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Integration of y=(1+0.25x^2)^0.5

  1. Jul 23, 2015 #1
    1. The problem statement, all variables and given/known data
    I am told to find out the length of he curve y=¼x^2

    2. Relevant equations
    The length of the curve should be

    ∫√(1+¼x^2)dx


    3. The attempt at a solution
    I substituted tanx and got ∫2/cos^3x dx

    So how could I continue and work out the solution? Should I use integration by part, or I had already done wrong?
     
  2. jcsd
  3. Jul 23, 2015 #2

    SammyS

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    Hello gaywaiha. Welcome to PF !

    Change ##\displaystyle \ \frac{1}{\cos^3(x)} \ ## to ##\ \sec^3(x) \ . ##
     
  4. Jul 23, 2015 #3
    How ? Instead substitute x = 2*tan(θ) . You will however have to use by-parts later .
     
  5. Jul 23, 2015 #4
    You've already done a couple of things wrong.

    1. What does the word Precalculus mean to you? I'm moving this to the Calculus and Beyond forum

    2. Your substitution should be x = 2 tan θ

    3. Show us your work

    Chet
     
  6. Jul 23, 2015 #5

    Ray Vickson

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    As stated, the problem has an easy solution that I am willing to repeat here: the length of the curve ##y = (1/4)x^2## is infinite. No detailed calculations needed!
     
  7. Jul 23, 2015 #6
    Guys thank you very much. I am sorry that I substituted 2tanx and forgot to type the 2. But now I finally found the answer by separate 1/cos^3x into (1/cosx)(tanx)' and work it out using integration by part.

    Thank you
     
  8. Jul 23, 2015 #7

    HallsofIvy

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    However, you still haven't addressed Chestermiller's point- until you have put some bounds on it, the length is infinite!
     
  9. Jul 23, 2015 #8
    Sorry about that. The interval is (0≤x≤2).
     
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