(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 .
Explanation
(a) \(\int_{1} ^{4} \frac{x(3x -2)}{2\sqrt{x}} \mathrm {d} x\)
= \(\int_{1} ^{4} \frac{1}{2} x(x^{-\frac{1}{2}})(3x - 2) \mathrm {d} x\)
= \(\frac{1}{2} \int_{1} ^{4} (3x^{\frac{3}{2}} - 2x^{\frac{1}{2}}) \mathrm {d} x\)
= \(\frac{1}{2} [\frac{3x^{\frac{5}{2}}}{\frac{5}{2}} - \frac{2x^{\frac{3}{2}}}{\frac{3}{2}}]_{1} ^{4}\)
= \(\frac{1}{2} [\frac{6x^{\frac{5}{2}}}{5} - \frac{4x^{\frac{3}{2}}}{3}]_{1} ^{4}\)
= \(\frac{1}{2} [\frac{6(4^\frac{5}{2})}{5} - \frac{4(4^\frac{3}{2})}{3}] - \frac{1}{2} [\frac{6(1^\frac{5}{2})}{5} - \frac{4(1^\frac{3}{2})}{3}]\)
= \(\frac{1}{2} [\frac{6 \times 32}{5} - \frac{4 \times 8}{3}] - \frac{1}{2} [\frac{6}{5} - \frac{4}{3}]\)
= \(\frac{1}{2}[38.4 - 10.67] - \frac{1}{2} [1.2 - 1.33]\)
= \(\frac{1}{2} [27.73 + 0.13]\)
= \(13.93\)
(b) \(2x^{2} + 2y^{2} - 8x + 5y - 10 = 0\)
\(x^{2} + y^{2} - 4x + \frac{5}{2}y = 5\)
\(x^{2} - 4x + [\frac{1}{2} (-4)]^{2} + y^{2} + \frac{5}{2}y + [\frac{1}{2} (\frac{5}{2})]^{2} = 5 + 4 + \frac{25}{16} = \frac{169}{16}\)
\((x - 2)^{2} + (y + \frac{5}{2})^{2} = \frac{169}{16}\)
(i) coordinates of the centre = \((2, -\frac{5}{4})\)
(ii) Radius = \(\sqrt{\frac{169}{16}} = \frac{13}{4}\)