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
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
Answer Details
\(\int_{2.4}^{3.6} \frac{1}{\sqrt{x^2 - 2}}\)dx using trapezium rule
| x | 2.4 | 2.6 | 2.8 | 3.0 | 3.2 | 3.4 | 3.6 |
| x\(^2\) - 2 | 3.76 | 4.76 | 5.84 | 7.00 | 8.24 | 9.56 | 10.96 |
| \(\sqrt{x^2 - 2}\) |
1.9391 | 2.1817 | 2.4166 | 2.6458 | 2.8705 | 3.0919 | 3.3106 |
| \(\frac{1}{\sqrt{x^2 - 2}}\) | 0.5157 | 0.4583 | 0.4138 | 0.3780 | 0.3484 | 0.3234 | 0.3021 |
| \(y_1\) | \(y_2\) | \(y_3\) | \(y_4\) | \(y_5\) | \(y_6\) | \(y_7\) |
= \(\frac{1}{2}\)(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.
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