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Question Number 208242 by alcohol last updated on 08/Jun/24
Commented by alcohol last updated on 08/Jun/24
pleasehelpmetranslateandsolve
Answered by mr W last updated on 09/Jun/24
∫abf(x)dx=∑n−1k=0∫x2kx2(k+1)f(x)dx≈∑n−1k=0x2(k+1)−x2k6[y2k+4y2k+1+y2(k+1)]≈b−a6n∑n−1k=0[y2k+4y2k+1+y2(k+1)]≈b−a6n[2(y0+y2+y4+...+y2n)+4(y1+y3+y5+...+y2n−1)−y0−y2n]≈b−a6n[y0+y2n+2(y2+y4+...+y2n−2)+4(y1+y3+y5+...+y2n−1)]ln2=∫12dxx≈2−16×2[11+12+2(46)+4(45+47)]≈0.69325(withn=2)or≈2−16×3[11+12+2(68+610)+4(67+69+611)]≈0.69317(withn=3)
Commented by mr W last updated on 09/Jun/24
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