#### To determine

**To:**

Sketch the region and find the enclosed area.

#### Answer

i) The sketch of the region:

ii) Area =12

#### 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=x3 and y=x

**3) Calculation:**

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

x3=x

x3-x=0

x(x2-1)=0

that is,

x=0 and x2-1=0

x=0 and x2=1

x=0, x=1 and x=-1

Thus, the points of intersection are at x=0, x=1 and x=-1. The region is sketched in the following figure.

The shaded region is a symmetric about x=0

The total area A=2A1, where A1 is the area under the curve from x=0 and x=1.

Here x≥x3 when 0≤x≤1. Therefore, let’s assume that

fx=x

gx=x3

Therefore, the required area is

A1=∫01x-(x3)dx

A1=∫01x-x3 dx

Compute the integral using the standard integration rule.

A1=x22-x4401

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

A1=122-144-0-0

A1=12-14

A1=14

So, from this, A is given by

A=2A1

A=214

A=12

**Conclusion:**

i) The sketch of the region:

ii) Area =12