Area bounded by a curve and arbitrary line

[PhizKid]

Homework Statement

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.

Homework Equations

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.

 

[Arkuski]

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]

 

[PhizKid]

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]

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..

 

[haruspex]

how do you solve for the values of m ?

[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.