Chapter 8.2, Problem 1E

To determine

To find the area of the shaded region.

Answer to Problem 1E

The required area of shaded region is, $\pi /2$ 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, $y={\cos }^{2}x.$

  

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,

  $\begin{array}{l}\text{ A}={\displaystyle {\int }_a^b\left[f\left(x\right)-g\left(x\right)\right]}dx\\ \text{ A}={\displaystyle {\int }_0^{\pi }\left[\left(1\right)-\left({\cos }^{2}x\right)\right]}dx\\ \text{ A}={\displaystyle {\int }_0^{\pi }\left[1-{\cos }^{2}x\right]}dx\\ \text{ A}={\displaystyle {\int }_0^{\pi }{\sin }^{2}x}dx\\ \text{ A}={\displaystyle {\int }_0^{\pi }\frac{1-\cos 2x}{2}dx}\left[{\text{sin}}^{2}x=\frac{1-\cos 2x}{2}dx\right]\\ \text{ A}=\frac{1}{2}{\displaystyle {\int }_0^{\pi }\left(1-\cos 2x\right)dx}\\ ={\left[\frac{1}{2}\left(x-\sin 2x\left(\frac{1}{2}\right)\right)\right]}_0^{\pi }\\ =\left[\frac{1}{2}\left(\pi -\sin 2\pi \left(\frac{1}{2}\right)\right)\right]-\left[\frac{1}{2}\left(0-\sin 2.0\left(\frac{1}{2}\right)\right)\right]\\ =\left[\frac{1}{2}\pi -\sin 2\pi \left(\frac{1}{4}\right)\right]-\left[0-0\left(\frac{1}{2}\right)\right]\\ =\left[\frac{1}{2}\pi -0\left(\frac{1}{4}\right)\right]-\left[0-0\left(\frac{1}{2}\right)\right]\\ =\left[\frac{1}{2}\pi -0\right]-\left[0-0\right]\\ =\frac{1}{2}\pi \\ =\frac{\pi }{2}\end{array}$

Hence, the required area of shaded region $\pi /2$ squared units.