Chapter 8.2, Problem 47E

To determine

To Find:

To find the area of the propeller shaped region enclosed between the given graphs $y=\frac{2x}{x^2+1}$ and $y=x^3$ .

Answer to Problem 47E

The area of the propeller shaped region enclosed between the given graph is $\frac{4\ln \left(2\right)-1}{2}$ .

Explanation of Solution

Given information:

The given functions are $y=\frac{2x}{x^2+1}$ and $y=x^3$ .

The red curve is the graph of $y=\frac{2x}{x^2+1}$ and the blue curve is the graph of $y=x^3$ .

  

These curves intersects at $\left(1,1\right)$ and $\left(-1,-1\right)$ . Since both the areas are equivalent, the integral runs from $x=0$ to $x=1$ and then multiply by two.

The area of the propeller shaped region is given by,

  $\begin{array}{l}A=2{\displaystyle \underset{0}{\overset{1}{\int }}\left(\left(\frac{2x}{x^2+1}\right)-x^3\right)dx}\\ A=2\left[{\displaystyle \underset{0}{\overset{1}{\int }}\left(\frac{2x}{x^2+1}\right)}dx-{\displaystyle \underset{0}{\overset{1}{\int }}{x}^{3}dx}\right]\end{array}$

Consider the first integral and substitute $u=x^2+1$ ,the integral runs from $u=1,u=2$

  $\begin{array}{l}=2\cdot {\displaystyle {\int }_1^2\frac{1}{2u}}du\\ =2\cdot \frac{1}{2}{\left[\ln \left|u\right|\right]}_1^2\\ ={\left[\ln \left|u\right|\right]}_1^2\\ =\ln \left(2\right)\end{array}$

Now consider second integral ${\displaystyle \underset{0}{\overset{1}{\int }}{x}^{3}dx}$ ,

  ${\displaystyle \underset{0}{\overset{1}{\int }}{x}^{3}dx}={\left[\frac{x^{3+1}}{3+1}\right]}_0^1={\left[\frac{x^4}{4}\right]}_0^1=\frac{1}{4}$

Hence,

  $\begin{array}{l}A=2\left[{\displaystyle \underset{0}{\overset{1}{\int }}\left(\frac{2x}{x^2+1}\right)}dx-{\displaystyle \underset{0}{\overset{1}{\int }}{x}^{3}dx}\right]\\ A=2\left[\ln \left(2\right)-\frac{1}{4}\right]\\ A=2\left[\frac{4\ln \left(2\right)-1}{4}\right]\\ A=\frac{4\ln \left(2\right)-1}{2}\end{array}$

Therefore, the area of the propeller shaped region enclosed between the given graph is $\frac{4\ln \left(2\right)-1}{2}$ .