The area bounded by the curve `y=sin^(2)x-2 sin x ` and the x-axis, where `x in [0, 2pi]`, is

Options

A. 4 sq. units
B. 8 sq. units
C. 16 sq. units
D. 20 sq. units

Answer

B

Explanation

`y=sin^(2)x-2 sin x=sinx(sinx-2)`
`y=0therefore sin x=0 therefore x=0, pi,2pi`
Also `sin x-2 lt0`
`therefore" "ygt0` if sin x lt 0
and y lt 0 if sin x gt 0
`therefore" Graph of the function is shown in the figure:"`

The required area = The shaded area
`=|int_(0)^(pi)(sin^(2)x-2sinx)dx|+int_(pi)^(2pi)(sin^(2)x-2sinx)dx`
`=|(1)/(2)int_(0)^(pi)(1-cos2x)dx-2int_(0)^(pi)sinxdx|+int_(pi)^(2pi)((1-cos2x)/(2)-2sin x)dx`
`=|(1)/(2)(x-(sin2x)/(2))+2cosx|_(0)^(pi)+[(1)/(2)(x-(sin2x)/(2))+2cosx]_(pi)^(2pi)`
`=|(1)/(2)pi-4|+(2pi-pi)/(2)+4`
`=(4-(pi)/(4))+4+(pi)/(2)=8" sq. units"`