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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)) |
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Answer» `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` |
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