The graph of f(x)=(3x−π)cos(1/2(x)) lies below the x-axis for −π < x < π/3 and above for π/3 < x < π. The graph of f(x)=(3x−π)cos(1/2(x)) lies below the x-axis for −π...

The graph of f(x)=(3x−π)cos(1/2(x)) lies below the x-axis for −π < x < π/3 and above for π/3 < x < π. The graph of f(x)=(3x−π)cos(1/2(x)) lies below the x-axis for −π < x < π/3 and above for π/3 < x < π. Use this to find the area enclosed by the graph and the x-axis between x=-π and x=π to 4 decimal places

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the area bounded by graph of f(x), x axis and `x=-pi, x=pi` , hence you should evaluate the definite integral of this function such that:

`int_(-pi)^pi (3x−pi)*cos(x/2) dx = int_(-pi)^(pi/3) (3x−pi)*cos(x/2) dx + int_(pi/3)^pi (3x−pi)*cos(x/2) dx`

You need to split the integral into two simpler integrals such that:

`int_(-pi)^(pi/3) (3x−pi)*cos(x/2) dx = int_(-pi)^(pi/3) 3x*cos(x/2) dx - int_(-pi)^(pi/3) pi*cos(x/2) dx `

You need to evaluate the definite integrals from the right such that:

`int_(-pi)^(pi/3) pi*cos(x/2) dx = pi*int_(-pi)^(pi/3) cos(x/2) dx = 2pi*sin(x/2)|_(-pi)^(pi/3)`

`pi*int_(-pi)^(pi/3) cos(x/2) dx = 2pi(sin pi/6 - sin (-pi/2))`

You need to remember that the sine function is odd, hence `sin (-pi/2) = -sin (pi/2)` `pi*int_(-pi)^(pi/3) cos(x/2) dx = 2pi(sin pi/6+ sin pi/2)`

`pi*int_(-pi)^(pi/3) cos(x/2) dx= 2pi(1/2 + 1)`

`pi*int_(-pi)^(pi/3) cos(x/2) dx= 3pi`

You should evaluate `int_(-pi)^(pi/3) 3x*cos(x/2) dx`  using parts such that:

`int udv = uv - int vdu`

`u = 3x =gt du = 3dx`

`dv = cos(x/2)dx =gt v = 2sin(x/2)`

`int_(-pi)^(pi/3) 3x*cos(x/2) dx = 6xsin(x/2)|_(-pi)^(pi/3) - 6int_(-pi)^(pi/3) sin(x/2) dx`

`int_(-pi)^(pi/3) 3x*cos(x/2) dx= (6xsin(x/2) + 12 cos(x/2))|_(-pi)^(pi/3)`

`int_(-pi)^(pi/3) 3x*cos(x/2) dx = 6(pi/3*sin(pi/6)- (pi)*sin(pi/2)+2 cos(pi/6) - 2cos(pi/2)) `

`int_(-pi)^(pi/3) 3x*cos(x/2) dx= 6(pi/6 - pi +sqrt3) = 6sqrt3 - 5pi`

`int_(-pi)^(pi/3) (3x−pi)*cos(x/2) dx = 6sqrt3 - 5pi - 3pi`

`int_(-pi)^(pi/3) (3x−pi)*cos(x/2) dx = 6sqrt3 - 8pi`

You need to evaluate `int_(pi/3)^(pi) (3x−pi)*cos(x/2) dx`  such that:

`int_(pi/3)^(pi) (3x−pi)*cos(x/2) dx = int_(pi/3)^(pi) 3x*cos(x/2) dx - int_(pi/3)^(pi) pi*cos(x/2) dx `

`int_(pi/3)^(pi) pi*cos(x/2) dx = 2pi*sin(x/2)|_(pi/3)^(pi)`

`int_(pi/3)^(pi) pi*cos(x/2) dx = 2pi*(sin pi/2 - sin pi/6)`

`int_(pi/3)^(pi) pi*cos(x/2) dx = 2pi*(1 - 1/2)`

`int_(pi/3)^(pi) pi*cos(x/2) dx = pi`

`int_(pi/3)^(pi) 3x*cos(x/2) dx= (6xsin(x/2) + 12 cos(x/2))|_(pi/3)^(pi)`

`int_(pi/3)^(pi) 3x*cos(x/2) dx = 6(pi*sin(pi/2) - (pi/3)*sin(pi/6) + 2 cos(pi) - 2cos(pi/6))` 

`int_(pi/3)^(pi) 3x*cos(x/2) dx = 6(pi - pi/6- 2 - sqrt3) `

`int_(pi/3)^(pi) 3x*cos(x/2) dx = 5pi - 12 - 6sqrt3`

`int_(pi/3)^(pi) (3x−pi)*cos(x/2) dx = 5pi - 12 - 6sqrt3 - pi`

`int_(pi/3)^(pi) (3x−pi)*cos(x/2) dx = 4pi - 12 - 6sqrt3`

`int_(-pi)^pi (3x−pi)*cos(x/2) dx = -(6sqrt3 - 8pi + 4pi - 12 - 6sqrt3)`

`int_(-pi)^pi (3x−pi)*cos(x/2) dx = 4pi + 12 = 24.5663`

Hence, evaluating the area bounded by the graph of function and x axis, between`x=-pi`  and `x=pi`  yields `int_(-pi)^pi (3x−pi)*cos(x/2) dx = 4pi + 12 = 24.5663.`

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