Area bounded by a curve and arbitrary line

[PhizKid]

Homework Statement

Find the values of m for y = mx that enclose a region with $y = \frac{x}{x^2 + 1}$ 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:

$mx = \frac{x}{x^{2} + 1} \\\\<br /> mx(x^{2} + 1) = x \\\\<br /> mx^{3} + (m - 1)x = 0 \\\\<br /> x(mx^{2} + (m - 1)) = 0 \\\\$

So when x = 0 and:

$mx^{2} + m - 1 = 0 \\\\<br /> x = \sqrt{\frac{m - 1}{m}}$

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

$m\sqrt{\frac{m - 1}{m}} = \frac{\sqrt{\frac{m - 1}{m}}}{\frac{m - 1}{m} + 1} \\\\<br /> m = \frac{1}{\frac{m - 1}{m} + 1} \\\\<br /> m - 1 + m = 1 \\\\<br /> 2m = 2 \\\\<br /> m = 1$

So at y = 1x or y = x, it intersects the graph $y = \frac{x}{x^2 + 1}$ 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:

$2 \cdot \int_{0}^{\sqrt{\frac{m - 1}{m}}} (\frac{x}{x^{2} + 1} - mx) dx$

Because it's symmetrical I guess. Evaluating:

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

So the area is:
$ln(\frac{m - 1}{m} + 1) - 2m + 1$ 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 $0<x<1$. When I solved for $m$, here is what I got:

$mx=\frac{x}{x^2+1}$
$m=\frac{1}{x^2+1}$
$m(x^2+1)=1$
$x^2+1=\frac{1}{m}$
$x^2=\frac{1}{m}-1$
$x^2=\sqrt{\frac{1}{m}-1}$

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

$ln(\frac{1}{m})+m-1$

 

[PhizKid]

Your process is right but you went wrong almost immediately. Yes, an area is enclosed only when $0<x<1$. When I solved for $m$, here is what I got:

$mx=\frac{x}{x^2+1}$
$m=\frac{1}{x^2+1}$
$m(x^2+1)=1$
$x^2+1=\frac{1}{m}$
$x^2=\frac{1}{m}-1$
$x^2=\sqrt{\frac{1}{m}-1}$

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

$ln(\frac{1}{m})+m-1$

Okay, I see my algebra error, but after you get $x=\sqrt{\frac{1}{m}-1}$, 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 ?

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