Question Number 208242 by alcohol last updated on 08/Jun/24
Commented by alcohol last updated on 08/Jun/24
$${please}\:{help}\:{me}\:{translate}\:{and}\:{solve} \\ $$
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)](https://www.tinkutara.com/question/Q208257.png)
$$\int_{{a}} ^{{b}} {f}\left({x}\right){dx} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\int_{{x}_{\mathrm{2}{k}} } ^{{x}_{\mathrm{2}\left({k}+\mathrm{1}\right)} } {f}\left({x}\right){dx} \\ $$$$\approx\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{{x}_{\mathrm{2}\left({k}+\mathrm{1}\right)} −{x}_{\mathrm{2}{k}} }{\mathrm{6}}\left[{y}_{\mathrm{2}{k}} +\mathrm{4}{y}_{\mathrm{2}{k}+\mathrm{1}} +{y}_{\mathrm{2}\left({k}+\mathrm{1}\right)} \right] \\ $$$$\approx\frac{{b}−{a}}{\mathrm{6}{n}}\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left[{y}_{\mathrm{2}{k}} +\mathrm{4}{y}_{\mathrm{2}{k}+\mathrm{1}} +{y}_{\mathrm{2}\left({k}+\mathrm{1}\right)} \right] \\ $$$$\approx\frac{{b}−{a}}{\mathrm{6}{n}}\left[\mathrm{2}\left({y}_{\mathrm{0}} +{y}_{\mathrm{2}} +{y}_{\mathrm{4}} +…+{y}_{\mathrm{2}{n}} \right)+\mathrm{4}\left({y}_{\mathrm{1}} +{y}_{\mathrm{3}} +{y}_{\mathrm{5}} +…+{y}_{\mathrm{2}{n}−\mathrm{1}} \right)−{y}_{\mathrm{0}} −{y}_{\mathrm{2}{n}} \right] \\ $$$$\approx\frac{{b}−{a}}{\mathrm{6}{n}}\left[{y}_{\mathrm{0}} +{y}_{\mathrm{2}{n}} +\mathrm{2}\left({y}_{\mathrm{2}} +{y}_{\mathrm{4}} +…+{y}_{\mathrm{2}{n}−\mathrm{2}} \right)+\mathrm{4}\left({y}_{\mathrm{1}} +{y}_{\mathrm{3}} +{y}_{\mathrm{5}} +…+{y}_{\mathrm{2}{n}−\mathrm{1}} \right)\right] \\ $$$$ \\ $$$$\mathrm{ln}\:\mathrm{2}=\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{dx}}{{x}} \\ $$$$\approx\frac{\mathrm{2}−\mathrm{1}}{\mathrm{6}×\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2}\left(\frac{\mathrm{4}}{\mathrm{6}}\right)+\mathrm{4}\left(\frac{\mathrm{4}}{\mathrm{5}}+\frac{\mathrm{4}}{\mathrm{7}}\right)\right] \\ $$$$\approx\mathrm{0}.\mathrm{69325}\:\:\:\:\left({with}\:{n}=\mathrm{2}\right) \\ $$$${or} \\ $$$$\approx\frac{\mathrm{2}−\mathrm{1}}{\mathrm{6}×\mathrm{3}}\left[\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2}\left(\frac{\mathrm{6}}{\mathrm{8}}+\frac{\mathrm{6}}{\mathrm{10}}\right)+\mathrm{4}\left(\frac{\mathrm{6}}{\mathrm{7}}+\frac{\mathrm{6}}{\mathrm{9}}+\frac{\mathrm{6}}{\mathrm{11}}\right)\right] \\ $$$$\approx\mathrm{0}.\mathrm{69317}\:\:\:\left({with}\:{n}=\mathrm{3}\right) \\ $$
Commented by mr W last updated on 09/Jun/24
