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Arc Length

  1. Jan 25, 2012 #1
    1. The problem statement, all variables and given/known data

    so I did

    2. Relevant equations

    s=∫ba√(1+[f'(x)]2dx

    3. The attempt at a solution
    y=2/3x3/2+7
    y'=x1/2 0≤x≤1
    =∫6√(1+(x1/2)2)dx
    =[2/3(1+x)3/2]7

    my answer came out wrong
     

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  2. jcsd
  3. Jan 26, 2012 #2

    Mark44

    Staff: Mentor

    Why do you have this inequality?
    What is that 7 doing?
     
  4. Jan 26, 2012 #3
    I saw it in an example for the inequality.. The 7 is for the a & b part I don't know how to place the 0 on the bottom.
     
  5. Jan 26, 2012 #4

    Mark44

    Staff: Mentor

    Your inequality is saying that the interval is [0, 1]. It's actually [0, 6].

    Here's how your integral looks using LaTeX.
    [tex]\int_0^6 \sqrt{1+x}~dx[/tex]

    Here's how to write that integral in LaTeX.

    [ tex]\int_0^6 \sqrt{1+x}~dx[ /tex]

    (The extra spaces in the tex and /tex tags prevent the browser from rendering the code inside.)

    Here's the antiderivative with limits of integration shown.
    [tex]\left. \frac{2}{3}(1 + x)^{3/2}\right|_0^6[/tex]

    The LaTeX for that.
    [ tex]\left. \frac{2}{3}(1 + x)^{3/2}\right|_0^6[ /tex]
     
  6. Jan 26, 2012 #5
    I got it correct :))
     

    Attached Files:

    • 29.jpg
      29.jpg
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      16.1 KB
      Views:
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  7. Jan 26, 2012 #6

    Mark44

    Staff: Mentor

    Good to know. That's what I got, too.
     
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