`y = tan(x), y = 2sin(x), (-pi/3)<=x<=(pi/3)` Sketch the region enclosed by the given curves and find its area.

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gsarora17 eNotes educator| Certified Educator

`y=tan(x) , y=2sin(x), -pi/3<=x<=pi/3`

Refer the attached image. Graph of y=tan(x) is plotted in blue color and graph of y=2sin(x) is plotted in red color.

Area of the region enclosed by the given curves(A)=`2int_0^(pi/3)(2sin(x)-tan(x))dx` (by symmetry)

`A=2[-2cos(x)-lnsec(x)]_0^(pi/3)`

`A=2((-2cos(pi/3)-lnsec(pi/3))-(-2cos(0)-lnsec(0)))`

`A=2((-2*(1/2)-ln2)-(-2-0))`

`A=2(1-ln2)`

 

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