By Riley, Hobson.

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0 →0 The limit is not ﬁnite and so the integral does not exist. 33 PRELIMINARY CALCULUS (f) Yet again, the integrand has an inﬁnity (at x = 1) and the limit test has to be applied 1 0 x dx = lim −(1 − x2 )1/2 z→1 (1 − x2 )1/2 z 0 = 0 + 1 = 1. This time the limit does exist; the integral is deﬁned and has value 1. 40 Show, using the following methods, that the indeﬁnite integral of x3 /(x + 1) is 1/2 J= 3 2 35 (5x − 6x2 + 8x − 16)(x + 1)1/2 + c. (a) Repeated integration by parts. (b) Setting x + 1 = u2 and determining dJ/du as (dJ/dx)(dx/du).

2m n(n − 1) 2m−1 y (x) = m nxm−1 + x + ··· 2! rm n(n − 1) · · · (n − r + 1) rm−1 x + ··· . + r! y(x) = 1 + nxm + Now, expanding the result given in part (a) gives y (x) = nmxm−1 (1 + xm )n−1 (n − 1)(n − 2) · · · (n − s) ms x + ··· s! mn(n − 1)(n − 2) · · · (n − s) ms+m−1 = nmxm−1 + · · · + x + ··· . s! e. s = r − 1. 12 Find the positions and natures of the stationary points of the following functions: (a) x3 − 3x + 3; (b) x3 − 3x2 + 3x; (c) x3 + 3x + 3; (d) sin ax with a = 0; (e) x5 + x3 ; (f) x5 − x3 .

We can then write the equation as c b a x2 + x + a a a x+ b 2a 2 − = 0, b2 + c = 0, 4a a x+ b 2a 2 = b2 − 4ac = λ2 . 4a Since b2 > 4ac and a > 0, λ is real, positive and non-zero. e. both roots are real and they are distinct; thus, the ‘if’ part of the proposition is established. 12 PRELIMINARY ALGEBRA Now assume that both roots are real, α and β say, with α = β. Then, aα2 + bα + c = 0, aβ 2 + bβ + c = 0. Subtraction of the two equations gives a(α2 − β 2 ) + b(α − β) = 0 ⇒ b = −(α + β)a, since α − β = 0.