Chapter 8.2, Problem 10E

To determine

To find the area of the shaded region.

Answer to Problem 10E

The required area of shaded region is, $\frac{5}{6}$ 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 function is, $y=x^2\text{ , }x+y=2$ .

  

Calculation: The required area of shaded region is obtained as,

The graph curves to find the relative position in the plane,

The graph is therefore,

  $\begin{array}{l}\text{ A}={\displaystyle {\int }_a^b\left[f\left(t\right)-g\left(t\right)\right]}dt\\ ={\displaystyle {\int }_0^1{x}^2}dx+{\displaystyle {\int }_1^{2}2-x}dx\left(\begin{array}{l}\text{To solve equation in term of x and }\\ \text{evaluate two seperate integrals}\end{array}\right)\\ ={\left[\frac{1}{3}{x}^3\right]}_0^1+{\left[2x-\frac{1}{2}{x}^2\right]}_1^2\\ \text{ =}\frac{1}{3}+\left(\left(4-2\right)-\left(2-\frac{1}{2}\right)\right)\\ \text{ =}\frac{1}{3}+2-\frac{3}{2}\\ \text{ =}\frac{5}{6}\end{array}$

Hence, the required area of shaded region is, $\frac{5}{6}$ squared units.