#### To determine

**To sketch:**

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

#### Answer

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

The graph with typical approximating rectangle is,

ii. The area of region bounded by given curve is A=92

#### Explanation

**1) Concept:**

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

ii. Identify top and bottom boundaries of region

iii. Formula-

Area of typical rectangle is,

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

Total area is

A= limn→∞∑i=1nyT-yB∆x = ∫ab(yT-yB) dx

**2) Given:**

y=x-22 and y=x

3) **Calculation:**

Given curves are, y=x-22 and y=x

First find the points of intersection of these two curve we shall the equations simultaneously that is we shall solve

x-22 =x

x2-4x-x+4=0

That is,

x2-5x+4=0

x=4, x=1

Therefore, the points of intersection of two curves is at, x=1, x=4

Graph of region enclosed by given curve is,

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

From graph the top and bottom curves are,

yT=x, and yB=x-22

The graph with typical rectangle is,

Area of typical rectangle is,

yT-yB∆x=x-x2+4x-4∆x=(-x2+5x-4)∆x

And the region lies between x=1 and x=4

So total area is,

A= ∫ab(yT-yB) dx

= ∫14(-x2+5x-4)dx

Integrating,

= -4x-x33+5x2214

By using upper and lower limit,

=-44-433+5242--41-133+5212

Simplifying,

-16-643+802+4+13-5 2

=-96-128+240+24+2-156

Simplifying,

A=276

A=92

Therefore, area of region bounded by given curve is A=92**Conclusion:**

Therefore

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

The graph with typical rectangle is,

ii. Area of region bounded by given curve is A=92