Chapter 6, Problem 14RE

To determine

To find :the total area between the curve and the $x$ -axis.

Answer to Problem 14RE

The total area between the curve and the $x$ -axis is: $2$

Explanation of Solution

Given information:

Given expression is : $y=\cos x,0\le x\le \pi$

Calculation:

If $f$ is continuous at every point of $[a,b],$ and if 1 $F$ is any anti-derivative of 3 $f$ on

  $[a,b],$ then

  ${{\displaystyle \int }}_a^{b}f(x)dx=F(b)-F(a)$

This part of the fundamental theorem is also called the integral evaluation theorem.

Proof:

Part 1 of the fundamental theorem tells us that an anti-derivative of $f$ exists, namely

  $G(x)={{\displaystyle \int }}_a^{x}f(t)dt$

Thus, if $F$ is any anti-derivative of $f$ then for some constant $F(x)=G(x)+C$ for some constant $C$

  $\begin{array}{cc}F(b)-F(a)& =[G(b)+C]-[G(a)+C]\\ & =G(b)-G(a)\\ & ={{\displaystyle \int }}_a^{b}f(t)dt-{{\displaystyle \int }}_a^{b}f(t)dt\\ & ={{\displaystyle \int }}_a^{b}f(t)dt-0\\ F(b)-F(a)& ={{\displaystyle \int }}_a^{b}f(t)dt\end{array}$

So,

  ${{\displaystyle \int }}_a^{b}f(x)dx=F(b)-F(a)$

The graph is:

  

Find the area of the region between curve and $x$ -axis by using above proof:

For $\left[0,\frac{\pi }{2}\right]$ function is positive and for $\left[\frac{\pi }{2},\pi \right]$ function is negative. So, for finding area we calculate two areas separately by using the integral value then add the two areas. For $[0,\pi ]$

  $f(x)=\cos x$

If $F$ any anti-derivative of $f$ then

  $F(x)=\sin x$

For $\left[0,\frac{\pi }{2}\right]$ function is positive. Thus

  $\begin{array}{cc}\text{area for}\left[0,\frac{\pi }{2}\right]& ={{\displaystyle \int }}_0^{\frac{\pi }{2}}(\cos x)dx\\ & =\left[\sin \frac{\pi }{2}\right]-[\sin 0]\\ & =1\end{array}$

For $\left[\frac{\pi }{2},\pi \right]$ function is negative. Thus,

  $\begin{array}{cc}\text{area for}\left[\frac{\pi }{2},\pi \right]& =-{{\displaystyle \int }}_{\frac{\pi }{2}}^{\pi }(\cos x)dx\\ & =-[\sin \pi ]+\left[\sin \frac{\pi }{2}\right]\\ & =-0+1\\ & =1\end{array}$

Total area for $[0,\pi ]=$ area for $\left[0,\frac{\pi }{2}\right]+$ area for $\left[\frac{\pi }{2},\pi \right]$

  $\begin{array}{l}=1+1\hfill \\ =2\hfill \end{array}$

Hence, the answer is : $2$