\frac{1}{y}\\ x=y-y^{-1}\\ \frac{dx}{dy}=1+y^{-2}\\ dx=(1+y^{-2})dy\\when \quad x=0\quad \\ y=\frac{1}{y}\\ y^2=1\\ y=\pm1 \qquad \text{this is a stumbling block, I only want one value} \\when\quad x=1\\ 1=y-y^{-1}\\ y=y^2-1\\y^2-y-1=0 \\y=\frac{1\pm\sqrt{5}}{2} \quad \text{two answers again :(} \)

\(\displaystyle\int_0^1 x+\frac{1}{x+\frac{1}{x+\frac{1}{x+....}}}dx\\ =\displaystyle\int_?^? y(1+y^{-2})dy\\ =\displaystyle\int_?^? y+y^{-1}dy\\ =\left[ \frac{y^2}{2}+lny \right ]_?^?\)

Now I have to think about those questions marks  

To be continued   :)

I can see Heureka's answer underneath ..... I'll have to think about it more but I will leave this here for now.

Maybe Heuarka, you might like to commnent on what I have done ??

Challenging - definitely

Boring - Never!

Thanks Heureka,

I was wondering how this would be done.  

I did think of substituting x=tan(z) but I didn't get very far with it.   

Now I shall take a good look at your answer :))

In this case it is the same thing but the calculator is using correct order of operation.

X/100*59/118   

(X/100)*(59/118)

assuming that the distribution is normal and the mean is 100 then the prob will be 88%

I love your illustrated thought process Hectictar :)

I used this site: