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CALCULUS HANDOUT 6 - RIEMANN-DARBOUX INTEGRAL - Definitions

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CALCULUS HANDOUT 6 - RIEMANN-DARBOUX INTEGRAL - Definitions CALCULUS HANDOUT 6 - RIEMANN-DARBOUX INTEGRAL - definitions THE RIEMANN DARBOUX INTEGRAL A partition P of the interval [a; b] is a finite set of points fx0; x1; ::; xng satisfying a = x0 < x1 < :: < xn = b. Consider a function f defined and bounded on [a; b]. n X Upper Darboux sum of f related to P : Uf (P ) = Mi(xi − xi−1) where Mi = sup f(x). i=1 xi−1≤x≤xi n X Lower Darboux sum of f related to P : Lf (P ) = mi(xi − xi−1) where mi = inf f(x). xi−1≤x≤xi i=1 Consider M = supff(x) j a ≤ x ≤ bg and m = infff(x) j a ≤ x ≤ bg. For any partition P of [a; b] we have: m(b − a) ≤ Lf (P ) ≤ Uf (P ) ≤ M(b − a). Lf = fLf (P ) j P is a partition of [a; b]g and Uf = fUf (P ) j P is a partition of [a; b]g are bounded sets. So Lf = sup Lf and Uf = inf Uf exist. Moreover, Lf ≤ Uf . A function defined and bounded on [a; b] is Riemann-Darboux integrable on [a; b] if Lf = Uf . Z b This common value is denoted by f(x) dx = Lf = Uf . a Properties of the Riemann-Darboux integral: If f and g are Riemann-Darboux integrable on [a; b] then all the integrals below exist and Z b Z b Z b (1) (α f(x) + β g(x)) dx = α f(x) dx + β g(x) dx for any α; β 2 R: a a a Z b Z c Z b (2) f(x) dx = f(x) dx + f(x) dx for any a ≤ c ≤ b. a a c Z b Z b (3) If f(x) ≤ g(x) on [a; b] then f(x) dx ≤ g(x) dx. a a Z b Z b (4) f(x) dx ≤ jf(x)j dx. a a Classes of Riemann-Darboux integrable functions: If f is continuous on [a; b], then f is Riemann-Darboux integrable on [a; b]. A function f is called piecewise continuous on [a; b] if there exists a partition P = fx0; x1; : : : ; xng of [a; b] and continuous functions fi defined on [xi−1; xi], such that f(x) = fi(x) for x 2 (xi−1; xi), i = 1; 2; : : : ; n. n Z b X Z xi A piecewise continuous function is Riemann-Darboux integrable and f(x) dx = fi(x) dx. a i=1 xi−1 The integral mean value theorem: If f and g are continuous on [a; b] and g(x) ≥ 0 for x 2 [a; b], then there exists c between a and b such Z b Z b that f(x) · g(x) dx = f(c) g(x) dx. a a The fundamental theorem of calculus: Z x If f is Riemann-Darboux integrable on [a; b] and F (x) = f(t) dt, then F is continuous on [a; b]. a Furthermore, if f is continuous on [a; b], then F is differentiable on [a; b] and F 0 = f. Any function Φ such that Φ0 = f is called a primitive (antiderivative) of f. Two primitives of the same function f differ by a constant. Z b If F is a primitive of f, then f(x) dx = F (b) − F (a). a Integration by parts: If the functions f and g are continuously differentiable on [a; b], then Z Z f(x) · g0(x) dx = f(x) · g(x) − f 0(x) · g(x) dx Z Z where f(x)g0(x)dx and f 0(x)g(x)dx represent the set of primitives of fg0 and f 0g, respectively. 1 b Z b Z b 0 0 Consequence: f(x) · g (x) dx = f(x) · g(x) − f (x) · g(x) dx a a a Change of variables: If the function g :[α; β] ! [a; b] is a continuously differentiable bijection having the property g(α) = a, g(β) = b and f :[a; b] ! R1 is continuous, then Z Z f(x) dx ◦ g = (f ◦ g)(t) · g0(t) dt Z Z where f(x)dx and (f ◦ g)(t) · g0(t) dt represent the set of primitives of f and (f ◦ g) · g0, respectively. Z g(β) Z β Consequence: f(x) dx = (f ◦ g)(t) · g0(t) dt g(α) α Improper integrals: Let f be a function bounded on [a; 1) and Riemann-Darboux integrable on [a; b] for every b > a. Z b Z 1 If lim f(x) dx exists, then the improper integral of first kind f(x) converges. b!1 a a Let f be a function defined on (a; b] and Riemann-Darboux integrable on [a + "; b], for " 2 (0; b − a). Z b Z b If lim f(x) dx exists, then the improper integral of second kind f(x) dx converges. "!0+ a+" a Comparison test for improper integrals of first kind: Let f and g be defined on [a; 1) and Riemann-Darboux integrable on [a; b] for every b > a. Z 1 Z 1 Suppose that 0 ≤ f(x) ≤ g(x) for all x ≥ a and g(x) dx converges. Then f(x) dx converges too. a a CALCULUS HANDOUT 6 - RIEMANN-DARBOUX INTEGRAL - solved examples Ex.1 Find the primitives of the following functions on their maximal domain of definition: p 3 3x + 11 a) f(x) = x x + 1 b) f(x) = p c) f(x) = 1 + x − 2 x2 − x − 6 Solution:a) Method I: Integration by parts method 0 0 p 2 3 h(x) = x ) h (x) = 1; g (x) = x + 1 ) g(x) = (x + 1) 2 3 3 +1 Z p 2 3 2 Z 3 2 3 2 (x + 1) 2 2 3 4 5 ) x x + 1dx = x(x + 1) 2 − (x + 1) 2 dx = x(x + 1) 2 − · = x(x + 1) 2 − (x + 1) 2 + C 3 3 3 3 3 3 15 + 1 2 2 3 2 2 3 2 2 2 3 3 2 = (x + 1) 2 x − (x + 1) + C = (x + 1) 2 x − x − + C = (x + 1) 2 x − + C 3 5 3 5 5 3 5 5 Method II: Change of variables method t = x + 1 , x = t − 1 ) dt = dx Z Z Z Z Z Z 2+ 1 1 +1 p p p p 1+ 1 1 t 2 t 2 2 5 2 3 ) x x + 1dx = (t − 1) tdt = t tdt − tdt = t 2 dt − t 2 dt = − = t 2 − t 2 1 1 5 3 2 + + 1 2 2 2 5 2 3 2 3 3 2 3 3 3 = (x + 1) 2 − (x + 1) 2 + C = (x + 1) 2 (x + 1) − 1 + C = (x + 1) 2 x + − 1 + C 5 3 3 5 3 5 5 2 3 3 2 = (x + 1) 2 x − + C 3 5 5 b) We consider the following change of variables p t = x − 2j2 , t2 = x − 2 , x = t2 + 2 ) dx = 2tdt Z 3 Z 3 Z 6t ) p dx = · 2tdt = dt 1 + x − 2 1 + t 1 + t 2 Consider the change of variables: u = 1 + t , t = u − 1 ) du = dt. Then Z 6t Z 6(u − 1) Z 1 p p dt = du = 6 du − du = 6u − ln juj = 6(1 + t) − ln j1 + tj = 6(1 + x − 2) − ln j1 + x − 2j + C 1 + t u u 3x + 11 3x + 11 4 1 c) We can show that = = − (check!). Then it results that x2 − 2 − 6 (x − 3)(x + 2) x − 3 x + 2 Z 3x + 11 Z 4 1 Z 1 Z 1 dx = − dx = 4 dx − dx = 4 ln jx − 3j − ln jx + 2j + C: x2 − x − 6 x − 3 x + 2 x − 3 x + 2 Ex.2 We compute the following integrals: π 2 Z Z2 a) ln xdx b) sin5 x cos3 xdx 1 0 Solution: a) Using the integration by parts method, we obtain 2 2 2 2 2 Z Z Z 2 Z 1 Z ln xdx = 1 · ln xdx = x0 · ln xdx = x ln x − x · dx = 2 ln 2 − 1 ln 1 − dx 1 x 1 1 1 1 1 2 = 2 ln 2 − 1 · 0 − x = 2 ln 2 − (2 − 1) = 2 ln 2 − 1: 1 b) We have π π π Z2 Z2 Z2 sin5 x cos3 xdx = sin5 x cos2 x cos xdx = sin5 x(1 − sin2 x) cos xdx 0 0 0 π Consider the change of variables u = sin x ) du = du = cos xdx. Then x = 0 ) u = 0, x = ) u = 1. 2 π 2 1 1 1 Z Z Z Z u6 1 u8 1 ) sin5 x(1 − sin2 x) cos xdx = u5(1 − u2)du = u5du − u7du = − 6 0 8 0 0 0 0 0 16 06 18 08 1 1 4 − 3 1 = − − + = − = = 6 6 8 8 6 8 24 24 Ex.3 We study the convergence of the following improper integrals: 0 1 2 1 Z 1 Z Z 1 Z 1 a) p dx b) cos xdx c) p dx d) dx 3 − x 2 − x x + ex −∞ 3 0 1 Solution: 0 0 Z 1 Z 1 p 0 p p p a) p dx = lim p dx = lim (−2 3 − x) = lim (−2 3 + 2 3 − t) = −2 3 + 1 = +1 3 − x t→−∞ 3 − x t→−∞ t t→−∞ −∞ t 0 Z 1 ) p dx is divergent 3 − x −∞ 1 t Z Z t b) cos xdx = lim cos xdx = lim sin x = lim (sin t − sin 3) = lim sin t − sin 3 t!1 t!1 3 t!1 t!1 3 3 1 Z As lim sin t does not exist (check!), it results that cos xdx is divergent.
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