(a) Using the substitution \(u = 5 - x^{2}\), evaluate \(\int_{1}^{2} \frac{x}{\sqrt{5 - x^{2}}} \mathrm {d} x\).
(b) If \(y = px^{2} + qx; \frac{\mathrm d y}{\mathrm d x} = 6x + 7\) and \(\frac{\mathrm d^{2} y}{\mathrm d x^{2}} = 6\), find the values of p and q.
(a) \(u = 5 - x^{2}\)
= \(\int_{1}^{2} \frac{x}{\sqrt{u}} dx\)
\(\frac{\mathrm d u}{\mathrm d x} = -2x\)
\(\mathrm {d} x = \frac{\mathrm d u}{-2x}\)
\(\implies \int_{1}^{2} \frac{x}{\sqrt{u}} \frac{\mathrm d u}{-2x}\)
= \(-\frac{1}{2} \int_{1}^{2} \frac{1}{\sqrt{u}} \mathrm {d} u\)
= \(-\frac{1}{2} \int_{1}^{2} u^{-\frac{1}{2}} \mathrm {d} u\)
= \(-\frac{1}{2} [2u^{1}{2}]|_{1}^{2}\)
= \(-\frac{1}{2} [2\sqrt{5 - x^{2}}]|_{1}^{2}\)
= \(-\frac{1}{2} [2 - 4]\)
= 1.
(b)\(y = px^{2} + qx\)
\(\frac{\mathrm d y}{\mathrm d x} = 2px + q = 6x + 7\)
\(\implies 2p = 6; p = 3\)
\(q = 7\)
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