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Throughout the sixteenth and seventeenth centuries, mathematicians constructed a wealth of latest rules yet had no longer conscientiously hired actual definitions, proofs, or methods to record and enforce them. despite the fact that, within the early nineteenth century, mathematicians started to realize the necessity to accurately outline their phrases, to logically turn out even seen rules, and to exploit rigorous tools of manipulation.
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The corresponding values of y (x) are −24, 36 and 72. Since y(±∞) = ∞, the lowest value of y is that corresponding to the lowest minimum, which can only be at x = 1 or x = −2, as y must be positive at a minimum. The values of y(x) at these two points are y(1) = 1 and y(−2) = −26, and so the lowest value taken is −26. 15 Show that y(x) = xa2x exp x2 has no stationary points other than x = 0, if √ √ exp(− 2) < a < exp( 2). Since the logarithm of a variable varies monotonically with the variable, the stationary points of the logarithm of a function of x occur at the same values of x as the stationary points of the function.
Using d(u/v)/dx = (vu − uv )/v 2 , and obtain 1 + sin x , f(x) = cos x cos x(cos x) − (1 + sin x)(− sin x) f (x) = cos2 x 1 + sin x = cos2 x f(x) . = cos x Thus, since sec x = f (x)/f(x), it follows that sec x dx = ln[ f(x) ] + c = ln 1 + sin x cos x + c = ln(sec x + tan x) + c. 37 By making the substitution x = a cos2 θ + b sin2 θ, evaluate the deﬁnite integrals J between limits a and b (> a) of the following functions: (a) [(x − a)(b − x)]−1/2 ; (b) [(x − a)(b − x)]1/2 ; (c) [(x − a)/(b − x)]1/2 .
Write it in such a form that x appears only in a term that is the square of a linear function of x. We then examine the overall sign of the terms that do not contain x; this determines the form of the integral. In case (iii) there is no such term. We write b2 − 4ac as ∆2 > 0, or 4ac − b2 as ∆ 2 > 0, as needed. (i) For ∆2 = b2 − 4ac > 0, dx J= a = 1 a x+ 2 b 2a b2 4a2 − − c a − c a dx x+ b 2 2a b 2a b 2a − − ∆2 4a2 ∆ 2a ∆ 2a = 1 a x+ ln a∆ x+ = 1 2ax + b − ∆ ln + k. ∆ 2ax + b + ∆ + +k (ii) For −∆ 2 = b2 − 4ac < 0, dx J= a = 1 a x+ b 2 2a b2 4a2 − dx x+ = 1 2a tan−1 a∆ = 2 tan−1 ∆ b 2 2a + x+ ∆2 4a2 b 2a ∆ 2a 2ax + b ∆ 30 +k + k.
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