`y = cos(pix), y = 4x^2 - 1` Sketch the region enclosed by the given curves and find its area.

`y=cos(pix) , y=4x^2-1`

Refer the attached image. Graph of cos(pix) is plotted in blue color and graph of y=4x^2-1 is plotted in red color.

From the graph , the curves intersect at x=`+-` 1/2.

Area enclosed by the curves A=`int_(-1/2)^(1/2)(cos(pix)-(4x^2-1))dx`

`A=2int_0^(1/2)(cos(pix)-4x^2+1)dx`

`A=2[sin(pix)/pi-4(x^3/3)+x]_0^(1/2)`

`A=2[sin(pix)/pi-(4x^3)/3+x]_0^(1/2)`

`A=2((sin(pi/2)/pi-4/3(1/2)^3+1/2)-(sin(0)/pi-4/3(0)^3+0))`

`A=2((1/pi-4/3(1/8)+1/2)-0)`

`A=2(1/pi-1/6+1/2)`

`A=2(1/pi+2/6)`

`A=2/pi+2/3`

`y=cos(pix) , y=4x^2-1`

Refer the attached image. Graph of cos(pix) is plotted in blue color and graph of y=4x^2-1 is plotted in red color.

From the graph , the curves intersect at x=`+-` 1/2.

Area enclosed by the curves A=`int_(-1/2)^(1/2)(cos(pix)-(4x^2-1))dx`

`A=2int_0^(1/2)(cos(pix)-4x^2+1)dx`

`A=2[sin(pix)/pi-4(x^3/3)+x]_0^(1/2)`

`A=2[sin(pix)/pi-(4x^3)/3+x]_0^(1/2)`

`A=2((sin(pi/2)/pi-4/3(1/2)^3+1/2)-(sin(0)/pi-4/3(0)^3+0))`

`A=2((1/pi-4/3(1/8)+1/2)-0)`

`A=2(1/pi-1/6+1/2)`

`A=2(1/pi+2/6)`

`A=2/pi+2/3`