#### To determine

**To sketch:**

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

#### Answer

A=323

#### 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=4-x2, y=0

**3) Calculation:**

The curves y=4-x2, y=0 are given by,

The region enclosed by the curve y=x3 , y=0, x=0 and x=1 is shaded in black.

To find the area of region enclosed by the given curves

From the above graph, the curve y=fx=4-x2 is continuous on -2,2, then

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

∫-22(4-x2) dx=F2-F-2…(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),

ddx4x11-x2+12+1=4-x2 That is

ddx4x-x33=4-x2

Therefore,

Fx=4x-x33

Substitute F(x) in (1) to get,

∫-22(4-x2) dx=F2-F(-2)

=42-233-4-2--233

=8-83--8+(-8)3

=8-83+8-83

=16-163

=48-163

=323

**Conclusion:**

Therefore, the required area is A=323