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Question

Evaluate, 21(3x21)dx as the limit of a sum

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Solution

Consider the problem

Given 21(3x21)dx

Consider x1+rh

where,

h=21n=1n

As, x1,r0

And x2,r1n

21(3x21)dx=limnnr=0h[3(1+rh)21]
=limnnr=11n[3(1+rn)21]=limnnr01n((3+3r2n2+6rn)1)=limnnr1n(2+3r2n2+6rn)=limn1n(nr=0(2)+nr=0(3r2n2)+nr=0(6r2))=limn1n(nr=0(2)+3n2nr=0(r2)+6nnr=0(r))=limn1n(2n+3n2(n(n+1)(2n+1)6)+6n(n(n+1)2))=limn(2nn+3n3(n(n+1)(2n+1)6)+6n2(n(n+1)2))=limn(2+1n3(n(n+1)(2n+1)2)+3n2(n(n+1)1))=limn(2+12(1)(1+1n)(2+1n)+3(1)(1+1n))=2+12limn((1+1n)(2+1n)+3limn(1+1n))=2+12×2+3×1=2+1+3=6

Hence, 21(3x21)dx=6



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