Chapter 7.2, Problem 20E

To determine

To calculate : The area of the region enclosed by the given line and curve.

Answer to Problem 20E

The required area is $1\frac{2}{3}$ .

Explanation of Solution

Given information :

The line and curve are: $y=\sqrt{\left|x\right|}$ and $5y=x+6$ .

Calculation :

Draw the graph of the given line and curve.

  

For calculating the area enclosed by the graphs first we need to calculate the points where the points intersect.

The curves intersect at three points:

  $x=-1,x=4\text{ and }x=9$ .

Because of the absolute value sign, break the integral up at $x=0$ also:

The area can be divided into three parts

First, ${\displaystyle {\int }_{-1}^0\left(\frac{x+6}{5}-\sqrt{-x}\right)}dx$

Second, ${\displaystyle {\int }_0^4\left(\frac{x+6}{5}-\sqrt{x}\right)}dx$

Third, ${\displaystyle {\int }_4^9\left(\sqrt{x}-\frac{x+6}{5}\right)}dx$

Hence, the total area is the sum of the above three areas.

  $\begin{array}{l}{\displaystyle {\int }_{-1}^0\left(\frac{x+6}{5}-\sqrt{-x}\right)}dx+{\displaystyle {\int }_0^4\left(\frac{x+6}{5}-\sqrt{x}\right)}dx+{\displaystyle {\int }_4^9\left(\sqrt{x}-\frac{x+6}{5}\right)}dx\\ ={\left[\frac{\frac{1}{2}{x}^2+6x}{5}+\frac{2}{3}{\left(-x\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}_{-1}^0+{\left[\frac{\frac{1}{2}{x}^2+6x}{5}-\frac{2}{3}{x}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}_0^4+{\left[\frac{2}{3}{x}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{\frac{1}{2}{x}^2+6x}{5}\right]}_4^9\\ =\left[0-\left(-\frac{11}{10}+\frac{2}{3}\right)\right]+\left[\left(\frac{32}{5}-\frac{16}{3}\right)-0\right]+\left[\left(18-\frac{189}{10}\right)-\left(\frac{16}{3}-\frac{32}{3}\right)\right]\\ =\frac{13}{30}+\frac{16}{15}+\frac{1}{6}\\ =\frac{5}{3}\\ =1\frac{2}{3}\end{array}$

Therefore, the required area is $1\frac{2}{3}$ .