#### To determine

**To sketch:**

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

#### Answer

A=43

#### 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=2x-x2, y=0

**3) Calculation:**

The curves y=2x-x2, y=0 are shown below

The region enclosed by the curve y=2x-x2 , y=0 is shaded in black

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

From above graph, the curve y=fx=2x-x2 bounded between x=0 and x=2, is continuous on 0,2, then

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

∫02(2x-x2) dx=F2-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),

ddx2x1+11+1-x2+12+1=2x-x2

ddx2x22-x33=2x-x2

Therefore,

Fx=2x22-x33=x2-x33

Substitute F(x) in (1) at x=0 and x=2,

∫02(2x-x2) dx=F2-F(0)

=22-233-02-033

=4-83-0+(0)3

=4-83

=12-83

=43

**Conclusion:**

Therefore, the required area is A=43