(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 ...
Question 1 Report
(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.
Answer Details
(a) Let \(u=5-x^2\Rightarrow \dfrac{du}{dx}=-2x\), so \(x\,dx=-\tfrac12\,du\). Limits: \(x=1\Rightarrow u=4\); \(x=2\Rightarrow u=1\).
\[\int_{1}^{2}\frac{x}{\sqrt{5-x^2}}\,dx=\int_{4}^{1}\frac{-\tfrac12\,du}{\sqrt{u}}=-\frac12\int_{4}^{1}u^{-1/2}\,du=-\frac12\Big[2\sqrt u\Big]_{4}^{1}=-\Big[\sqrt u\Big]_{4}^{1}.\] \[=-\big(\sqrt1-\sqrt4\big)=-(1-2)=1.\]The value of the integral is 1.
(b) With \(y=px^2+qx\), \(\dfrac{dy}{dx}=2px+q\) and \(\dfrac{d^2y}{dx^2}=2p\).
Comparing \(\dfrac{dy}{dx}=6x+7\): \(2p=6\Rightarrow p=3\) and \(q=7\). The second derivative \(2p=6\) confirms this.
So \(p=3,\ q=7\).
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