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

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

Vedclass · Accepted Answer

$0.5222$

Vedclass · Answer

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

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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