Chapter 6.1, Problem 45E

To determine

To Find: The value of $m$ for which the line $y=mx$ and the curve $y=\frac{x}{\left(x^2+1\right)}$ enclose the region and the find the area of the region.

Answer to Problem 45E

The value of the area is $m-\ln m-1$ and the required value of $0<m<1$ in the first quadrant and the third quadrant.

Explanation of Solution

Given:

The given equation is,

  $y=\frac{x}{\left(x^2+1\right)}$

Calculation:

Consider that $y=mx$ and the curve $y=\frac{x}{\left(x^2+1\right)}$ enclose the region when the curves intersect each other at least for 2 points.

The curves intersect each other when,

  $\begin{array}{l}mx=\frac{x}{\left(x^2+1\right)}\\ x\left(m{x}^2+m-1\right)=0\\ x=10,\sqrt{\frac{1-m}{m}}\end{array}$

The value of $x=10,\sqrt{\frac{1-m}{m}}$ is defined for the region when $0<m<1$ .

For $y=mx$ and $y=\frac{x}{x^2+1}$ enclose the region when $0<m<1$ for $m=1$ and there is the first one point of the intersection $\left(0,0\right)$ . Since, the value of $y=mx$ and $y=\frac{x}{x^2+1}$ are the odd function. This shows that the area of the enclosed by the curves will be symmetric about the origin as shown in Figure 1

  

Figure 1

The area enclosed by $y=mx$ and $y=\frac{x}{x^2+1}$ for $0<m<1$ is,

  $\begin{array}{c}A=2{\displaystyle {\int }_0^{\frac{\sqrt{1-m}}{m}}\left(\frac{x}{x^2+1}-mx\right)}dx\\ =2{\displaystyle {\int }_0^{\frac{\sqrt{1-m}}{m}}\left(\frac{x}{x^2+1}\right)}-2{\displaystyle {\int }_0^{\frac{\sqrt{1-m}}{m}}mxdx}\end{array}$

Solve for $2{\displaystyle {\int }_0^{\frac{\sqrt{1-m}}{m}}\left(\frac{x}{x^2+1}\right)}$ as,

Substitute $t^2+1=t$ then $2xdx=dt$ or $xdx=\frac{1}{2}dx$ . For $x=0$ and $t=1$ then,

  $\begin{array}{c}x=\sqrt{\frac{1-m}{m}}\\ t=\frac{1-m}{m}+1\\ =\frac{1}{m}\end{array}$

Then,

  $\begin{array}{l}{\displaystyle {\int }_0^{\frac{\sqrt{1-m}}{m}}\left(\frac{x}{x^2+1}\right)}=\frac{1}{2}{\displaystyle {\int }_1^{\frac{1}{m}}\frac{dt}{t}}\\ \frac{1}{2}{\displaystyle {\int }_1^{\frac{1}{m}}\frac{dt}{t}}=\frac{1}{2}{\left[\ln t\right]}_1^{\frac{1}{m}}\\ \frac{1}{2}{\displaystyle {\int }_1^{\frac{1}{m}}\frac{dt}{t}}=\frac{1}{2}\ln \frac{1}{m}\end{array}$

The next step is to evaluate $2{\displaystyle {\int }_0^{\frac{\sqrt{1-m}}{m}}mxdx}$ as,

  $\begin{array}{c}{\displaystyle {\int }_0^{\frac{\sqrt{1-m}}{m}}mxdx}=m{\displaystyle {\int }_0^{\frac{\sqrt{1-m}}{m}}xdx}\\ =m{\left[\frac{x^2}{2}\right]}_0^{\frac{\sqrt{1-m}}{m}}\\ =\frac{1-m}{2}\end{array}$

Then, the area of the figure is,

  $\begin{array}{c}A=2\left(\frac{1}{2}\right)\ln \frac{1}{m}-2\left(\frac{1-m}{2}\right)\\ =m+\ln \frac{1}{m}-1\\ =m-\ln m-1\end{array}$