#### To determine

**To:**

Sketch the region and find the enclosed area

#### Answer

i) The sketch of the region

ii) Area =34

#### 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=1x2, y=x and y=18x

**3) Calculation:**

The point of intersection occurs when both the equations are equal to each other. So intersection of y=1x2 and y=x is given by

1x2=x

x3=1

that is,

at x=1

Intersection of y=1x2 and y=18x is given by

1x2=18x

x3=8

that is at x=2

Intersection point of y=x and y=18x is given by

x=18x

that is, x=0

Thus, the points of intersection are at x=2 and x=3 and x=0. The region is sketched in the following figure.

Here A=A1+A2

Where A1 is the area between the curves from 0≤x≤1

And A2 is the area between the curves from 1≤x≤2

Between x=0 and x=1 the area is bound by curves y=x and y=18x

Here x≥18x when 0≤x≤1. Therefore, let’s assume that

fx=x

gx=18x

Therefore, the required area is

A1=∫01x-18xdx

A1=∫01x-18x dx

A1=∫0178x dx

Compute the integral using the standard integration rule.

A1=78·x2201

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

A1=78·122-78·0

A1=716

A1=716

Now 1x2≥18x when 1≤x≤2. Therefore, let’s assume that

fx=1x2

gx=18x

Therefore, the required area is

A2=∫121x2-18xdx

A2=∫121x2-18xdx

A2=∫12x-2-18xdx

Compute the integral using the standard integration rule.

A2=-1x-18·x2212

A2=-12-18·222--11-18·122

A2=-12-14+1+116

A2=516

A2=516

So, from this, A is given by:

A=A1+A2

A=716+516

A=34

**Conclusion:**

i) The sketch of the region

ii) Area =34