CALCULUS HANDOUT 6 - RIEMANN-DARBOUX - definitions

THE RIEMANN DARBOUX INTEGRAL

A partition P of the [a, b] is a finite set of points {x0, x1, .., xn} 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 = sup{f(x) | a ≤ x ≤ b} and m = inf{f(x) | a ≤ x ≤ b}. For any partition P of [a, b] we have: m(b − a) ≤ Lf (P ) ≤ Uf (P ) ≤ M(b − a). Lf = {Lf (P ) | P is a partition of [a, b]} and Uf = {Uf (P ) | P is a partition of [a, b]} are bounded sets. So Lf = sup Lf and Uf = inf Uf exist. Moreover, Lf ≤ Uf .

A 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 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 ∈ 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 ≤ |f(x)| 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 = {x0, x1, . . . , xn} of [a, b] and continuous functions fi defined on [xi−1, xi], such that f(x) = fi(x) for x ∈ (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 ∈ [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 : 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 : 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, ∞) and Riemann-Darboux integrable on [a, b] for every b > a. Z b Z ∞ If lim f(x) dx exists, then the of first kind f(x) converges. b→∞ a a Let f be a function defined on (a, b] and Riemann-Darboux integrable on [a + ε, b], for ε ∈ (0, b − a). Z b Z b If lim f(x) dx exists, then the improper integral of second kind f(x) dx converges. ε→0+ a+ε a Comparison test for improper integrals of first kind: Let f and g be defined on [a, ∞) and Riemann-Darboux integrable on [a, b] for every b > a. Z ∞ Z ∞ 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: √ 3 3x + 11 a) f(x) = x x + 1 b) f(x) = √ c) f(x) = 1 + x − 2 x2 − x − 6 Solution:a) Method I: Integration by parts method

0 0 √ 2 3 h(x) = x ⇒ h (x) = 1; g (x) = x + 1 ⇒ g(x) = (x + 1) 2 3

3 +1 Z √ 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 √ √ √ √ 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 √ t = x − 2|2 ⇔ t2 = x − 2 ⇔ x = t2 + 2 ⇒ dx = 2tdt

Z 3 Z 3 Z 6t ⇒ √ 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 √ √ dt = du = 6 du − du = 6u − ln |u| = 6(1 + t) − ln |1 + t| = 6(1 + x − 2) − ln |1 + x − 2| + 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 |x − 3| − ln |x + 2| + 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 ∞ 2 ∞ Z 1 Z Z 1 Z 1 a) √ dx b) cos xdx c) √ dx d) dx 3 − x 2 − x x + ex −∞ 3 0 1 Solution: 0 0 Z 1 Z 1 √ 0 √ √ √ a) √ dx = lim √ dx = lim (−2 3 − x) = lim (−2 3 + 2 3 − t) = −2 3 + ∞ = +∞ 3 − x t→−∞ 3 − x t→−∞ t t→−∞ −∞ t 0 Z 1 ⇒ √ dx is divergent 3 − x −∞ ∞ t Z Z  t  b) cos xdx = lim cos xdx = lim sin x = lim (sin t − sin 3) = lim sin t − sin 3 t→∞ t→∞ 3 t→∞ t→∞ 3 3 ∞ Z As lim sin t does not exist (check!), it results that cos xdx is divergent. t→∞ 3 2 t Z 1 Z 1 √ t  √ √  √ c) √ dx = lim √ dx = lim − 2 − x = lim − 2 − t + 2 = 2 2 − x t→2− 2 − x t→2− 0 t→2− 0 0 2 2 Z 1 √ Z 1 We have obtained that √ dx = 2, therefore √ dx is convergent. 2 − x 2 − x 0 0

3 ∞ Z 1 d) We study the convergence of the integral dx. x + ex 1 1 1 x > 1 ⇒ < = e−x x + ex ex ∞ t Z Z t 1 e−xdx = lim e−xdx = lim −e−x = lim −e−t + e−1 = t→∞ t→∞ 1 t→∞ e 1 1 ∞ Z 1 As e−xdx = < ∞, therefore it is convergent, appplying the comparison test for improper integrals of first e 1 ∞ Z 1 kind, we obtain that dx is convergent. x + ex 1

CALCULUS HANDOUT 6 - RIEMANN-DARBOUX INTEGRAL - exercises

1. Find the primitives of the following functions on their maximal domains of definition: 1 1 1 − x2 1. f(x) = x2 + 2x + 3 11. f(x) = √ + √ 23. f(x) = √ x 3 x2 (1 + x2) 1 + x2 1 √ √ 2. f(x) = x + 3 2 1 x 12. f(x) = x x + 2x x 24. f(x) = 3. f(x) = sin x + cos x 13. f(x) = 1 + cos 3x 1 + x4 2 1 14. f(x) = ex cosh(2x) x 4. f(x) = √ 25. f(x) = 4 1 − 4x2 1 + x 15. f(x) = |x| 2 1 x 5. f(x) = √ 1 26. f(x) = 4 − x2 16. f(x) = √ √ 1 + x12 1 + x + x + 1 x 2 2 e 6. f(x) = + x earctan x 27. f(x) = 2 2 e2x + e−2x sin x cos x 17. f(x) = p 1 (1 + x2)3 1 7. f(x) = 28. f(x) = sin2 x · cos2 x 18. f(x) = sin(ln x) 1 + |1 − ex| √ 1 x |x| 8. f(x) = 19. f(x) = e 29. f(x) = x2 + 4 arcsin x 20. f(x) = e 1 + |x| 1 1 9. f(x) = x4 arctan x 30. f(x) = 4x2 + 1 21. f(x) = 3 + cos x 1 + x2 1 √ 1 10. f(x) = 31. f(x) = x2 − 1 22. f(x) = tan x 1 + sin2 x 2. Compute the following integrals: x−1 Z 1 e x+1 Z 1 x + 1 1. f(x) dx where f(x) = 3 7. f(x) dx where f(x) = 4 2 0 (1 + x) 0 x + x + 1 π 3 Z 2 Z √ 2. f(x) dx where f(x) = ln(sin x) 8. f(x) dx where f(x) = x + 1 + (1 − 2x)[x − 1] 0 0 π  Z 2 1 x 3. f(x) dx where f(x) = ln(cos x) Z 1  e , x ≤ 0 9. f(x) dx where f(x) = √ 2 0 1 + x − 1 Z 1 −1  , x > 0 4. f(x) dx where f(x) = x ln(sin πx) x  0 3 x Z 1  − + e , x ≤ 0 Z 1 x earctan x  √ 4 √ 5. f(x) dx where f(x) = 10. f(x) dx where f(x) = p 2 3 x + 4 − 4 0 (1 + x ) −1  , x > 0  x Z 3 1 6. f(x) dx where f(x) = √ 2 2 (1 + x) 1 + x + x 3. Find recurrence formulae for the following integrals:

π π π 1 1 Z 2 Z 2 Z 4 Z p Z 1. sinn x dx 2. sin2n x dx 3. tgnx dx 4. x2n 1 − x2 dx 5. xn sin πx dx 0 0 0 0 0

4 4. Study the convergence of the following improper integrals (a, b > 0, p, q ∈ R): π ∞ ∞ 2 b Z 1 Z 1 Z 1 Z 1 1. dx 8. √ dx 15. √ dx 21. dx 1 + x2 x sin x (b − x)p 0 0 0 a ∞ ∞ ∞ ∞ Z 1 Z e−x Z arctan x Z 1 2. dx 9. dx 16. dx 22. dx x2 1 + x2 x xp ln x 1 0 0 2 ∞ 1 ∞ ∞ Z 1 Z e−x Z sin2 x Z ln x 3. √ dx 10. √ dx 17. dx 23. dx x x x2 xp 1 0 0 1 ∞ ∞ ∞ ∞ Z Z 1 Z 1 Z x ln x 4. sin x dx 11. dx 18. dx 24. dx xp (ln x)ln x (1 + x2)p −∞ a 2 0 1 π 1 Z b 2 Z 1 Z 1 Z ln(sin x) p 1 5. √ dx 12. dx 19. √ dx 25. ln dx x xp x x 0 0 0 0 1 ∞ ∞ Z Z ∞ Z 2 1 −x2 Z 1 sin x 6. √ dx 13. e dx 20. dx 26. p dx 1 − x2 xp(1 + xq) x 0 −∞ 0 0 1 ∞ Z 1 Z 1 7. dx 14. dx x 1 + | sin x| 0 0

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