(a) Using the trapezium rule with seven ordinates, evaluate \(\int_{0}^{3} \frac{\mathrm d x}{x^{2} + 1}\), correct to two decimal places.
(b) Using matrix method, solve \(-2x + y = 3; - x + 4y = 1\).
(a) \(\int_{0}^{3} \frac{\mathrm d x}{x^{2} + 1}\)
| \(x\) | 0 | 0.5 | 1.0 | 1.5 | 2.0 | 2.5 | 3 |
| \(x^{2} + 1\) | 1 | 1.25 | 2.0 | 3.25 | 5.0 | 7.25 | 10.0 |
| \(\frac{1}{x^{2} + 1}\) | 1 | 0.8 | 0.5 | 0.308 | 0.2 | 0.138 | 0.1 |
h = 0.5
\(y_{1} = 1; y_{2} = 0.8; y_{3} = 0.5; y_{4} = 0.308; y_{5} = 0.2; y_{6} = 0.138; y_{7} = 0.1\)
\(y_{1} + y_{7} = 1 + 0.1 = 1.1\)
\(2(y_{2} + ... + y_{6}) = 2(0.8 + 0.5 + 0.308 + 0.2 + 0.138) = 2(1.946) = 3.892\)
\(\int_{0}^{3} \frac{\mathrm d x}{x^{2} + 1} = \frac{1}{2} \times 0.5 (1.1 + 3.892) = 0.25(4.992)\)
= \(1.248 \approxeq 1.25 (\text{to 2 decimal place})\)
(b) -2x + y = 3
-x + 4y = 1
\(det = \begin{vmatrix} -2 & 1 \\ -1 & 4 \end{vmatrix} = -7\)
\(det_{1} = \begin{vmatrix} 3 & 1 \\ 1 & 4 \end{vmatrix} = 11\)
\(det_{2} = \begin{vmatrix} -2 & 3 \\ -1 & 1 \end{vmatrix} = 1\)
\(\therefore x = \frac{det_{1}}{det} = \frac{11}{-7} = -1\frac{4}{7}\)
\(\therefore y = \frac{det_{2}}{det} = \frac{1}{-7}\)
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