Integral problem.(arc length)

1. May 25, 2005

Vadim

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

Vadim

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

Vadim

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

$$y = e^x$$

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

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

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

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

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

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

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