TS EAMCET · Maths · Definite Integration
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
- C 0.5222
- D 0.2555
Answer & Solution
Correct Answer
(C) 0.5222
Step-by-step Solution
Detailed explanation
Here, \(\quad h=\frac{6-2}{4}=1\) Let, \(\quad y=\frac{1}{x^2-x}\) At \(\quad x_0=2, y_0=\frac{1}{2^2-2}=\frac{1}{4-2}=\frac{1}{2}\) \(x_1=3, y_1=\frac{1}{3^2-3}=\frac{1}{6}\) \(x_2=4, y_2=\frac{1}{4^2-4}=\frac{1}{12}\) \(x_3=5, y_3=\frac{1}{5^2-5}=\frac{1}{20}\)…
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