Q87 / 160 · AP EAMCET 2003 · Mathematics

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

  • A
    0.3222
  • B
    0.2333
  • Correct answer
    C
    0.5222
  • D
    0.2555

Solution

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 $\int_2^6 \frac{1}{x^2-x} d x=\frac{h}{3}\left[y_0+y_4+4\left(y_1+y_3\right)+2 \cdot y_2\right]$ $=\frac{1}{3}\left[\frac{1}{2}+\frac{1}{30}+4\left(\frac{1}{6}+\frac{1}{20}\right)+2 \cdot \frac{1}{12}\right]$ $=\frac{1}{3}\left[\frac{16}{30}+4 \cdot \frac{26}{120}+\frac{1}{6}\right]$ $=\frac{1}{3}\left\lfloor\frac{16}{30}+\frac{26}{30}+\frac{5}{30}\right\rceil$ $=\frac{16+26+5}{90}=\frac{47}{90}=0.5222$

Hence, the correct answer is option: 'C'.

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