# what is the area under the curve f(x)=(3x-pi)cos(0.5x) on interval (-pi,pi). For interval (-pi,pi/3) function lies below x axis, (pi/3,pi)- above x axis.

## Expert Answers

`f(x)= (3x-pi)(cos(0.5x)`

`==> int f(x)= int (3x-pi)(cos(0.5x)) dx`

`==> int f(x)= 3int xcos(0.5x) dx - pi int cos(0.5x) dx`

`==> int f(x)= 3( 1/(0.5)^2 cos(0.5x) +x/0.5 sin(0.5x)) - pi(sin0.5x)/x `

`==> int f(x)= 12cos(0.5x) + 6xsin(0.5x) - (pisin(0.5x))/x`

`==> int_((-pi)^(pi/3)) f(x)= 12cos(pi/6) + 2pisin(pi/6) - (pi(sin(pi/6))/(pi/3)) `

`- ( 12cos(-pi/2) + 2pisin(-pi/2) - (pisin(-pi/2))/(-pi))`

`==>12(sqrt3/2) + 2pi(1) -3(1) - (12(0)-2pi(1) - (1))`

`==> 6sqrt3 + 2pi - 3 + 2pi + 1 `

`==> 6sqrt3 + 4pi - 2`

`==> A = 2(6sqrt3 + 4pi -2) = 12sqrt3 + 8pi - 4`

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