#### To determine

**To:**

Sketch the region and find the enclosed area

#### Answer

i) The sketch of the region.

ii) Area =23+π3

#### Explanation

**1) Concept:**

Formula:

The area A of the region bounded by the curves y=f(x), 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=1-cosx, 0≤x≤π

**3) Calculation:**

The point of intersection occurs when both the equations are equal to each other, that is,

cosx=1-cosx

2cosx=1

cosx=12

x=π3 when 0≤x≤π

Thus from the graph the points of intersection is at x=0, x=π3 and x=π. The region is sketched in the following figure.

Here, A=A1+A2

Where A1 is the area between the curves from 0≤x≤π3

And A2 is the area between the curves from π3≤x≤π

Here,cosx≥1-cosx when 0≤x≤π3. Therefore,

fx=cosx

gx=1-cosx

Therefore, the required area is

A1=∫0π3cosx-1-cosxdx

A1=∫0π3cosx-1+cosx dx

A1=∫0π32cosx-1 dx

Compute the integral using the standard integration rule.

A1=2sinx-x0π3

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

A1=2sinπ3-π3-2sin0-0

A1=2·32-π3

A1=3-π3

A1=3-π3

Now, 1-cosx≥cosx when π3≤x≤π. Therefore,

fx=1-cosx

gx=cosx

Therefore, the required area is

A2=∫π3π1-cosx-cosxdx

A2=∫π3π1-2cosxdx

Compute the integral using the standard integration rule.

A2=x-2sinxπ3π

A2=π-2sinπ-π3-2sinπ3

A2=π-0-π3+2·32

A2=2π3+3

So, from this , A is given by:

A=A1+A2

A=3-π3+2π3+3

A=23+π3

**Conclusion:**

i) The sketch of the region

ii) Area =23+π3