Question

Find the values of m for y = mx that enclose a region with y = x / (x^2 + 1) and find the area of this bounded region.

So I set the two functions equal to each other to solve for x in terms of m:

\[mx = \frac{x}{x^{2} + 1} \\\\
mx(x^{2} + 1) = x \\\\
mx^{3} + (m - 1)x = 0 \\\\
x(mx^{2} + (m - 1)) = 0 \\\\\]
So when x = 0 and:

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

Solve: \[m\sqrt{\frac{m - 1}{m}} = \frac{\sqrt{\frac{m - 1}{m}}}{\frac{m - 1}{m} + 1} \\\\
m = \frac{1}{\frac{m - 1}{m} + 1} \\\\
m - 1 + m = 1 \\\\
2m = 2 \\\\
m = 1\]

Answer 1

So at y = 1x or y = x, it intersects the graph y = 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:

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

Answer 2

Because it's symmetrical I guess. Evaluating:

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

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

Not sure where I went wrong here.

Answer 3

When solving for x in your first post, you should get x = 0 (which you did) and
\[\color{red}{x^2=\frac{1-m}{m}}, \text{ where you got}\\
x^2=\frac{m-1}{m}\]
I'm not sure if that changes the rest of your work. I just suspected there might be some mistake in the algebra, and there was! I'll see if it changes anything.

Answer 4

So, it looks like the area should be
\[\ln\left(\frac{1-m}{m}+1\right)+m-1\]

Answer 5

Okay, after that correction, the only solution I get is x = 0. I can't do the integration because I can't solve for any boundaries for which there is a region bounded by the two functions..

Answer 6

I'm not sure I understand... What's stopping you from getting three roots to
\[mx=\frac{x}{x^2+1}?\]

Answer 7

How do I solve for the values of m? After I fixed my algebra error, I can't seem to find the boundaries anymore

Answer 8

No, you solve for x, and that gives you the limits of integration. You did that just fine in your first post (minus the minor mistake). The integral should be
\[2\int_{0}^{\sqrt{\frac{1-m}{m}}} \left(\frac{x}{x^2+1}-mx\right)\;dx\], and I've already posted the area in an earlier post.

Answer 9

But for what values of 'm' is this applicable? Is it just the domain of the area?

Answer 10

Think about what values of m make the upper limit of integration undefined.
\[\sqrt\frac{1-m}{m}\]
If m=1, you have zero area.
If m>1, you have a complex number.
If m<0, again, you have a complex number.
If m=0, it's undefined. I'm pretty sure the domain for m remains the same (0 < m < 1).

Answer 11

let
\[
f\left(x\right)=\frac{x}{x^{2}+1},\, g\left(x\right)=mx
\]


If g(x) encloses a region with f(x), they intersect when

\begin{eqnarray*}
\frac{x}{x^{2}+1} & = & mx\\
x & = & mx^{3}+mx\\
mx^{3}+(m-1)x & = & 0
\end{eqnarray*}


Besides the solution of x = 0, the equation needs to have more than
1 solution (otherwise the area isn't enclosed, because it's still
open at 1 side). After factorising we get:
\[
x(mx^{2}+m-1)=0
\]


So mx\textasciicircum{}2 + m - 1 must have 1 or more solutions which
aren't 0, which means that

\[
\frac{0\pm\sqrt{-4m\left(m-1\right)}}{2m}=\frac{\pm\sqrt{-4m^{2}+4m}}{2m}\neq0
\]


which gives us
\[
\sqrt{-4m^{2}+4m}\neq0,\mbox{ so }-4m^{2}+4m>0
\]


-4m^2 + 4m is a parabola that opens downward, and
intersects the x-axis at m = 0 and m = 1, so when 0 < m < 1, y = mx encloses a region with x/(x^2 + 1) and you can calculate the area using the integral you already posted.

Answer 12

\textascircuum{} = ^, I don't know why it did that