#### To determine

**To:**

Sketch the region and find the enclosed area.

#### Answer

i) The sketch of the region

ii) Area =83

#### 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=x2 and y=4x-x2

**3) Calculation:**

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

x2=4x-x2

2x2-4x=0

2xx-2=0

Thus, the points of intersection is at x=0 and x=2. The region is sketched in the following figure.

Here,4x-x2≥x2 when 0≤x≤2. Therefore, let’s assume that

fx=4x-x2

gx=x2

Therefore, the required area is

A=∫024x-x2-x2dx

A=∫024x-x2-x2 dx

After simplifying this, it becomes

A=∫02(4x-2x2)dx

Using the sum and difference rule and the constant multiple rule of integral, it becomes

A=4∫02x dx-2∫02(x2)dx

Compute the integral using the standard integration rule.

A=4x2202-2x3302

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

A=4222-022-2233-033

A=42+0-283+0

A=4·2-2·83

A=8-163

A=83

**Conclusion:**

i) The sketch of the region

ii) Area =83