Question 1 Report
Using the trapezium rule with seven ordinates, evaluate, correct to three decimal places, \(\int_{2.4}^{3.6} \frac{1}{\sqrt{x^2 - 2}}\)dx
\(\int_{2.4}^{3.6} \frac{1}{\sqrt{x^2 - 2}}\)dx using trapezium rule
\(\sqrt{x^2 - 2}\)
\(y_7\)
= \(\frac{1}{2}\)(h)[[(\(y_1\) + \(y_7\)] + 2[\(y_2\) + \(y_3\) + \(y_4\) + \(y_5\) + \(y_6\)]]
= \(\frac{1}{2}\)(0.2)[0.5157 + 0.3021 ] + 2[0.4583 + 0.4138 + 0.3780 + 0.3484 + 0.3234]
= (0.1)[0.8178 + 3.8440] = 0.46618 ≈ 0.466 to 3 dp.
Answer Details
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