#### To determine

**To sketch:**

The region enclosed by the given curve and find the area of the region.

#### Answer

i. The graph of the region enclosed by the given curves is

The graph with the typical approximating rectangle is

iii. The area of the region bounded by the given curve is

A=3π28-1

#### Explanation

**1) Concept:**

i. The intersection of two curve is obtained by solving the simultaneous equation of the curve

ii. Identify the top and the bottom boundaries of the region.

iii. Formula-

The area of a typical rectangle is

yT-yB∆x, where yT = top boundary curve and yB= bottom boundary curve.

The total area is

A= limn→∞∑i=1nyT-yB∆x = ∫ab(yT-yB) dx

**2) Given:**

y=x and y=sinx, x=π/2, x=π

3) **Calculation:**

The given curves are y=x ,y=sinx, x=π/2 and x=π

The graph of the region enclosed by the given curve is

Find the area by integrating with respect to x(since, we found the top curve and the bottom curve from the graph and the restriction on x is given, so it is easy to find the area by integrating with the respective x)

From the graph the top and the bottom curves are

yT=x, and yB=sinx

The graph with the typical rectangle is

The area of a typical rectangle is

yT-yB∆x=(x-sinx)∆x

And the region lies between x=π2 and x=π

So, the total area is

A= ∫ab(yT-yB) dx

= ∫π2π (x-sinx)dx

Integrating,

= x22+cosxπ2π

By using the upper and the lower limits,

=π22+cosπ-π28+cosπ2

= π22-1- π28-0

Simplifying,

=π22-π28-1

A=3π28-1

Therefore, the area of the region bounded by the given curve is

A=3π28-1

**Conclusion:**

i. The graph of the region enclosed by the given curves is

The graph with the typical approximating rectangle is

The area of the region bounded by the given curve is

A=3π28-1