(a) Evaluate \(\int_{1} ^{2} \frac{x}{\sqrt{5 - x^{2}}} \mathrm {d} x\) (b)(i) Evaluate: \(\begin{vmatrix} 2 & -3 & 1 \\ 0 & 1 & -2 \\ 1 & 2 & -3 \end{vmatr...
Question 1 Report
(a) Evaluate \(\int_{1} ^{2} \frac{x}{\sqrt{5 - x^{2}}} \mathrm {d} x\)
(b)(i) Evaluate: \(\begin{vmatrix} 2 & -3 & 1 \\ 0 & 1 & -2 \\ 1 & 2 & -3 \end{vmatrix}\)
(ii) Using your answer in b(i), solve the simultaneous equations :
\(2x - 3y + z = 10\)
\(y - 2z = -7\)
\(x + 2y - 3z = -9\)
Answer Details
(a) Let \(u = 5 - x^2\), so \(du = -2x\,dx\) and \(x\,dx = -\tfrac12\,du\).
\[\int \frac{x}{\sqrt{5 - x^2}}\,dx = -\frac12\int u^{-1/2}\,du = -\sqrt{u} = -\sqrt{5 - x^2}\] \[\int_{1}^{2}\frac{x}{\sqrt{5 - x^2}}\,dx = \Big[-\sqrt{5 - x^2}\Big]_1^2 = -\sqrt{1} + \sqrt{4} = -1 + 2 = 1\](b)(i) Expanding along the first row:
\[\begin{vmatrix} 2 & -3 & 1 \\ 0 & 1 & -2 \\ 1 & 2 & -3 \end{vmatrix} = 2(1\cdot(-3) - (-2)\cdot2) + 3(0\cdot(-3) - (-2)\cdot1) + 1(0\cdot2 - 1\cdot1)\] \[= 2(1) + 3(2) + 1(-1) = 2 + 6 - 1 = 7\](ii) The coefficient matrix of the system is exactly this determinant, so \(\Delta = 7\). By Cramer's rule (replacing each column with \((10, -7, -9)^T\)):
\[\Delta_x = \begin{vmatrix} 10 & -3 & 1 \\ -7 & 1 & -2 \\ -9 & 2 & -3 \end{vmatrix} = 14,\quad \Delta_y = \begin{vmatrix} 2 & 10 & 1 \\ 0 & -7 & -2 \\ 1 & -9 & -3 \end{vmatrix} = -7,\quad \Delta_z = \begin{vmatrix} 2 & -3 & 10 \\ 0 & 1 & -7 \\ 1 & 2 & -9 \end{vmatrix} = 21\] \[x = \frac{14}{7} = 2,\quad y = \frac{-7}{7} = -1,\quad z = \frac{21}{7} = 3\]\(x = 2,\ y = -1,\ z = 3\). Check: \(2(2) - 3(-1) + 3 = 10\) and \((-1) - 2(3) = -7\).
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