#### To determine

**To sketch:**

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

#### Answer

A=163

#### 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=x, y=0, x=4

**3) Calculation:**

The curves y=x , y=0 and x=4 are given by,

The region enclosed by the curve y=x , y=0, x=0 and x=4 is shown by the black region

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

From the above graph, the curve y=fx=x is bounded between x=0 and x=4, and is continuous on 0,4, then

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

∫04x dx=F4-F0…..(1)

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

ddxFx=fx…..(2)

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

ddxx12+112+1=x=x12

ddxx3232=x=x12

Therefore,

Fx=x3232=23x32

Substitute F(x) in (1) at x=0 and x=4,

∫04x dx=F4-F(0)

=23432-23032

=23(432-0)

=23(4·4-0)

=23(4·2-0)

=2·83

=163

**Conclusion:**

Therefore, the required area is A=163