#### To determine

**To:**

Sketch the region and find the enclosed area

#### Answer

i) The sketch of the region

ii) Area =473-91232

#### 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=2x3 and y=18x2

**3) Calculation:**

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

2x3=18x2

2x=1383x6

1024x-x6=0

x(1024-x5)=0

that is,

x=0 and 1024-x5=0

x=0 and x5=1024

So x=0 and x=4;

Thus, from the graph, the points of intersection are at x=0, x=4. The region is sketched in the following figure.

Here, A=A1+A2

Where A1 is the area between the curves from 0≤x≤4

And A2 is the area between the curves from 4≤x≤6

Here,2x3≥18x2 when 0≤x≤4. Therefore, let’s assume that

fx=2x3

gx=18x2

Therefore, the required area is

A1=∫042x3-18x2dx

A1=∫042x3-18x2 dx

Compute the integral using the standard integration rule.

A1=23·x4343-18·x3304

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

A1=3·23·4434-18·433-0-0

A1=3·2*4434-6424=3·51234-6424

A1=3·84-6424

A1=8024=103

Now 18x2≥2x3 when 4≤x≤6. Therefore

fx=18x2

gx=2x3

Therefore, the required area is

A2=∫4618x2-2x3dx

Compute the integral using the standard integration rule.

A2=18·x33-23·x434346

A2=18·633-23·64343-18·433-23·44343

A2=21624-3·23·6634-6424-3·2·4434

A2=21624-18 1234-6424+244

A2=21624-108 12324-6424+14424

A2=373-91232

So, from this , A is given by:

A=A1+A2

A=103+373-91232

A=473-91232

**Conclusion:**

i) The sketch of the region

ii) Area =473-91232