∫20x[x2+42]dx= where [.] or the greatest integer function
A
4
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B
5
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C
3
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D
6
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Solution
The correct option is B5 ∫20x[x2+42]dx=∫20x[x22+2]dx ∫20x{[x22]+2}dx =∫√20x.0dx+∫2√2x.1dx+4 ....(∵xϵ[0,√2)⇒[x22]=0 and xϵ[√2,2)⇒[x22]=1 (x21)2√2+4=(2−1)+4=5 Ans: B