# Homework Help: Area - Riemann Sum/Integration problem

1. Jan 13, 2006

### Xcron

This problem is in the section of my book called "The Natural Logarithmic Function" so I am guessing that I would have to use a natural log somewhere...but anyways...the problem says:

For what values of $$m$$ do the line $$y=mx$$ and the curve $$y=\frac{x}{(x^2+1)}$$ enclose a region? Find the area of the region.

For some reason...I related this problem to an optimization problem because I thought if I could find the line with the greatest slope that is tangent to the curve, then I would know the values for which $$m$$ can be used with...but that didn't seem to use any natural logs or anything..

2. Jan 13, 2006

### TD

It may be a good idea to find the intersection points. If there aren't any intersection points, there's no enclosed region.
It can also be useful to graph the problem for a few values of m, this gives you a better insight in the question.

3. Jan 13, 2006

### Xcron

Well, from graphing the problem...it seems like $$y=x$$ may be the right bound of the interval for which $$y=mx$$ and the curve enclose a region...but I would need to show a calculation for this...

I tried: $$mx= \frac{x}{(x^2+1)} m=\frac{1}{x^2+1}$$ but that only gives me the same equation of the curve or perhaps it gives me the equation for the bounds of the interval for which $$m$$ can be used with..

Seems like $$m$$ would belong to $$[0,1)$$...

4. Jan 13, 2006

### TD

I think you're on the right track:

$$m = \frac{1} {{x^2 + 1}} \Leftrightarrow m\left( {x^2 + 1} \right) = 1 \Leftrightarrow mx^2 + m - 1 = 0$$

Now, solving this would give the intersection points (which you'll be needing for you integration limits, I assume), but for the moment we're only interested in when there are solutions (i.e. when they intersect and thus enclose a region). You can express this by saying that the discriminant of this quadratic equation in x has to be positive, this will give a condition on m.

5. Jan 14, 2006

### Xcron

Ahh....well, sorry to say this but I am still somewhat confused as to how that equation would be solved...two variables...need to solve for $$m$$ but that would just give us $$\frac{1}{x^2+1}$$ ... hmmm...

6. Jan 14, 2006

### siddharth

For the the straight line $$y=mx$$ and the curve $$y=\frac{x}{x^2+1}$$ to enclose some finite region, there should be atleast 2 points of intersection.
One point is (0,0). To find the other points, you need to solve
$$mx^2 + m -1 =0$$.
So for what values of 'm' does a solution to the above equation exist? That will give you the values of 'm' for which a region is enclosed.
To find the area, find the points of intersection in terms of 'm'. You don't need to solve for both the variables, you only need to find x in terms of 'm'.

7. Jan 14, 2006

### Xcron

Ohhhhhh, now I understand. I had understood the part about the 2 points of intersection but was trying to figure out which variable we solve that equation for and I now realize that we find the point on the curve using the intersecting line...which is why we solve for the x in terms of m. I got it, hehe.

so to continue...I would then continue to do:
$$\int_{0}^{\sqrt{\frac{1-m}{m}}} (\frac{x}{x^2+1}-mx)dx$$

Is this right?..m would also be a constant, right?

Last edited: Jan 14, 2006
8. Jan 14, 2006

### siddharth

Yeah, that is almost right. What about the other point of intersection (ie, $-\sqrt\frac{1-m}{m}$? so the area will be twice your integral.

9. Apr 10, 2008

$$mx^2 + m -1 =0$$