# Integral problem.(arc length)

1. May 25, 2005

my problem is to find the arc length of $$y=e^x$$ between 0 and 1

what i've got is $$\int_{0}^{1} \sqrt{1+(e^x)^2}dx$$ which i then substitute $$u=e^x$$ giving $$\int_{1}^{e} \frac{\sqrt{1+u^2}}{u^2}u du$$ which i then sub in v=$$\sqrt{1+u^2}$$ giving $$\int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v^2}{(v^2-1)^2} dv = \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v}{(v^2-1)} dv = \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v}{(v-1)(v+1)} dv$$ which then needs to be integrated using partial fractions, and this is where i run into problems

i get $$\frac{1}{2} \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{1}{v-1}+\frac{1}{v+1}dv$$ which results in my getting an answer of 1.

the book on the other hand gets $$\int_{\sqrt{2}}^{\sqrt{1+e^2}}1+ \frac{\frac{1}{2}}{v-1}-\frac{\frac{1}{2}}{v+1}dv$$ but i cant see how they got the 1, or the negative sign on the second fraction

2. May 25, 2005

### dextercioby

I think you have a problem there.EDIT;It doesn't work.Well,when u got rid of the square in the variable "v".

Daniel.

3. May 25, 2005

### arildno

This equality is wrong:
$$\int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v^2}{(v^2-1)^2} dv = \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v}{(v^2-1)} dv$$

4. May 25, 2005

yeah, i am seeing that...the book jumps from $$\int_{1}^{e} \frac{\sqrt{1+u^2}}{u^2}u du$$ to $$\int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v}{(v^2-1)} dv$$ and (without checking) i figured $$\frac{x^2}{y^2}=\frac{x}{y}$$ my mistake. now i'm really confused, because i see how they get $$\int_{1}^{e} \frac{\sqrt{1+u^2}}{u^2}u du$$ but dont know where to go from there now.

5. May 25, 2005

### dextercioby

$$\int_{1}^{e} \sqrt{\frac{1}{u^{2}}+1} \ du$$

Make the sub.

$$\frac{1}{u} =\sinh t$$

Daniel.

6. May 25, 2005

### GCT

From here, $$\int_{1}^{e} \frac{\sqrt{1+u^2}}{u} du$$[/quote]

try the substitution $$u=1/t$$

7. May 25, 2005

can anyone see what the book did(i would like to get it solved the same way they did it.) up to the point where it subs in v i can see what they did, but when i sub in v i get $$\int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v^2}{(v^2-1)^2} dv$$ and dont know how they got the roots of it.
and even then i dont see how they got the 1 and the negative in the partial fraction.

8. May 25, 2005

### dextercioby

GCT,you're not being original...

Daniel.

9. May 26, 2005

### GCT

10. May 27, 2005

### OlderDan

If the book makes that jump, it is a typographical error. It should be

$$\int_{1}^{e} \frac{\sqrt{1+u^2}}{u^2}u du = \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v^2}{(v^2-1)} dv$$

And their result follows. The u in this equation is just the original y, and v is as you defined it. Taking it from the top

$$y = e^x$$

$$dy = e^x dx = ydx \Rightarrow dx = \frac{{dy}}{y}$$

$$ds = \sqrt {dx^2 + dy^2 } = \left( {\sqrt {\frac{1}{{y^2 }} + 1} } \right)dy = \frac{{\sqrt {1 + y^2 } }}{y}dy\left( { = \frac{{\sqrt {1 + y^2 } }}{{y^2 }}ydy} \right)$$

$$v = \sqrt {1 + y^2 } \Rightarrow dv = \frac{1}{{2v}}2ydy = \frac{{ydy}}{v} \Rightarrow dy = \frac{v}{y}dv$$

$$v^2 = 1 + y^2 \Rightarrow y^2 = v^2 - 1$$

$$ds = \frac{{\sqrt {1 + y^2 } }}{y}dy = \frac{v}{y} \cdot \frac{v}{y}dv = \frac{{v^2 }}{{y^2 }}dv = \frac{{v^2 }}{{v^2 - 1}}dv$$

$$ds = \left( {1 + \frac{1}{{v^2 - 1}}} \right)dv = \left( {1 + \frac{1}{{v - 1}} \cdot \frac{1}{{v + 1}}} \right)dv = \left( {1 + \frac{{\frac{1}{2}}}{{v - 1}} - \frac{{\frac{1}{2}}}{{v + 1}}} \right)dv$$