Chapter 5, Problem 16RE

To determine

To find: The between curve and x -axis.

Answer to Problem 16RE

The area A between curves $y=\cos x$ and x -axis in the interval $0\le x\le \pi$ is 2 units.

Explanation of Solution

Given data: The curve is $y=\cos x,0\le x\le \pi$ .

Method/Formula used:

The area between x -axis and the curve $y=f\left(x\right)$ in the interval $a\le x\le b$ is given by $A={\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$ , here the curve meet x -axis at points $\left(a,0\right)$ and $\left(b,0\right)$ .

Calculation:

The given curve is $y=\cos x$ whose graph is shown in Fig.1. The curve meet x -axis at points $\left(0,0\right)$ , $\left(\pi /2,0\right)$ and $\left(\pi ,0\right)$ .

  

Therefore, the area A between curves $y=\cos x$ and x -axis between points $\left(0,0\right)$ and $\left(\pi ,0\right)$ is

  $A=\left|{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\cos xdx}\right|+\left|{\displaystyle \underset{\pi /2}{\overset{\pi }{\int }}\cos xdx}\right|$

  $\begin{array}{c}A={\left|\sin x\right|}_0^{\pi /2}+{\left|\sin x\right|}_{\pi /2}^{\pi }\\ =\left|\sin \left(\pi /2\right)-\sin \left(0\right)\right|+\left|\sin \left(\pi \right)-\sin \left(\pi /2\right)\right|\\ =\left|1-0\right|+\left|0-1\right|\leftarrow \sin \left(0\right)=0,\sin \left(\pi /2\right)=1,\sin \left(\pi \right)=0\\ =2\end{array}$

Thus, the area A between curves $y=\cos x$ and x -axis in the interval $0\le x\le \pi$ is 2 units.