#### 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=36

#### 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

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=x2-4x and y=2x

3) **Calculation:**

Given curves are, y=x2-4x and y=2x

First find the intersection of these two curve by solving simultaneous equation

x2-4x=2x

x2-4x-2x=0

That is,

x2-6x=0

xx-6=0

x=0 or x=6

Therefore, the points of intersection of two curves is at, x=0,x= 6

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=2x, and yB=x2-4x

The graph with typical rectangle is,

Area of typical rectangle is,

yT-yB∆x=2x-x2+4x∆x=(-x2+6x)∆x

And the region lies between x=0 and x=6

So total area is,

A= ∫ab(yT-yB) dx

= ∫06(-x2+6x)dx

Integrating,

= -x33+6x2206

By using upper and lower limit,

=-633+6262-0

Simplifying,

=-2163+2162 =108-72

Simplifying,

A=36

Therefore, area of region bounded by given curve is A=36

**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=36