By K F Riley; M P Hobson; S J Bence
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Extra resources for Mathematical methods for physics and engineering : a comprehensive guide [STUDENT SOLNS]
Example text
D) 0 1 If u and v are functions of x, the general formula for integration by parts is b a b uv dx = [ uv ] ba − u v dx. a Any given integrand w(x) has to be written as w(x) = u(x)v (x) with v (x) chosen so that (i) it can be integrated explicitly, and (ii) it results in a u that has u no more complicated than u itself. There are usually several possible choices but the one that makes both u and v as simple as possible is normally the best. (a) Here the obvious choice at the first stage is u(x) = x2 and v (x) = sin x.
But ∞ Γ(1) = e−x dx = 1. 0 Hence Γ(n + 1) = n!. (a)(ii) Applying the recurrence relation derived above, Γ 7 2 = 5 3 1 2 2 2 1 2 Γ = 15 8 √ π. (b) With this general definition of a factorial, we have − 23 ! = Γ − 12 = 1 1 −2 Γ 1 2 √ = −2 π. 43 By integrating by parts twice, prove that In as defined in the first equality below for positive integers n has the value given in the second equality: π/2 In = sin nθ cos θ dθ = 0 n − sin(nπ/2) . n2 − 1 Taking sin nθ as u and cos θ as v and noting that with this choice u = −n2 u 35 PRELIMINARY CALCULUS and v = −v, we expect that after two integrations by parts we will recover (a multiple of) In .
From x3 + y 3 − 12x − 8y − 16 = 0 we obtain, by implicit differentiation, that 3x2 + 3y 2 y − 12 − 8y = 0 ⇒ y = 3x2 − 12 . 8 − 3y 2 Clearly y = 0 at x = ±2. At x = 2, 8 + y 3 − 24 − 8y − 16 = 0 ⇒ y = 0. However, at x = −2, −8 + y 3 + 24 − 8y − 16 = 0, with one solution y = 0. Thus the point (−2, 0) lies on the curve and y = 0 there. It follows that the curve touches the x-axis at that point. 31 Find the indefinite integrals J of the following ratios of polynomials: (a) (b) (c) (d) (x + 3)/(x2 + x − 2); (x3 + 5x2 + 8x + 12)/(2x2 + 10x + 12); (3x2 + 20x + 28)/(x2 + 6x + 9); x3 /(a8 + x8 ).