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Q.

The integral I=∫02 x2dx ([t] denotes the greatestinteger less than or equal to t) is equal to:

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a

5−23

b

5−2−3

c

6−2−3

d

3−2

answer is B.

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Detailed Solution

∫02 x2dx=∫01 x2dx+∫12 x2dx+∫23 x2dx+∫32 x2dx=∫01 0dx+∫12 1dx+∫23 2dx+∫32 3dx=0+(2−1)+2(3−2)+3(2−3)=−2−3+5
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