`int_0^(pi/2)(|sin(x) - cos(2x)|)dx` Evaluate the integral and interpret it as the area of a region. Sketch the region.

`int_0^(pi/2)` (|sin(x)-cos(2x)|)dx

Since `sin(x)-cos(2x)<=0,[0,pi/6]`

and `sin(x)-cos(2x)>=0,[pi/6,pi/2]`

So, the integral can be split as,

`=int_0^(pi/6)(-(sin(x)-cos(2x)))dx+int_(pi/6)^(pi/2)(sin(x)-cos(2x))dx`

`=int_0^(pi/6)(cos(2x)-sin(x))dx+int_(pi/6)^(pi/2)(sin(x)-cos(2x))dx`

`=[1/2sin(2x)+cos(x)]_0^(pi/6)+[-cos(x)-1/2sin(2x)]_(pi/6)^(pi/2)`

`=(1/2sin(pi/3)+cos(pi/6)-(1/2sin(0)+cos(0))+(-cos(pi/2)-1/2sin(pi))-(-cos(pi/6)-1/2sin(pi/3))`

`=(1/2*sqrt(3)/2+sqrt(3)/2)-(1)+(0)-(-sqrt(3)/2-1/2*sqrt(3)/2)`

`=(sqrt(3)/4+sqrt(3)/2-1+sqrt(3)/2+sqrt(3)/4)`

`=(3/2sqrt(3)-1)`

`~~1.598`

Graph is attached. Integral is the sum of the region from (0 to pi/6) and (pi/6 to pi/2).

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`int_0^(pi/2)` (|sin(x)-cos(2x)|)dx

Since `sin(x)-cos(2x)<=0,[0,pi/6]`

and `sin(x)-cos(2x)>=0,[pi/6,pi/2]`

So, the integral can be split as,

`=int_0^(pi/6)(-(sin(x)-cos(2x)))dx+int_(pi/6)^(pi/2)(sin(x)-cos(2x))dx`

`=int_0^(pi/6)(cos(2x)-sin(x))dx+int_(pi/6)^(pi/2)(sin(x)-cos(2x))dx`

`=[1/2sin(2x)+cos(x)]_0^(pi/6)+[-cos(x)-1/2sin(2x)]_(pi/6)^(pi/2)`

`=(1/2sin(pi/3)+cos(pi/6)-(1/2sin(0)+cos(0))+(-cos(pi/2)-1/2sin(pi))-(-cos(pi/6)-1/2sin(pi/3))`

`=(1/2*sqrt(3)/2+sqrt(3)/2)-(1)+(0)-(-sqrt(3)/2-1/2*sqrt(3)/2)`

`=(sqrt(3)/4+sqrt(3)/2-1+sqrt(3)/2+sqrt(3)/4)`

`=(3/2sqrt(3)-1)`

`~~1.598`

Graph is attached. Integral is the sum of the region from (0 to pi/6) and (pi/6 to pi/2).

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