Find the area enclosed the curve y=sin x and the X-axis between `x=0 and x=pi`.

Options

A. `(7)/(2)` sq. units
B. `(7)/(4)+sqrt(3)` sq. units
C. `(7sqrt(3))/(4)` sq. units
D. `7-(sqrt(3))/(4)` sq. units

Answer

A

Explanation

We have `y=sin 2x, y=sqrt(3) sin x`
Solving, we get `sin 2x=sqrt(3) sin x`
`therefore" "2sin x cos x = sqrt(3) sin x`
`therefore" "sin x=0 or cos x =sqrt(3)//2`
`therefore" "x=0, pi or x=pi//6`

`therefore" Required area "`
`=overset(pi//6)underset(0)int(sin 2x-sqrt(3)sin x)dx+overset(pi)underset(pi//6)int(sqrt(3)sin x - sin 2 x) dx `
`=[-(cos 2x)/(2)+sqrt(3)cos x]_(0)^(pi//6)+[-sqrt(3) cos x +(cos 2x)/(2)+]_(pi//6)^(pi)`
`=[(7)/(4)-sqrt(3)]+[sqrt(3)+(7)/(4)]`
`=(7)/(2)`