Chapter 8.2, Problem 58E

To determine

To Find: The area of the region enclosed by the curves

Answer to Problem 58E

The area of the region enclosed by the curves $=m-\ln (m)-1$

Explanation of Solution

Given information:

Two curves

  $y=\frac{x}{x^2+1}$ and $y=mx$ , $0<m<1$

Concept used: If $f(x)$ and $g(x)$ are continuous with $f(x)\ge g(x)$ throughout $[a,b]$ , then the area between the curves $y=f(x)$ and $y=g(x)$ from a to b is the integral of $[f(x)-g(x)]$ from a to b

  $A={\displaystyle \underset{a}{\overset{b}{\int }}[f(x)-g(x)]}dx$

Calculation:

Two curves

  $y=\frac{x}{x^2+1}$ and $y=mx$ , $0<m<1$

  $\begin{array}{l}\frac{x}{x^2+1}=mx\\ x=-\sqrt{\frac{1-m}{m}},\frac{\sqrt{1-m}}{m}\end{array}$

  

On the interval $\left[-\sqrt{\frac{1-m}{m}},0\right]$ , $mx\ge \frac{x}{x^2+1}$ and on interval $\left[0,\sqrt{\frac{1-m}{m}}\right]$ , $\frac{x}{x^2+1}\ge mx$

Therefore, the area is

  $\begin{array}{l}A={\displaystyle \underset{-\sqrt{\frac{1-m}{m}}}{\overset{0}{\int }}\left[mx-\frac{x}{x^2+1}\right]}dx+{\displaystyle \underset{0}{\overset{\sqrt{\frac{1-m}{m}}}{\int }}\left[\frac{x}{x^2+1}-mx\right]}dx\\ A=m-\ln (m)-1\end{array}$

The area of the region enclosed by the curves $=m-\ln (m)-1$