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Problem Application of Integrals(Arc Legth) - I'm stuck

  1. Feb 8, 2013 #1
    So i'm stuck resolving this integral...

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

    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. jcsd
  3. Feb 8, 2013 #2

    Mark44

    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
  4. Feb 8, 2013 #3
    Sorry, you're right the funcion is
    [tex]f(x)=e^x[/tex]
    between 0 and 1
     
  5. Feb 8, 2013 #4
    Wouldn't you need to parametrize the function first? I.e, let ##\vec\gamma(t) = (t, e^t)##?
     
  6. Feb 8, 2013 #5
    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

    [itex]L=\int \sqrt{(\frac{dy}{dx})^{2} + 1}[/itex]
     
  7. Feb 8, 2013 #6

    Mark44

    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)
     
  8. Feb 8, 2013 #7
    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##.
     
  9. Feb 8, 2013 #8
    I dont need one i just say
    [tex]t=e^{u}+1 ; dt=e^{u}du[/tex] and so [tex]du=\frac{dt}{e^{u}}=\frac{dt}{t-1}[/tex]
     
  10. Feb 8, 2013 #9
    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.
     
  11. Feb 8, 2013 #10
    Yes, i'v already done that and then i integrated by part the integral [itex]\int\frac{v}{(v-1)(v+1)}[/itex] and i come up with this [itex]\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[/itex]

    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...
     
  12. Feb 8, 2013 #11
    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.
     
  13. Feb 8, 2013 #12
    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...
     
  14. Feb 8, 2013 #13
    ##\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?
     
  15. Feb 8, 2013 #14
    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.
     
  16. Feb 8, 2013 #15
    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##
     
  17. Feb 8, 2013 #16
    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##!
     
  18. Feb 8, 2013 #17
    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
     
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