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Question

The value of limn1n2{sin2π4n+2sin22π4n++nsin24π4n} is

A
122π+4π2
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B
141π+2π2
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C
14+1π2π2
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D
122π
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Solution

The correct option is B 141π+2π2
Reimann sum
limn1nnr=0f(rn)=10f(x)dx.
limn1n2{sin2π4n+2sin22π4n++n sin2nπ4n}
=limn1n{1nsin2π4n+2nsin22π4n+3nsin23π4n++nn sin2 nπ4n}
=limn1nnr=1(rnsin2π4n×r+0)
=limn1nnr=0(rn) sin2(π4×(rn))
=10x sin2πx4dx=10x⎜ ⎜1cosπx22⎟ ⎟dx
=12[x221010xcos(πx2)]
=12⎢ ⎢ ⎢x2210⎪ ⎪ ⎪⎪ ⎪ ⎪xsinπx2π2∣ ∣ ∣1010sinπx2π2dx⎪ ⎪ ⎪⎪ ⎪ ⎪⎥ ⎥ ⎥=141π+2π2

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