## anonymous 5 years ago Find the arc length of the graph of the function from x=0 to x=2. F(x) = ln(x+1).

1. anonymous

hello

2. anonymous

sjin, the arc length of a function is given by$s=\int\limits_{x_1}^{x_2}\sqrt{1+(y')^2}dx$after considering a differential of arc length using Pythagoras' Theorem. The problem here will be with the integration.

3. anonymous

$y=\log (x+1) \rightarrow y'=\frac{1}{x+1}$so$s=\int\limits_{0}^{2}\sqrt{1+\left( \frac{1}{x+1} \right)^2}dx=\int\limits_{0}^{2}\frac{\sqrt{(x+1)^2+1}}{x+1}dx$

4. anonymous

You're going to have to make a few substitutions along the way to solve this. You should start with $u=x+1$to obtain$\int\limits_{u_1}^{u_2}\frac{\sqrt{u^2+1}}{u}du$then$w=u^2+1$to obtain$\frac{1}{2}\int\limits_{w_1}^{w_2}\frac{w^{1/2}}{w-1}dw$and finally, $\zeta = w^{1/2}$to obtain$\int\limits_{\zeta_1}^{\zeta_2}\frac{\zeta^2}{\zeta^2-1}d \zeta$The last integrand may be written$\frac{\zeta^2}{\zeta^2-1}=\frac{\zeta^2-1+1}{\zeta^2 -1}=1+\frac{1}{\zeta^2-1}=1+\frac{1}{2}\frac{1}{\zeta -1}-\frac{1}{2}\frac{1}{\zeta +1}$

5. anonymous

I would have banged out a sinh sub, but I haven't looked too closely.

6. anonymous

After solving the final integral form, and back-substituting, you obtain,$s=\sqrt{(x+1)^2+1}+\frac{1}{2}\log \frac{\sqrt{(x+2)^2+1}-1}{\sqrt{(x+1)^2+1}+1}|_0^2$

7. anonymous

with$s=\sqrt{10}-\sqrt{2}+\frac{1}{2}\log \frac{\sqrt{10}-1}{\sqrt{10}+1}-\frac{1}{2}\log \frac{\sqrt{2}-1}{\sqrt{2}+1}$

8. anonymous

As INewton mentioned, there are other avenues for substitution.

9. anonymous

thank you so much!

10. anonymous

np

11. anonymous

Indeed, I'm pretty impressed you typed all that LaTeX on what is a pretty poor engine on this site.

12. anonymous

I'm impressed too. I feel exhausted.