Using trapezoidal rule, approximate \(\int_{1}^{2}\frac{1}{x}dx\) with n = 5

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Solution:

When n = 5, and b = 2, we have Δx = (2-1)/5 = 0.2, and so the Trapezoidal Rules gives

\(\int^2_1\frac{1}{x}dx = T_{5}\)

= 0.2 / 2 [f(1) + 2f(1.2) + 2f(1.4) + 2f(1.6) + 2f(1.8) + f(2)]

\(= 0.1 \times \left ( \frac{1}{1} + \frac{2}{1.2} + \frac{2}{1.4} + \frac{2}{1.6} + \frac{2}{1.8} + \frac{1}{2} \right )\)

= 0.0695635

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