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Basic intergrate question x 3

  1. Nov 11, 2009 #1
    1. The problem statement, all variables and given/known data
    http://u1.imgupload.co.uk/1257897600/5a0e_image.bmp


    2. Relevant equations
    above


    3. The attempt at a solution
    1)
    needs some hints here.
    may i treat e^2x as (e^x)2 and then apply the formula to get the answers?

    2)
    http://u1.imgupload.co.uk/1257897600/bd3b_1.png

    i didnt write a full attempt but it is the flow of my thought.
    did my attempt correct? or in fact there are other easier way to get the answers?

    3)
    http://u1.imgupload.co.uk/1257897600/96d0_2.bmp
    this one i just wonder whether my work is correct or not. (shown in pic)

    thx a lots.:blushing:
     
  2. jcsd
  3. Nov 11, 2009 #2

    Gib Z

    User Avatar
    Homework Helper

    1) Assuming the formula is for the standard form [itex] \int \frac{1}{\sqrt{x^2-1}} dx[/itex], no you can not treat it that simply. You could try letting u=e^x.

    2) Your method will work. I've tried some things and that seems to be the most elementary.

    3) That is pretty much the correct evaluation of the "net area" but strictly the integral does not converge at all because the integrand is not well defined over the interval of integration.
     
  4. Nov 11, 2009 #3

    Mark44

    Staff: Mentor

    For 3, as Gib Z already noted, the integrand is undefined at a point in the interval. For this problem you need to split the integral into two improper integrals:
    [tex]\lim_{a \rightarrow 1^-} \int_0^a \frac{dx}{x - 1}~+~\lim_{b \rightarrow 1^+} \int_b^3 \frac{dx}{x - 1}[/tex]
    Evaluate each integral and take the limit. There's no guarantee that either improper integral will converge, though.
     
  5. Nov 11, 2009 #4
    actually how can i know when the question is asking me using improper integral? or it's needed to decide by myself?

    and for 2), i worked out the answer and found it's damn long and complex. so how can i check whether im correct?

    1)
    http://u1.imgupload.co.uk/1257897600/492b_aaa.png
    please judge

    thx again.
     
  6. Nov 11, 2009 #5

    Mark44

    Staff: Mentor

    I think this is wrong, which you can confirm by differentiating your final result. How did you go from this integral:
    [tex]\int \frac{du}{\sqrt{u^4 - 4u^2}}[/tex]
    to this?
    [tex]ln(u^2 + \sqrt{u^4 - 4u^2}) + C[/tex]

    I would have left the integral like this:
    [tex]\int \frac{du}{u\sqrt{u^2 - 4}}[/tex]
    and worked toward a trig substitution, with sec([itex]\theta[/itex]) = u/2.
     
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