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The value of 02xx2+1 dx, where [x] is 

the greatest integer less than or equal to x is

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a
2
b
8/3
c
4
d
none of these

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detailed solution

Correct option is D

For x∈[0, 2], x2+1∈[1, 5] we must break ⌊0, 2⌋=⌊0,1⌋∪[1,2]∪[2,3]∪[3, 2].Hence ∫02 xx2+1dx=∫01 xx2+1dx+∫12 xx2+1dx+∫23 xx2+1dx+∫32 xx2+1dx =∫01 xdx+∫12 x2dx+∫23 x3dx+∫32 x4dx =12+1323/2−1+14[9−4]+1532−35/2 =46960+1323/2−1535/2.


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