`y = cos(x), y = sin(2x), x = 0, x = pi/2` Sketch the region enclosed by the given curves and find its area.

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`y=cos(x) , y=sin(2x) , x=0 , x=pi/2`

Refer the attached image, y=cos(x) is plotted in red color and y=sin(2x) is plotted in blue color.

From graph,

cos(x) is above sin(2x) from 0 to pi/6

sin(2x) is above cos(x) fron pi6 to pi/2

Area of the region enclosed by the given curves A=`int_0^(pi/6)(cos(x)-sin(2x))dx+int_(pi/6)^(pi/2)((sin(2x)-cos(x))dx`

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

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

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

`A=(1/2+1/4)-(0+1/2)+(1/2-1)-(-1/4-1/2)`

`A=1/4-1/2+1/4+1/2`

`A=1/2`

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