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Area bounded by a curve and arbitrary line

  1. Feb 17, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the values of m for y = mx that enclose a region with [itex]y = \frac{x}{x^2 + 1}[/itex] and find the area of this bounded region.

    2. Relevant equations

    3. The attempt at a solution
    So I set the two functions equal to each other to solve for x in terms of m:

    [itex]mx = \frac{x}{x^{2} + 1} \\\\
    mx(x^{2} + 1) = x \\\\
    mx^{3} + (m - 1)x = 0 \\\\
    x(mx^{2} + (m - 1)) = 0 \\\\[/itex]

    So when x = 0 and:

    [itex]mx^{2} + m - 1 = 0 \\\\
    x = \sqrt{\frac{m - 1}{m}}[/itex]

    Now to solve for m, replace [itex]mx = \frac{x}{x^{2} + 1}[/itex] with [itex]x = \sqrt{\frac{m - 1}{m}}[/itex]:

    [itex]m\sqrt{\frac{m - 1}{m}} = \frac{\sqrt{\frac{m - 1}{m}}}{\frac{m - 1}{m} + 1} \\\\
    m = \frac{1}{\frac{m - 1}{m} + 1} \\\\
    m - 1 + m = 1 \\\\
    2m = 2 \\\\
    m = 1[/itex]

    So at y = 1x or y = x, it intersects the graph [itex]y = \frac{x}{x^2 + 1}[/itex] exactly. Looking at the graph, if the slope is less steep, it looks like it will create a region. I don't know how to mathematically show this. So I would guess between 0 < m < 1 (not inclusive I guess?) it will create a region. And for the area, the integral I set up is:

    [itex]2 \cdot \int_{0}^{\sqrt{\frac{m - 1}{m}}} (\frac{x}{x^{2} + 1} - mx) dx[/itex]

    Because it's symmetrical I guess. Evaluating:

    [itex]2 \cdot \int_{0}^{\sqrt{\frac{m - 1}{m}}} (\frac{x}{x^{2} + 1} - mx) dx \\\\
    ln(x^2 + 1) - [m(x^{2} + 1)] \\\\
    ln(\frac{m - 1}{m} + 1) - [m(\frac{m - 1}{m} + 1)][/itex]

    So the area is:
    [itex]ln(\frac{m - 1}{m} + 1) - 2m + 1[/itex] for 0 < m < 1

    Not sure where I went wrong here.
     
  2. jcsd
  3. Feb 17, 2013 #2
    Your process is right but you went wrong almost immediately. Yes, an area is enclosed only when [itex]0<x<1[/itex]. When I solved for [itex]m[/itex], here is what I got:

    [itex]mx=\frac{x}{x^2+1}[/itex]
    [itex]m=\frac{1}{x^2+1}[/itex]
    [itex]m(x^2+1)=1[/itex]
    [itex]x^2+1=\frac{1}{m}[/itex]
    [itex]x^2=\frac{1}{m}-1[/itex]
    [itex]x^2=\sqrt{\frac{1}{m}-1}[/itex]

    From there, proceed the exact way you did and you should get the following solution for the area:

    [itex]ln(\frac{1}{m})+m-1[/itex]
     
  4. Feb 17, 2013 #3
    Okay, I see my algebra error, but after you get [itex]x=\sqrt{\frac{1}{m}-1}[/itex], how do you solve for the values of m and get the boundaries 0 < x < 1? I can't do it anymore after I've fixed my error..
     
  5. Feb 17, 2013 #4

    haruspex

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    [itex]x^2=\frac{1}{m}-1[/itex] allows you to write an inequality for [itex]\frac{1}{m}-1[/itex]. Simplify that, considering the cases m > or < 0 separately.
     
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