#### To determine

**To:**

Sketch the region and find the enclosed area.

#### Answer

i) The sketch of the region

ii) Area =72

#### 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=12-x2 and y=x2-6

**3) Calculation:**

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

12-x2=x2-6

18=2x2

x2=9

Using the square root property

x=3 or x=-3

Thus, the points of intersection are at x=-3 and x=3. The region is sketched in the following figure.

Here, 12-x2≥x2-6 when -3≤x≤3. Therefore, let’s assume that

fx=12-x2 and gx=x2-6

Therefore, the required area is

A=∫-3312-x2-x2-6dx

A=∫-3312-x2-x2+6 dx

After simplifying this, it becomes

A=∫-33(18-2x2)dx

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

A=18∫-33dx-2∫-33(x2)dx

Compute the integral using the standard integration rule.

A=18x-33-2x33-33

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

A=183--3-2333--333

A=183+3-2(9+9)

A=18·6-2·18

A=72

**Conclusion:**

i) The sketch of the region

ii) Area =72