The value of the integral ∫31√3+x3dx lies in the interval
A
(1,3)
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B
(2,30)
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C
(4,2√30)
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D
none of these
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Solution
The correct option is C(4,2√30) As f(x)=√3+x3 is monotonically increasing ∀xϵ[1,3] For 1<x<3⇒√3+13<√3+x3<√3+33⇒2<√3+x3<√30⇒∫312dx≤∫31√3+x3dx≤∫31√30dx⇒2[x]31≤∫31√3+x3dx≤√30[x]31⇒4≤∫31√3+x3dx≤2√30