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Exact arc length of ln x

  1. Feb 28, 2012 #1
    1. The problem statement, all variables and given/known data

    Use the integration tables to find the exact arc length of the curve
    f(x)=ln x 1≤x≤e

    Reference the table number formula used

    Then approx. your answer and compare that to the approx. "straight line distance" between 2 points

    coordinates of two points (1,0) (e,1)

    2. Relevant equations
    ∫sqrt(1+(f'(x))^2)

    distance formula I'm guessing?


    3. The attempt at a solution
    y = ln(x)
    y' = (1/x)

    L = ∫ (1 to e) [ sqrt(1 + (1/x)^2) ] dx
    L = ∫ (1 to e) [ sqrt(1 + (1/x)^2) ] dx

    = ∫(1 to e) [ sqrt(1 + (1/x^2)) ] dx
    = ∫ (1 to e) [ sqrt( (x^2 + 1) / x^2) ) ] dx
    =∫ (1 to e) [(1/x)*sqrt(x^2 + 1) ] dx

    Integral # 28
    ∫ [(1/u)sqrt(a^2 + u^2)] du
    = sqrt(a^2 + u^2) - a*ln | [a + sqrt(a^2 + u^2)] / u] | + C

    In this integral, a = 1 and u = x

    Int (1 to e) [(1/x)*sqrt(x^2 + 1) ] dx =
    sqrt(1 + x^2) - ln | [1 + sqrt(1 + x^2)] / x] | (1 to e)

    =
    sqrt(1 + e^2) - ln | [1 + sqrt(1 + e^2)] / e] |
    =
    sqrt(2) - ln | [1 + sqrt(2)]] |

    0.53

    straight line =1.98

    Is this right?
     
  2. jcsd
  3. Feb 28, 2012 #2

    SammyS

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    Isn't a straight line the shortest distance between two pints?

    You only evaluated the anti-derivative at x = e, not at x = 1.
     
  4. Feb 28, 2012 #3
    Yes, so the distance formula would be right?

    And I evauluated it again, the answer should be ≈ 2.

    But, is the integration correct?
     
  5. Feb 28, 2012 #4

    SammyS

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    It is if the integral table is correct. You appear to have used it correctly.
     
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