Introduction to Riemnn integrl Chpter 6. Riemnn Integrl Won-Kwng Prk Deprtment of Mthemtics, The College of Nturl Sciences Kookmin University Second semester, 2015 1 / 41
Introduction to Riemnn integrl Prtition Definition If I := [, b] is closed bounded intervl in R, then prtition of I is finite, ordered set P = {x 0, x 1,, x n 1, x n} of points in I such tht = x 0 < x 1 < < x n 1 < x n = b. The points of P re used to divide I onto non-overlpping subintervls I 1 := [x 0, x 1 ], I 2 := [x 1, x 2 ],, I n := [x n 1, x n]. 2 / 41
Introduction to Riemnn integrl Upper nd lower sum Definition Let f : I R be bounded on I := [, b] nd P = {x 0, x 1,, x n 1, x n} be prtition of I. 1 The upper sum of f for the prtition P is the sum U(f, P) = n M i [f ](x i x i 1 ), i=1 where M i [f ] := sup {f (x) : x [x i 1, x i ]} for i = 1, 2,, n. 2 The lower sum of f for the prtition P is the sum L(f, P) = n m i [f ](x i x i 1 ), i=1 where m i [f ] := inf {f (x) : x [x i 1, x i ]} for i = 1, 2,, n. For convenience, we will often bbrevite x i x i 1 to x i. 3 / 41
Introduction to Riemnn integrl Upper nd lower sum Corollry Let f : I R be bounded on I := [, b] nd P be prtition of I. Then L(f, P) U(f, P). 4 / 41
Introduction to Riemnn integrl Refinement Definition If Q nd P re prtitions of [, b] nd P Q, we sy tht Q is refinement of P. 5 / 41
Introduction to Riemnn integrl Refinement Theorem Let f : I R be bounded on I := [, b] nd P = {x 0, x 1,, x n 1, x n} be prtition of I. If Q is refinement of P then following rrngement holds: L(f, P) L(f, Q) U(f, Q) U(f, P). 6 / 41
Introduction to Riemnn integrl Refinement Corollry Let f : I R be bounded on I := [, b] nd P nd Q re prtitions of I then L(f, Q) U(f, P). 7 / 41
Introduction to Riemnn integrl Refinement Theorem Let L(f ) := sup {L(f, P) : P is prtition of I} U(f ) := inf {U(f, P) : P is prtition of I} then L(f ) U(f ). 8 / 41
Introduction to Riemnn integrl Upper nd lower integrl Definition Let f : I R be bounded on I := [, b]. 1 The upper integrl of f on I is defined by f (x)dx := U(f ) = inf {U(f, P) : P is prtition of I}. 2 The lower integrl of f on I is defined by f (x)dx := L(f ) = sup {L(f, P) : P is prtition of I}. 9 / 41
Introduction to Riemnn integrl Riemnn integrl Definition Let f : I R be bounded on I := [, b]. If L(f ) = U(f ), we sy tht f is Riemnn integrble (or integrble) on I, nd in this cse we define the Riemnn integrl of f on I, denoted f (x)dx or f to be the common vlue f (x)dx = f (x)dx. 10 / 41
Introduction to Riemnn integrl Riemnn integrl Exmple Let f : [0, 1] R be Dirichlet s discontinuous function defined by { 1 if x is rtionl f (x) := 0 if x is irrtionl. We clim tht f isn t Riemnn integrble. 11 / 41
Introduction to Riemnn integrl Riemnn integrl Theorem (Riemnn s condition) Let f : I R be bounded on I := [, b]. Then f is Riemnn integrble if nd only if for every ε > 0, there exists prtition P of I such tht U(f, P) L(f, P) < ε. 12 / 41
Properties of Riemnn Integrl Monotone function Theorem If f : [, b] R is monotone on [, b] then f is Riemnn integrble on [, b]. 13 / 41
Properties of Riemnn Integrl Monotone function Exmple Let f : [0, 1] R be function defined by 0 if x = 0 f (x) := 1 1 if n n + 1 < x 1 n for n N. Then since f is monotone, f is Riemnn integrble on [0, 1]. 14 / 41
Properties of Riemnn Integrl Continuous function Theorem If f : [, b] R is continuous on [, b] then f is Riemnn integrble on [, b]. 15 / 41
Properties of Riemnn Integrl Continuous function Exmple Let f : [0, 1] R be function defined s x sin 1 if x 0 f (x) = x 0 if x = 0. Then since f is continuous on [0, 1], f is Riemnn integrble on [, b]. Exmple Let h : [0, 1] R be Thome s function defined by h(x) := 1 n if x = m n 0 if x is irrtionl or x = 0. Then h is Riemnn integrble on [0, 1]. for m, n N nd gcd(m, n) = 1 16 / 41
Properties of Riemnn Integrl Linerity Theorem Let f, g : [, b] R be Riemnn integrble functions. 1 For α R, αf is Riemnn integrble nd 2 f + g is Riemnn integrble nd αf (x)dx = α f (x)dx. (f + g)(x)dx = f (x)dx + g(x)dx. 17 / 41
Properties of Riemnn Integrl Linerity Corollry Let f, g : [, b] R be Riemnn integrble functions. Then for α, β R, (αf + βg)(x)dx = α f (x)dx + β g(x)dx. 18 / 41
Properties of Riemnn Integrl Order preserving Theorem Let f, g : [, b] R be Riemnn integrble functions. 1 If f (x) 0 for ll x [, b] then 2 If f (x) g(x) for ll x [, b] then f (x)dx 0. f (x)dx g(x)dx. 19 / 41
Properties of Riemnn Integrl Order preserving Exmple Let f (x) = 0 nd g(x) = x for x [ 1, 3]. Then 3 however f (x) > g(x) for x [ 1, 0). f (x)dx = 0 < 4 = 3 1 1 g(x)dx 20 / 41
Properties of Riemnn Integrl Additivity Theorem Let f : [, b] R be function nd c (, b). If f is Riemnn integrble for closed subintervls [, c] nd [c, b] of [, b] then f is Riemnn integrble on [, b] nd f (x)dx = c f (x)dx + c f (x)dx. 21 / 41
Properties of Riemnn Integrl Composite function Theorem Let f : [, b] R be Riemnn integrble function on I = [, b] nd g : [c, d] R be continuous function on [c, d]. If f (I) [c, d] then g f is Riemnn integrble function. 22 / 41
Properties of Riemnn Integrl Composite function Corollry If f : [, b] R be Riemnn integrble function on I = [, b] then f n is Riemnn integrble. Corollry Let f : [, b] R be Riemnn integrble function on I = [, b] then f is Riemnn integrble nd f (x)dx f (x) dx. 23 / 41
Properties of Riemnn Integrl Composite function Theorem (Intermedite vlue theorem for integrls) Let f be continuous function on [, b], then for t lest one x [, b] we hve f (x) = 1 b f (t)dt. 24 / 41
The Fundmentl Theorem of Clculus Fundmentl theorem of clculus: first form Theorem (Fundmentl theorem of clculus: first form) Let f : [, b] R is differentible on [, b] nd f is Riemnn integrble on [, b] then f (x)dx = f (b) f (). 25 / 41
The Fundmentl Theorem of Clculus Fundmentl theorem of clculus: first form Exmple If f (x) = 1 2 x 2 for ll x [, b] then f (x) = x for ll x [, b]. Further, f is continuous so it is Riemnn integrble on [, b]. Therefore, the fundmentl Theorem implies tht xdx = f (b) f () = 1 2 (b2 2 ). Exmple If g(x) = Tn 1 x for ll x [, b] then g (x) = (x 2 + 1) 1 for ll x [, b]. Further, g is continuous so it is Riemnn integrble on [, b]. Therefore, the fundmentl Theorem implies tht 1 x 2 + 1 dx = g(b) g() = Tn 1 (b) Tn 1 (). 26 / 41
The Fundmentl Theorem of Clculus Fundmentl theorem of clculus: first form Exmple If h(x) = 2 x for ll x [0, b] then h is continuous on [0, b] nd h (x) = ( x) 1 for ll x (0, b]. Since h is not bounded on (0, b], it isn t Riemnn integrble on [0, b] no mtter how we define h(0). Therefore, the fundmentl Theorem does not pply. 27 / 41
The Fundmentl Theorem of Clculus Indefinite integrl Definition (Indefinite integrl) If f : [, b] R is Riemnn integrble on [, b] then the function defined by F(x) := x f (t)dt for x [, b] is clled the indefinite integrl of f with bsepoint. 28 / 41
The Fundmentl Theorem of Clculus Uniform continuity Theorem If f : [, b] R is Riemnn integrble on [, b] then, indefinite integrl F is uniformly continuous on [, b]. Definition (Lipschitz function) Let f : D R be function. If there exists constnt K > 0 such tht f (x) f (y) K x y for ll x, y D, then f is sid to be Lipschitz function or to stisfy Lipschitz condition on D. Theorem If f : D R is Lipschitz function, then f is uniformly continuous on D. 29 / 41
The Fundmentl Theorem of Clculus Fundmentl theorem of clculus: second form Theorem (Fundmentl theorem of clculus: second form) Let f : [, b] R is Riemnn integrble on [, b] nd continuous t point c [, b]. Then the indefinite integrl F is differentible t c nd F (c) = f (c). Proof. Suppose tht c [, b) nd consider the right-hnd derivtive of F t c. 30 / 41
The Fundmentl Theorem of Clculus Differentibility Theorem If f is continuous on [, b], then the indefinite integrl F is differentible on [, b] nd F (x) = f (x) for ll x [, b]. 31 / 41
The Fundmentl Theorem of Clculus Differentibility Exmple If f (x) := sgn(x) on [ 1, 1], then f is Riemnn integrble nd hs the indefinite integrl F(x) := x 1 with the bsepoint 1. However, since F (0) does not exist, F is not n ntiderivtive of f on [ 1, 1]. Exmple For x [0, 3], if we define F(x) := x 0 [t]dt then lthough f (x) = [x] is discontinuous on [0, 3], F is continuous on [0, 3]. 32 / 41
The Fundmentl Theorem of Clculus Substitution Theorem (Substitution theorem) Let J := [, b] nd let g : J R hve continuous derivtive on J. If f : I R is continuous on n intervl I contining g(j) then f (g(t)) g (t)dt = g(b) g() f (x)dx. Proof. Since g is differentible on J, g(j) is closed bounded intervl. Since f g nd g re continuous J, (f g)g is continuous on J so tht it is Riemnn integrble on J. 33 / 41
The Fundmentl Theorem of Clculus Integrtion by prts Theorem (Integrtion by prts) Let f, g be differentible on [, b] nd f, g re Riemnn integrble on [, b]. Then [ ] b f (x)g (x)dx = f (x)g(x) f (x)g(x)dx. 34 / 41
The Fundmentl Theorem of Clculus Tylor s theorem with the reminder Theorem (Tylor s theorem with the reminder) Suppose tht f, f,, f (n), f (n+1) exist on [, b] nd tht f (n+1) is Riemnn integrble on [, b]. Then we hve f (b) = f () + f ()(b ) + f () 2! where the reminder R n is given by R n = 1 n! (b ) 2 + + f (n) () (b ) n + R n (1) n f (n+1) (t) (b t) n dt. (2) 35 / 41
Improper Integrls Improper integrl Definition (Improper integrl) Let f : [, b] R be function nd c (, b). 1 Assume tht f is Riemnn integrble on [, c]. If the limit c lim c b f (x)dx exists in R, we sy tht f is improper integrble nd denote its limit f (x)dx = c lim c b f (x)dx. 2 Similrly, ssume tht f is Riemnn integrble on [c, b]. If the limit lim c + c f (x)dx exists in R, we sy tht f is lso improper integrble nd denote its limit f (x)dx = lim f (x)dx. c + c 36 / 41
Improper Integrls Improper integrl Exmple Let f (x) := x 1 3 for x (0, 1]. Since f is unbounded on (0, 1], f is not Riemnn integrble. However, for every c (0, 1), 1 c x 1 3 3 dx = 2 (1 c 2 3 3 ) nd lim c 0+ 2 (1 c 2 3 3 ) = 2. Hence, f is improper integrble on (0, 1] nd 1 0 x 1 3 dx = lim c 0+ 1 c x 1 3 dx = 3 2. Exmple Let g(x) := x 1 for x (0, 1]. Then for every c (0, 1), 1 c 1 dx = ln c nd lim ( ln c) =. x c 0+ 1 1 Hence, g is not improper integrble on (0, 1] nd dx =. 0 x 37 / 41
Improper Integrls Improper integrl Definition If f : [, b] R is Riemnn integrble on [, b] for every b > nd if the limit lim b exists in R, then the improper integrl f (x)dx f (x)dx = lim b is defined to be this limit. Similrly, one cn define f (x)dx = lim f (x)dx f (x)dx. 38 / 41
Improper Integrls Improper integrl Exmple Let f (x) := 1 1 + x 2. Then f is well-defined nd bounded on [0, ). Moreover f is Riemnn integrble on [0, b] for every b > 0 since f is continuous on [0, ). Since 0 1 1 + x 2 dx = Tn 1 (b) Tn 1 (0) = Tn 1 (b) nd lim b Tn 1 (b) = π 2, we cn obtin 0 1 b 1 dx = lim 1 + x 2 b 0 1 + x dx = π 2 2. Exmple Since 0 f (x)dx = 1 0 1 1 dx + dx x 1 x f = x 1/2 is not improper integrble on (0, ). nd 1 1 x dx =, 39 / 41
Improper Integrls Improper integrl Theorem Let f, g : [, ) R. For every b >, f nd g re Riemnn integrble on [, b]. Then if for x, 0 f (x) g(x) nd g is improper integrble on [, ) then f is improper integrble on [, ) nd f (x)dx g(x)dx. 40 / 41
Improper Integrls Improper integrl Theorem Let f : [, b] R is Riemnn integrble on [, b] for every b >. Then if there exists positive rel number M stisfies f (x) dx M then f nd f re improper integrble on [, ). 41 / 41