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If $[2,6]$ is divided into four intervals of equal length, then the approximate value of $\int_2^6 \frac{1}{x^2-x} d x$ using Simpson's rule, is
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The correct answer is:
0.5222
Here, $\quad h=\frac{6-2}{4}=1$
Let, $\quad y=\frac{1}{x^2-x}$
At $\quad x_0=2, y_0=\frac{1}{2^2-2}=\frac{1}{4-2}=\frac{1}{2}$
$x_1=3, y_1=\frac{1}{3^2-3}=\frac{1}{6}$
$x_2=4, y_2=\frac{1}{4^2-4}=\frac{1}{12}$
$x_3=5, y_3=\frac{1}{5^2-5}=\frac{1}{20}$
$x_4=6, y_4=\frac{1}{6^2-6}=\frac{1}{30}$
By Simpson's rule
$\int_2^6 \frac{1}{x^2-x} d x=\frac{h}{3}\left[y_0+y_4+4\left(y_1+y_3\right)+2 \cdot y_2\right]$
$=\frac{1}{3}\left[\frac{1}{2}+\frac{1}{30}+4\left(\frac{1}{6}+\frac{1}{20}\right)+2 \cdot \frac{1}{12}\right]$
$=\frac{1}{3}\left[\frac{16}{30}+4 \cdot \frac{26}{120}+\frac{1}{6}\right]$
$=\frac{1}{3}\left\lfloor\frac{16}{30}+\frac{26}{30}+\frac{5}{30}\right\rceil$
$=\frac{16+26+5}{90}=\frac{47}{90}=0.5222$
Let, $\quad y=\frac{1}{x^2-x}$
At $\quad x_0=2, y_0=\frac{1}{2^2-2}=\frac{1}{4-2}=\frac{1}{2}$
$x_1=3, y_1=\frac{1}{3^2-3}=\frac{1}{6}$
$x_2=4, y_2=\frac{1}{4^2-4}=\frac{1}{12}$
$x_3=5, y_3=\frac{1}{5^2-5}=\frac{1}{20}$
$x_4=6, y_4=\frac{1}{6^2-6}=\frac{1}{30}$
By Simpson's rule
$\int_2^6 \frac{1}{x^2-x} d x=\frac{h}{3}\left[y_0+y_4+4\left(y_1+y_3\right)+2 \cdot y_2\right]$
$=\frac{1}{3}\left[\frac{1}{2}+\frac{1}{30}+4\left(\frac{1}{6}+\frac{1}{20}\right)+2 \cdot \frac{1}{12}\right]$
$=\frac{1}{3}\left[\frac{16}{30}+4 \cdot \frac{26}{120}+\frac{1}{6}\right]$
$=\frac{1}{3}\left\lfloor\frac{16}{30}+\frac{26}{30}+\frac{5}{30}\right\rceil$
$=\frac{16+26+5}{90}=\frac{47}{90}=0.5222$
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