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If f (x) = min [ x ^(2), sin ""(x)/(2), ...

If `f (x) = min [ x ^(2), sin ""(x)/(2), (x -2pi) ^(2)],` the area bounded by the curve `y=f (x),` x-axis, `x=0 and x=2pi` is given by
Note: `x_(1)` is the point of intersection of the curves `x ^(2) and sin ""(x)/(2),x _(2)` is the point of intersection of the curves sin `""x/2 and (x-2pi)^(2))`

A

`int_(0)^(x _(1))(sin ""(x)/(2)) dx + int _(x_(1))^(pi) x ^(2) dx + int _(pi)^(x _(2)) (x-2pi)^(2) dx + int _(x _(2))^(2pi) (sin ""(x)/(2)) dx`

B

`int _(0) ^(x_(1)) x ^(2) dx +int _(x _(1))^(x _(3))(sin""(x)/(2)) dx + int _( x_(2)) ^(2pi) (x-2pi)^(2) dx,` where `x _(1) in (0, (pi)/(3)) and x _(2) in ((5pi)/(3), 2pi)`

C

`int _(0)^(x_(1)) x ^(2) dx + int _(x _(1)) ^(x_(2))sin ((x )/(2)) dx + int _( x_(2))^(2pi) (x-2pi) ^(2)dx,` where `x _(1) in ((pi)/(3), (pi)/(2)) and x _(2) in ((3pi)/(2) , 2pi)`

D

`int _(0)^(x _(1))x ^(2) dx + int _(x _(1)) ^(x_(2)) sin ((x )/(2)) dx + int _( x_(2))^(2pi) (x-2pi) ^(2) dx`, where `x _(1) in ((pi)/(2), (2pi)/(3)) and x _(2) in (pi, 2pi)`

Text Solution

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The correct Answer is:
B
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