Chapter 8.2, Problem 15E

To determine

To find the area of the region enclosed by the lines and curves.

Answer to Problem 15E

The area of regions enclosed by the lines and curves is $\frac{32}{3}$ squared units.

Explanation of Solution

Concept used: Area between curves.

If f and g are continuous with $f\left(x\right)\ge g\left(x\right)$ throughout [a,b], then the area between

The curves $y=f\left(x\right)\text{ and }y=g\left(x\right)$ from a to b is the integral of $\left[f-g\right]$ from ato b.

  $\begin{array}{l}A={\displaystyle {\int }_a^b\left[f\left(x\right)-g\left(x\right)\right]}dx\\ \end{array}$

Given information:

The given equations is, $y=x^2-2\text{ , }y=2$

Calculation: The area of the region enclosed by the lines and curves is obtained as,

  $\begin{array}{l}f\left(x\right)=x^2-2,f\left(x\right)=2\\ f\left(x\right)=f\left(x\right)\therefore x^2-2=2\therefore x^2=4\\ {\text{ X}}^2=4\\ \text{ X}=\pm 2\left(\text{solve for the limits}\right)\\ \text{ A}={\displaystyle {\int }_a^b\left[f\left(t\right)-g\left(t\right)\right]}dt\\ ={\displaystyle {\int }_{-2}^{2}2-x^2+2dx}\\ ={\displaystyle {\int }_{-2}^{2}4-x^{2}dx}\\ ={\left[4x-\frac{x^3}{3}\right]}_{-2}^2\\ =\left[4\left(2\right)-\frac{{\left(2\right)}^3}{3}\right]-\left[4\left(-2\right)-\frac{{\left(-2\right)}^3}{3}\right]\\ =\left(8-\frac{8}{3}\right)-\left(-8-\frac{-8}{3}\right)\\ =\left(\frac{24}{3}-\frac{8}{3}\right)+\left(\frac{24}{3}-\frac{8}{3}\right)\\ =\left(\frac{16}{3}\right)+\left(\frac{16}{3}\right)\\ =\frac{32}{3}\end{array}$

Hence, the area of regions enclosed by the lines and curves is $\frac{32}{3}$ squared units.