#### To determine

**To sketch:**

The region enclosed by given curves and to calculate its area.

#### Answer

A=14

#### Explanation

**1) Concept:**

i) Fundamental theorem of Calculus, Part 2

If f is continuous on a,b, then

∫abfxdx=Fb-F(a)

where F is any antiderivative of f, that is, a function F such that F'=f

ii) Power rule for antiderivative:

ddxxn+1n+1=xn

**2) Given:**

y=x3, y=0, x=1

**3) Calculation:**

The curves y=x3, y=0 and x=1 are given by,

The region enclosed by the curve y=x3 , y=0, x=0 and x=1 is shown bythe black region

To find the area of the region enclosed by the given curves

From the above graph, the curve y=fx=x3 bounded between x=0 and x=1 is continuous on 0,1, then

By using concept i) (Fundamental theorem of Calculus, Part 2),

∫01x3 dx=F1-F0…(1)

Where F is antiderivative of f, that is, a function F such that F'(x)=f(x). That is

ddxFx=fx…(2)

By using concept ii) (power rule of antiderivative),

ddxx3+13+1=x3

ddxx44=x3

Therefore,

Fx=x44=14x4

Substitute F(x) in (1) at x=0 and x=1,

∫01x3 dx=F1-F(0)

=1414-1404

=14(1-0)

=14

**Conclusion:**

Therefore, the required area is A=14