My Math Forum Area bounded by a curve and arbitrary line

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 February 17th, 2013, 08:22 AM #1 Member   Joined: Aug 2012 Posts: 88 Thanks: 0 Area bounded by a curve and arbitrary line 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: $mx^{2} + m - 1 = 0 \\\\ x = \sqrt{\frac{m - 1}{m}}$ Now to solve for m, replace $mx= \frac{x}{x^{2} + 1}$ with $x= \sqrt{\frac{m - 1}{m}}$: $m\sqrt{\frac{m - 1}{m}} = \frac{\sqrt{\frac{m - 1}{m}}}{\frac{m - 1}{m} + 1} \\\\ m = \frac{1}{\frac{m - 1}{m} + 1} \\\\ m - 1 + m = 1 \\\\ 2m = 2 \\\\ m = 1$ 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: $2 \cdot \int_{0}^{\sqrt{\frac{m - 1}{m}}} (\frac{x}{x^{2} + 1} - mx) dx$ 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.
 February 17th, 2013, 08:34 AM #2 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: Area bounded by a curve and arbitrary line First you have to have the following : $x=\pm\sqrt{\frac{1-m}{m}}$
February 17th, 2013, 08:45 AM   #3
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Re: Area bounded by a curve and arbitrary line

Quote:
 Originally Posted by zaidalyafey First you have to have the following : $x=\pm\sqrt{\frac{1-m}{m}}$
Oh, that makes sense. So if I plug it back into the original equation I should get 1 = 1. So then I'm just going in circles and not sure what to do

 February 17th, 2013, 09:08 AM #4 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: Area bounded by a curve and arbitrary line Ok , then tell me if you get the final solution right .

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### what values of m y=mx and y=x/(x^2 1) enclose a region

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