#### To determine

**To:**

Sketch the region and find the enclosed area.

#### Answer

i) The sketch of the region is

ii) Area =4π

#### Explanation

**1) Concept:**

Formula:

The area A of the region bounded by the curves y=fx, y=g(x) and the lines x=a and x=b is

A= ∫abfx-gxdx

fx-gx=fx-gx when fx≥g(x)gx-fx when gx≥f(x)

**2) Given:**

y=cosx and

y=2-cosx, 0≤x≤2π

**3) Calculation:**

When cosx=2-cosx, we have that cosx=1.

Thus, the points of intersection are at x=0 and x=2π. The region is sketched in the following figure.

Here 2-cosx≥cosx when 0≤x≤2π. Therefore, let’s set

fx=2-cosx

gx=cosx

Therefore, the required area is

A=∫02π2-cosx-cosxdx

A=∫02π2-cosx-cosx dx

A=∫02π2-2cosx dx

Using the sum and difference rule and the constant multiple rule of integral, it becomes

A=2∫02πdx-2∫02πcosxdx

Compute the integral using the standard integration rule.

A=2x02π-2sinx02π

Evaluate the integral by plugging the upper and the lower limits of integration.

A=22π-0-2sin2π-sin0

A=22π-20+0

A=2·2π-2·0

A=4π

**Conclusion:**

i) The sketch of the region is

ii) Area =4π