#### To determine

**To:**

Sketch the region and find the enclosed area.

#### Answer

i) The sketch of the region

ii) Area =2π+23

#### 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=cosπx and y=4x2-1

**3) Calculation:**

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

cosπx=4x2-1

As -1≤cosθ≤1

So at point of intersection,

cosπx= 4x2-1

That is 4x2-cosπx-1=0

By quadratic formula thus x=±0-4(4)(-cosπx-1)2(4)=±16(cosπx+1)8=±(cosπx+1)2 Thus by trial and error (in view of the graph) points of intersection are at x=-12 and x=12. The region is sketched in the following figure. We may also directly use the graph generated by some graphing utility to find the points of intersections.

Here,cosπx≥4x2-1 when -12≤x≤12. Therefore, let’s assume that

fx=cosπx

gx=4x2-1

Therefore, the required area is

A=∫-1212cosπx-4x2-1dx

A=∫-1212cosπx-4x2+1 dx

Compute the integral using the standard integration rule.

A=sinπxπ-4x33+x-1212

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

A=sinπ2π-41233+12-sin-π2π-4-1233-12

A=1π-16+12+1π-16+12

A=2π+23

**Conclusion:**

i) The sketch of the region

ii) Area =2π+23