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Question Number 104359 by bemath last updated on 21/Jul/20
solve this using Riemann  sum f(x)=2x ; [0,4] for n=4
solvethisusingRiemannsumf(x)=2x;[0,4]forn=4
Answered by Ar Brandon last updated on 21/Jul/20
∫_0 ^4 2xdx=lim_(n→∞) ((4−0)/n)Σ_(k=1) ^n 2(0+((4k)/n))=lim_(n→∞) (4/n)Σ_(k=1) ^n ((8k)/n)                  =lim_(n→∞) ((32)/n^2 )∙((n(n+1))/2)=lim_(n→∞) 16(1+(1/n))=16
042xdx=limn40nnk=12(0+4kn)=limn4nnk=18kn=limn32n2n(n+1)2=lim16n(1+1n)=16

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