The value of the integral, 3∫1[x2−2x−2]dx, where [x] denotes the greatest integer less than or equal to x, is
A
−4
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B
−5
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C
−√2−√3−1
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D
−√2−√3+1
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Solution
The correct option is C−√2−√3−1 I=3∫1[x2−2x+1−3]dx ⇒I=3∫1−3dx+3∫1[(x−1)2]dx Put x−1=t⇒dx=dt I=(−6)+2∫0[t2]dt ⇒I=−6+1∫00dt+√2∫11dt+√3∫√22dt+2∫√33dt ⇒I=−6+(√2−1)+2√3−2√2+6−3√3 ⇒I=−1−√2−√3