# Homework Help: Problem Application of Integrals(Arc Legth) - I'm stuck

1. Feb 8, 2013

### SclayP

So i'm stuck resolving this integral...

$\int^{1}_{0} \sqrt{e^{2x} + 1} \, dx$
$$u=2x ; du=2dx$$
$\frac{1}{2}\int^{2}_{0} \sqrt{e^u + 1} \, du$
$$t=e^u+1 ; dt=e^u du$$
$\frac{1}{2}\int^{e^2 + 1}_{2} \frac{\sqrt{t}}{t-1} \, dt$
$$v=\sqrt{t} ; dv=\frac{1}{2\sqrt{t}}$$
$\int \frac{v^3}{v^2 -1} \, dv$

And here i dont know how to proceed, i red i have to do long division but i dont know why... please help me.

Thanks.

Last edited by a moderator: Feb 8, 2013
2. Feb 8, 2013

### Staff: Mentor

Please show us the function whose arc length you need to find. If you have made a mistake in setting up the integral, then there's no point in trying to figure out how to integrate it. In other words, please provide the original problem statement.

Last edited: Feb 8, 2013
3. Feb 8, 2013

### SclayP

Sorry, you're right the funcion is
$$f(x)=e^x$$
between 0 and 1

4. Feb 8, 2013

### Karnage1993

Wouldn't you need to parametrize the function first? I.e, let $\vec\gamma(t) = (t, e^t)$?

5. Feb 8, 2013

### SclayP

In the first place i wouldn't know how to do it, and in the other hand i dont think it's necesary i mean, i think the way i did it it's correct but i'm stuck in that step. I used the next formula

$L=\int \sqrt{(\frac{dy}{dx})^{2} + 1}$

6. Feb 8, 2013

### Staff: Mentor

SclayP,
Your work looks OK, but I didn't check it. Assuming that it actually is OK, here's a link to a wiki topic on long division - http://en.wikipedia.org/wiki/Polynomial_long_division

You want to write v3/(v2 - 1) as a polynomial plus a proper rational expression. It will turn out to v + (proper rational expression)

7. Feb 8, 2013

### Karnage1993

Well essentially, your formula is the magnitude of the derivative of $\vec\gamma(t)$, which is what we'll use. Nevertheless, you seem to be having trouble using substitution. For example, on your second substitution, you don't have a $e^u$ to borrow to create a $dt$.

8. Feb 8, 2013

### SclayP

I dont need one i just say
$$t=e^{u}+1 ; dt=e^{u}du$$ and so $$du=\frac{dt}{e^{u}}=\frac{dt}{t-1}$$

9. Feb 8, 2013

### Karnage1993

Ahh, I see. I don't use this technique that often so it wasn't immediately obvious to me.

Okay, let's work on $\displaystyle\int \frac{v^3}{v^2 - 1} dv$.

Using long division, you can rewrite $\displaystyle\frac{v^3}{v^2 - 1}$ as $\displaystyle v + \frac{v}{v^2 - 1}$. This should make integrating much easier.

10. Feb 8, 2013

### SclayP

Yes, i'v already done that and then i integrated by part the integral $\int\frac{v}{(v-1)(v+1)}$ and i come up with this $\frac{v^2}{2} |^{\sqrt{e^2 +1}}_{\sqrt{2}} + \frac{1}{2} \int^{\sqrt{e^2 +1}}_{\sqrt{2}}\frac{1}{v-1} + \frac{1}{v+1}\, dv$

So when i integrate 1/(v-1) and 1/(v+1) it's left ln|v-1| and ln|v+1| but then when i evaluate the function the resault it's wrong (i know because i check with the calculator the first integral)...I don't know what the problem is...

11. Feb 8, 2013

### Karnage1993

You don't need to split the fraction $\displaystyle\frac{v}{v^2 -1}$. Just let $w = v^2 - 1$ and $\displaystyle \frac{1}{2} dw = v dv$. In this case, you borrow the $v$ to form the $dw$.

$\displaystyle\frac{1}{2} \int \frac{1}{w} \ dw = \frac{1}{2} ln(v^2 - 1)$.

From this point, simply substitute back for $v, t, u$ and solve the integral from 0 to 1. You may have made a mistake during the changing of the limits.

12. Feb 8, 2013

### SclayP

well yes you are right be we ended up with the same resault because ln(x^2 -1) equals ln[(x-1)(x+1)] equals ln(x-1) + ln(x+1)....

but knowing that we came to the same and assuming it's all correct why is the lenght not the same that the one that give my calculator...

13. Feb 8, 2013

### Karnage1993

\begin{align*}\displaystyle \frac{v^2}{2} + \frac{1}{2} ln(v^2 - 1) & = \frac{t}{2} + \frac{1}{2} ln(t - 1)\\ & = \frac{e^u + 1}{2} + \frac{1}{2} ln(e^u + 1 - 1)\\ & = \frac{e^{2x} + 1}{2} + \frac{1}{2} ln(e^{2x})\\ & = \frac{e^{2x} + 1}{2} + \frac{1}{2} |2x| \\ & = \frac{e^{2x} + 1}{2} + |x| \end{align*}

Evaluating from 0 to 1, we have:

$\displaystyle \frac{e^{2} + 1 - e^{1} - 1}{2} + 2 = \frac{e^{2} - e^{1}}{2} + 2$

Is this what you get?

14. Feb 8, 2013

### SclayP

Yes, that is what I GET but calculator says that the resault it is 2...and what we get by doing it "by hand" it 4.19 so, there is a problem in the proceedure.

15. Feb 8, 2013

### SithsNGiggles

I haven't looked over your work, but this can also be done via the following substitution:
$e^x=\tan u\\ e^x \;dx=\sec^2u\;du\\ \tan u\;dx=\sec^2u\;du\\ dx=\dfrac{\sec^2u}{\tan u}du$

16. Feb 8, 2013

### Karnage1993

Okay, I've went through your substitutions and noticed the error.

$\displaystyle v = \sqrt{t}, dv = \frac{1}{2\sqrt{t}}\ dt$
$\displaystyle 2\sqrt{t}\ dv = dt$
$\displaystyle 2v \ dv = dt$

Which means, $\displaystyle\frac{1}{2} \int \frac{\sqrt{t}}{t - 1} \ dt = \frac{1}{2} \int \frac{v}{v^2 - 1} 2v \ dv = \int \frac{v^2}{v^2 - 1}\ dv$.

So the $v^3$ should be a $v^2$!

17. Feb 8, 2013

### SclayP

Oh! hahaha i can't believe it was that, and i can't believe i didnt notice while doing it...well thank you for you help...

PD: I know it's late but...sorry for my english