# Sketch the region enclosed by the given curves. y = 4 cos 2x, y = 4 − 4 cos 2x, 0 ≤ x ≤ π/2 Find its area.

*print*Print*list*Cite

### 1 Answer

You need to find the points of intersection of curves, hence you should solve the equation `4 cos 2x = 4 - 4 cos 2x` such that:

`4 cos 2x = 4 - 4 cos 2x`

`8 cos 2x = 4 =gt 2 cos 2x = 1 =gt cos 2x = 1/2`

`2x = pi/3, 2pi/3, 4pi/3,...`

Notice that the values `2pi/3, 4pi/3,...` fall outside the interval `[0,pi/2], ` hence, the limits of integration are 0 and `pi/3` .

You need to select a value for x in interval `[0,pi/3]` such that:

`x = pi/4 =gt 8 cos 2*pi/4 - 4 = 8*cos pi/2 - 4 = -4 lt 0`

Hence, you need to integrate the function `4 - 8 cos 2x ` to find the area enclosed by the given curves such that:

`int_0^(pi/3) (4 - 8 cos 2x) dx = int_0^(pi/3) 4 dx- int_0^(pi/3) 8 cos 2x dx`

`int_0^(pi/3) (4 - 8 cos 2x) dx = (4x - (sin 2x)/2)|_0^(pi/3)`

`int_0^(pi/3) (4 - 8 cos 2x) dx = 4pi/3 - (sin(2pi/3))/2 - 0 + (sin 0)/2`

You need to use the double angle formula to find `sin 2pi/3` such that:

`sin 2pi/3 = 2 sin (pi/3)*cos (pi/3)`

`sin 2pi/3 = 2*sqrt3/2*1/2`

`sin 2pi/3 = sqrt3/2`

`int_0^(pi/3) (4 - 8 cos 2x) dx = 4pi/3 -sqrt3/4`

`int_0^(pi/3) (4 - 8 cos 2x) dx = (16pi - 3sqrt3)/12`

`int_0^(pi/3) (4 - 8 cos 2x) dx = (18.849 - 5.196)/12`

`int_0^(pi/3) (4 - 8 cos 2x) dx = (13.653)/12`

`int_0^(pi/3) (4 - 8 cos 2x) dx ~~ 1.137`

**Hence, evaluating the area enclosed by the given curves yields 1.137.**