If [2,6] is divided into four intervals of eq | vedclass

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(C) Given the interval $[2,6]$ is divided into $n=4$ sub-intervals of equal length.$h = \frac{6-2}{4} = 1$.Let $f(x) = \frac{1}{x^2-x}$.The values of

Vedclass · Accepted Answer

$0.5222$

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(C) Given the interval $[2,6]$ is divided into $n=4$ sub-intervals of equal length.$h = \frac{6-2}{4} = 1$.Let $f(x) = \frac{1}{x^2-x}$.The values of $x$ are $x_0=2, x_1=3, x_2=4, x_3=5, x_4=6$.The corresponding values of $y = f(x)$ are:$y_0 = f(2) = \frac{1}{4-2} = \frac{1}{2} = 0.5$$y_1 = f(3) = \frac{1}{9-3} = \frac{1}{6} \approx 0.1667$$y_2 = f(4) = \frac{1}{16-4} = \frac{1}{12} \approx 0.0833$$y_3 = f(5) = \frac{1}{25-5} = \frac{1}{20} = 0.05$$y_4 = f(6) = \frac{1}{36-6} = \frac{1}{30} \approx 0.0333$Using Simpson's rule: $\int_2^6 f(x) dx \approx \frac{h}{3} [y_0 + y_4 + 4(y_1 + y_3) + 2y_2]$$= \frac{1}{3} [\frac{1}{2} + \frac{1}{30} + 4(\frac{1}{6} + \frac{1}{20}) + 2(\frac{1}{12})]$$= \frac{1}{3} [\frac{16}{30} + 4(\frac{10+3}{60}) + \frac{1}{6}]$$= \frac{1}{3} [\frac{16}{30} + \frac{52}{60} + \frac{5}{30}] = \frac{1}{3} [\frac{32+52+10}{60}] = \frac{94}{180} = \frac{47}{90} \approx 0.5222$.

If $[2,6]$ is divided into four intervals of equal length,then the approximate value of $\int_2^6 \frac{1}{x^2-x} dx$ using Simpson's rule is

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