Chapter 5.4, Problem 47E

To determine

To find: The area between a given curve and the x -axis in the given interval.

Answer to Problem 47E

The total area between the curve $y=2-x$ , $0\le x\le 3$ and x -axis is 2.5 units.

Explanation of Solution

Given data: The given curve is $y=2-x$ and the interval $0\le x\le 3$ .

Method/Formula used:

The area between a curve $y=f\left(x\right)$ and x -axis in the interval $a\le x\le b$ is given by

  $A={\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$ … (1)

Calculations:

The given curve is $y=2-x$ and the interval $0\le x\le 3$ is shown in Fig. (1).

  

From the Fig. 1, clearly, the total area A between the curve and x -axis is

  $A=\left|A_1\right|+\left|A_2\right|$ ... (2)

From the result (1),

  $\begin{array}{c}\left|A_1\right|=\left|{\displaystyle \underset{0}{\overset{2}{\int }}\left(2-x\right)dx}\right|\\ =\left|{\left[2x-\frac{x^2}{2}\right]}_0^2\right|\\ =\left|\left(2\left(2\right)-\frac{{\left(2\right)}^2}{2}\right)-\left(2\left(0\right)-\frac{{\left(0\right)}^2}{2}\right)\right|\\ =2\end{array}$

Similarly,,

  $\begin{array}{c}\left|A_2\right|=\left|{\displaystyle \underset{2}{\overset{3}{\int }}\left(2-x\right)dx}\right|\\ =\left|{\left[2x-\frac{x^2}{2}\right]}_0^2\right|\\ =\left|\left(2\left(3\right)-\frac{{\left(3\right)}^2}{2}\right)-\left(2\left(2\right)-\frac{{\left(2\right)}^2}{2}\right)\right|\\ =\left|1.5-2\right|\\ =0.5\end{array}$ , therefore,

Substitute 2 for $\left|A_1\right|$ and 0.5 for $\left|A_2\right|$ in result (2),

  $\begin{array}{c}A=2.0+0.5\\ =2.5\end{array}$

Therefore, the total area between the curve $y=2-x$ , $0\le x\le 3$ and x -axis is 2.5 units.