(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.
Explanation
(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} [2sqrt{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)