If I=x∫02t2[t]dt,x∈R+, where [⋅] denotes the greatest integer function, then the value of I is
A
1ln2([x]+2{x}−1)
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B
1ln2([x]+2{x})
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C
1ln2([x]−2{x})
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D
1ln2([x]+2{x}+1)
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Solution
The correct option is A1ln2([x]+2{x}−1) Let x∈[n,n+1),n∈I
Now, I=x∫02t2[t]dt⇒I=x∫02{t}dt⇒I=n∫02{t}dt+x∫n2{t}dt⇒I=n1∫02{t}dt+x∫n2{t}dt⇒I=n1∫02tdt+x∫n2t−ndt⇒I=n[2tln2]10+[12n2tln2]xn⇒I=nln2+12nln2(2x−2n)⇒I=nln2+1ln2(2x−n−1)∴I=[x]+2{x}−1ln2