Chapter 6, Problem 2P

To determine

To find: The number $c$ such that the area of the shaded regions are equal.

Answer to Problem 2P

  $c=\frac{32}{27}$

Explanation of Solution

Given information :

Horizontal line $y=c$ is intersecting the curve $y=8x-27x^3$ and the area $R_1=R_2$

  

Formula used :

  ${\displaystyle \underset{a}{\overset{b}{\int }}{x}^n}dx={{\displaystyle \left[\frac{x^{n+1}}{n+1}\right]}}_a^b$

When $y=c$ is intersecting the curve $y=8x-27x^3$ , at intersecting point both the functional values are equal.

Calculation :

  

Given line is $y=c\mathrm{...........}\left(i\right)$

Given curve is $y=8x-27x^3$ ..................(ii)

At the intersecting points of (i) and (ii) ,

  $8x-27x^3=c$ ....................(iii)

The region $R_1$ limes between $x=0\text{ and }x=a$ and $R_2$ limes between $x=a\text{ and }x=b$

We have $R_1=R_2\mathrm{................}\left(i\right)$

So, ${\displaystyle \underset{0}{\overset{a}{\int }}\left[c-\left(8x-27x^3\right)\right]}dx={\displaystyle \underset{a}{\overset{b}{\int }}\left[\left(8x-27x^3\right)-c\right]dx}$

  $\Rightarrow {{\displaystyle \left[cx-\frac{8x^2}{2}+27\frac{x^4}{4}\right]}}_0^a={{\displaystyle \left[\frac{8x^2}{2}-27\frac{x^4}{4}-cx\right]}}_a^b$

Now, substituting $c$ by $8x-27x^3$ , we get,

  $\begin{array}{l}{{\displaystyle \left[8x^2-27x^4-\frac{8x^2}{2}+27\frac{x^4}{4}\right]}}_0^a={{\displaystyle \left[\frac{8x^2}{2}-27\frac{x^4}{4}-8x^2+27x^4\right]}}_a^b\\ \Rightarrow {{\displaystyle \left[4x^2-\frac{81}{4}{x}^4\right]}}_0^a={{\displaystyle \left[-4x^2+\frac{81}{4}{x}^4\right]}}_0^a\\ \Rightarrow \cancel{4a^2}-\cancel{\frac{81}{4}{a}^4}=-4b^2+\cancel{4a^2}+\frac{81}{4}{b}^4-\cancel{\frac{81}{4}{a}^4}\\ \Rightarrow b^2\left(\frac{81}{4}{b}^2-4\right)=0\\ \Rightarrow \frac{81}{4}{b}^2-4=0\left[\text{as }b\ne 0\right]\\ \Rightarrow b^2=\frac{16}{81}\\ \Rightarrow b=\frac{4}{9}\end{array}$

When $x=b$ , the equation (iii) be

  $\begin{array}{c}c=f\left(b\right)\\ =8b-27b^3\\ =8\times \frac{4}{9}-27\times \frac{64}{9\times 9\times 9}\\ =\frac{32}{9}-\frac{64}{27}\\ =\frac{96-64}{27}=\frac{32}{27}\end{array}$