(a) Evaluate : \(\int_{1} ^{4} \frac{x(3x - 2)}{2\sqrt{x}} \mathrm {d} x\) (b) The equation of a circle is given by \(2x^{2} + 2y^{2} - 8x + 5y - 10 = 0\). ...
Question 1 Report
(a) Evaluate : \(\int_{1} ^{4} \frac{x(3x - 2)}{2\sqrt{x}} \mathrm {d} x\)
(b) The equation of a circle is given by \(2x^{2} + 2y^{2} - 8x + 5y - 10 = 0\). Find the :
(i) coordinates of the centre ; (ii) radius of the circle .
Answer Details
(a) Simplify the integrand first. With \(\sqrt{x} = x^{1/2}\),
\[\frac{x(3x-2)}{2\sqrt{x}} = \frac{3x^2 - 2x}{2x^{1/2}} = \frac{3}{2}x^{3/2} - x^{1/2}.\]
Integrate:
\[\int\left(\frac{3}{2}x^{3/2} - x^{1/2}\right)dx = \frac{3}{2}\cdot\frac{x^{5/2}}{5/2} - \frac{x^{3/2}}{3/2} = \frac{3}{5}x^{5/2} - \frac{2}{3}x^{3/2}.\]
Evaluate from \(1\) to \(4\) (using \(4^{5/2}=32,\ 4^{3/2}=8\)):
\[\left[\frac{3}{5}(32) - \frac{2}{3}(8)\right] - \left[\frac{3}{5} - \frac{2}{3}\right] = \left(\frac{96}{5} - \frac{16}{3}\right) - \left(-\frac{1}{15}\right) = \frac{208}{15} + \frac{1}{15} = \frac{209}{15}.\]
So the integral is \(\dfrac{209}{15} = 13\tfrac{14}{15} \approx 13.93\).
(b) Divide the circle equation by 2 to get unit leading coefficients:
\[x^2 + y^2 - 4x + \tfrac{5}{2}y - 5 = 0.\]
Comparing with \(x^2 + y^2 + 2gx + 2fy + c = 0\): \(2g=-4\Rightarrow g=-2\), \(2f=\tfrac52\Rightarrow f=\tfrac54\), \(c=-5\).
(i) Centre \((-g, -f) = \left(2,\, -\tfrac{5}{4}\right)\).
(ii) Radius \(r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + \tfrac{25}{16} + 5} = \sqrt{\tfrac{169}{16}} = \tfrac{13}{4} = 3.25\).
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