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Question-208242




Question Number 208242 by alcohol last updated on 08/Jun/24
Commented by alcohol last updated on 08/Jun/24
please help me translate and solve
pleasehelpmetranslateandsolve
Answered by mr W last updated on 09/Jun/24
∫_a ^b f(x)dx  =Σ_(k=0) ^(n−1) ∫_x_(2k)  ^x_(2(k+1))  f(x)dx  ≈Σ_(k=0) ^(n−1) ((x_(2(k+1)) −x_(2k) )/6)[y_(2k) +4y_(2k+1) +y_(2(k+1)) ]  ≈((b−a)/(6n))Σ_(k=0) ^(n−1) [y_(2k) +4y_(2k+1) +y_(2(k+1)) ]  ≈((b−a)/(6n))[2(y_0 +y_2 +y_4 +...+y_(2n) )+4(y_1 +y_3 +y_5 +...+y_(2n−1) )−y_0 −y_(2n) ]  ≈((b−a)/(6n))[y_0 +y_(2n) +2(y_2 +y_4 +...+y_(2n−2) )+4(y_1 +y_3 +y_5 +...+y_(2n−1) )]    ln 2=∫_1 ^2 (dx/x)  ≈((2−1)/(6×2))[(1/1)+(1/2)+2((4/6))+4((4/5)+(4/7))]  ≈0.69325    (with n=2)  or  ≈((2−1)/(6×3))[(1/1)+(1/2)+2((6/8)+(6/(10)))+4((6/7)+(6/9)+(6/(11)))]  ≈0.69317   (with n=3)
abf(x)dx=n1k=0x2kx2(k+1)f(x)dxn1k=0x2(k+1)x2k6[y2k+4y2k+1+y2(k+1)]ba6nn1k=0[y2k+4y2k+1+y2(k+1)]ba6n[2(y0+y2+y4++y2n)+4(y1+y3+y5++y2n1)y0y2n]ba6n[y0+y2n+2(y2+y4++y2n2)+4(y1+y3+y5++y2n1)]ln2=12dxx216×2[11+12+2(46)+4(45+47)]0.69325(withn=2)or216×3[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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