Chapter 6.4, Problem 41E

To determine

To find the area of the region between the given curve and x-axis.

Answer to Problem 41E

  $\frac{3}{2}$ .

Explanation of Solution

Given:

  $y=2-x,0\le x\le 3$

Calculation:

Required area is,

  $\begin{array}{l}y=2-x,0\le x\le 3\\ {\displaystyle \underset{0}{\overset{3}{\int }}ydx}={\displaystyle \underset{0}{\overset{3}{\int }}\left(2-x\right)dx}\\ ={\left[2x-\frac{x^2}{2}\right]}_0^3\\ =\left[2\left(3\right)-\frac{{\left(3\right)}^2}{2}\right]-\left[2\left(0\right)-\frac{{\left(0\right)}^2}{2}\right]\\ =6-\frac{9}{2}\\ =\frac{3}{2}\end{array}$