DEFINITE INTEGRALS. f(x)dx exists. Note that, together with the definition of definite integrals, definitions (2) and (3) define b


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1 DEFINITE INTEGRALS JOHN D. MCCARTHY Astrct. These re lecture notes for Sections 5.3 nd Section 5.3 Definition 1. f is integrle on [, ] if f(x)dx exists. Definition 2. If f() is defined, then f(x)dx. = 0. Definition 3. If f is integrle on [, ], then Note tht, together with the definition of definite integrls, definitions (2) nd (3) define f(x)dx whenever nd re contined in some intervl I contined in the domin of f or when = nd is in the domin of f. Theorem 1. The Existence of Definite Integrls If f is continuous on [, ], then f is integrle on [, ]. Proof. The proof of this theorem uses methods not covered in MTH 132. Theorem 2. Order of Integrtion If f is integrle on n intervl I nd, I, then Proof. First, consider the cse where <. Then, y definition, Secondly, consider the cse where =. Then, y definition (2), 0 nd 0 = 0 nd, hence, Thirdly, consider the cse where >. Then, y definition (3), f(x)dx nd, hence, ( f(x)dx) = Thus, in ll cses, Theorem 3. Zero Width Intervl If f() is defined, then f(x) = 0. Proof. By definition (2), 0. Theorem 4. Constnt Multiple If f is integrle on n intervl I nd, I, then k k Dte: Novemer 28,
2 2 J. MCCARTHY Proof. First, consider the cse where <. Let P : x 0 <... < x n e prtition of [, ] nd c i [x i 1, x i ], 1 i n. Let: (1) S. = nd: (2) R. = so tht: kf(c i )( (x)) i f(c i )( (x)) i (3) nd: k lim S (4) It follows from equtions (1) nd (2) tht: lim R. (5) S = kr. (6) Hence, y equtions (3), (4), nd (5): k lim S = lim kr = k lim R = k Secondly, consider the cse where =. Then, y definition (2), k k 0 nd k k k0 = 0. Hence, k k Thirdly, consider the cse where >. Then y the first cse nd definition (3), k k k k Corollry 5. Additive Inverse If f is integrle on n intervl I nd, I, then Proof. By Theorem 4: (7) ( 1) ( 1) Theorem 6. Sum If f nd g re integrle on n intervl I nd, I, then (f(x) + g(x))dx = g(x)dx.
3 DEFINITE INTEGRALS 3 Proof. First consider the cse where <. Let P : x 0 <... < x n e prtition of [, ] nd c i [x i 1, x i ], 1 i n. Let: (8) R. = (f(c i ) + g(c i ))( (x)) i, (9) S. = nd: (10) T. = so tht: f(c i )( (x)) i, g(c i )( (x)) i (11) (12) nd: (f(x) + g(x))dx = lim R, lim S, (13) g(x)dx = lim T. It follows from equtions (8), (9), nd (10) tht: (14) R = S + T. (15) Hence, y the sum rule for limits nd equtions (11), (12), (13), nd (14): (f(x) + g(x))dx = lim R = lim S + T = lim S + = g(x)dx. lim T Secondly, consider the cse where =. Then, y definition (2), (f(x) + g(x))dx = (f(x) + g(x))dx = 0 nd g(x)dx = g(x)dx = = 0. Hence, (f(x) + g(x))dx = g(x)dx. Thirdly, consider the cse where >. Then, y the first cse nd definition (3), (f(x) + g(x))dx = (f(x) + g(x))dx = ( g(x)dx) = g(x)dx. Corollry 7. Difference If f nd g re integrle on n intervl I nd, I, then (f(x) g(x))dx = f(x)dx g(x)dx.
4 4 J. MCCARTHY Proof. It follows from Theorems 4 nd 6 tht: (16) (f(x) g(x))dx = (f(x) + ( 1)g(x))dx = = ( 1)g(x)dx = f(x)dx ( 1) g(x)dx. g(x)dx Theorem 8. Additivity If f is integrle on n intervl I nd,, c I, then Proof. Note tht there re severl cses to consider sed upon whether (i) <, =, or > ; (ii) < c, = c, or > c; nd (iii) < c, = c, or > c. Cse 1: < < c. Let P : x 0 <... < x m e prtition of [, ] nd Q : y 0 <... < y n e prtition of [, c]; c i [x i 1, x i ], 1 i m; nd d j [y j 1, y j ], 1 j n. Then let: (17) S. = nd: (18) T. = so tht: m f(c i )( (x)) i, f(d j )( (y)) j, j=1 (19) nd: lim S (20) lim T. Q 0 Now suppose tht ɛ is positive rel numer. By equtions (19) nd (20), there exist positive rel numers α nd β such tht if P < α nd Q < β, then: (21) S nd: (22) T f(x)dx < ɛ f(x)dx < ɛ. Note tht P Q : x 0 <... < x m = = y 0 <... < y n is prtition of [, c]. Let:
5 DEFINITE INTEGRALS 5 (23) R. = S + T = m f(c i )( (x)) i + f(d j )( (y)) j. Let δ = min{α, β}. Since α nd β re positive rel numers, δ is positive rel numer. Now suppose tht P Q < δ. Note tht P Q = mx{ P, Q }. Hence, P P Q < δ α nd Q P Q < δ β. It follows from equtions (21), (22), nd (23) tht: (24) R (. This implies tht: S f(x)dx) = (S f(x)dx + T (25) lim P Q 0 R = j=1 f(x)dx) + (T f(x)dx < ɛ + ɛ = 2ɛ. f(x)dx) Note tht R is Riemnn sum pproximtion for f(x)dx corresponding to the prtition P Q of [, c]. Thus: (26) lim R. P Q 0 It follows from equtions (25) nd (26) tht We shll now deduce the remining cses from Cse 1 nd the definitions of definite integrls. Cse 2: =. In this cse, y definition (2), 0. Hence, since = : (27) 0 + Henceforth, we ssume tht. Cse 3 = c. In this cse, y definition (2), 0. Moreover, y Theorem 2, f(x)dx, so tht 0. It follows tht Henceforth, we ssume tht c. Cse 4: = c. In this cse, y definition (2), (28) 0. Hence, since = c: Henceforth, we ssume tht c. 0 =
6 6 J. MCCARTHY Cse 5 < c <. By ssumption, f(x)dx, f(x)dx, nd f(x)dx exist. Hence, y Theorem 2, f(x)dx, c f(x)dx, nd f(x)dx exist. It follows from Cse 1 tht: (29) Tht is to sy: (30) nd, hence: c f(x)dx f(x)dx (31) Cse 6 < < c. By ssumption, f(x)dx, f(x)dx, nd f(x)dx exist. Hence, y Theorem 2, f(x)dx, f(x)dx, nd f(x)dx exist. It follows from Cse 1 tht: (32) Tht is to sy: (33) nd, hence: f(x)dx (34) Cse 7 c < <. By ssumption, f(x)dx, f(x)dx, nd f(x)dx exist. Hence, y Theorem 2, c f(x)dx, f(x)dx, nd c f(x)dx exist. It follows from Cse 1 tht: (35) Tht is to sy: c c (36) nd, hence: f(x)dx
7 DEFINITE INTEGRALS 7 (37) Cse 8 c < <. By ssumption, f(x)dx, Hence, y The f(x)dx, nd f(x)dx exist. orem 2, c f(x)dx, f(x)dx, nd f(x)dx exist. It follows from Cse 1 tht: c (38) Tht is to sy: c c (39) nd, hence: f(x)dx f(x)dx (40) Thus, in ll cses: (41) Theorem 9. Constnts If,, c R, then cdx = c( ). Proof. First consider the cse where <. Suppose tht P : x 0 <... < x n is prtition of [, ] nd c i [x i 1, x i ], 1 i n. Then: (42) R. = Thus: (43) c( (x)) i = c ( (x)) i = c( ). lim R = lim c( ) = c( ). Secondly, consider the cse where =. Then, y definition (2), cdx = cdx = 0 nd c( ) = c( ) = c0 = 0. Hence, cdx = c( ). Thirdly, consider the cse where >. Then, y the first cse nd definition (3), cdx = cdx = c( ) = c( ). Theorem 10. Domintion If f nd g re integrle on [, ] nd f(x) g(x) for ll x [, ], then f(x)dx g(x)dx.
8 8 J. MCCARTHY Proof. Let A = f(x)dx nd B = g(x)dx, which exist since f nd g re integrle on [, ]. Let ɛ e positive rel numer. By the definition of definite integrl, there exist positive rel numers α nd β such tht if: (44) P : = x 0 <... < x m = nd: (45) Q : = y 0 <... < y n = re prtitions of [, ] with P < α nd Q < β nd: (46) c i [x i 1, x i ], 1 i m nd: (47) d j [y j 1, y j ], 1 j n, then: m (48) f(c i )( (x)) i A < ɛ nd: (49) f(d j )( (y)) j B < ɛ. j=1 Choose positive integer N such tht 1/N < α nd 1/N < β. Let m = n = N; (50) x i = + (( )i/n), 0 i N; (51) y i = + (( )i/n), 0 i N; (52) c i = x i, 1 i N; nd: (53) d i = y i, 1 i N. Note tht with these choices P = 1/N < α nd Q = 1/N < β. It follows from equtions (48) nd (49) tht: (54) A nd: N f(c i )( (x)) i < ɛ
9 DEFINITE INTEGRALS 9 N (55) g(d i )( (y)) i B < ɛ. Note tht c i = d i, 1 i N. Since f(x) g(x) for ll x [, ], it follows tht: (56) f(c i ) g(d i ), 1 i N. Note tht: (57) ( (x)) i = ( (y)) i = 1/N > 0, 1 i N. It follows from equtions (54), (55), (56), nd (57) tht: N N (58) A < f(c i )( (x)) i + ɛ g(d i )( (y)) i + ɛ < (B + ɛ) + ɛ = B + 2ɛ. This proves tht A < B + 2ɛ for every positive rel numer ɛ. Tht is to sy, A B. Theorem 11. If f(x) 0 on [, ], then f(x)dx 0. Proof. By Theorems 9 nd 10: (59) 0 = 0 ( ) = 0dx Theorem 12. MxMin Inequlity If f hs mximum vlue mxf nd minimum vlue minf on [, ], then: (60) (minf) ( ) Proof. By the definition of minf nd mxf: f(x)dx (mxf) ( ). (61) minf f(x) mxf for ll x [, ]. It follows from Theorems 9 nd 10 tht: (62) (minf) ( ) = (minf)dx f(x)dx (mxf)dx = (mxf) ( ).
10 10 J. MCCARTHY 2. Section 5.4 Theorem 13. The Men Vlue Theorem for Definite Integrls If f is continuous on [, ], then there exists c [, ] such tht: (63) f(c) = 1 Proof. Since f is continuous on [, ], it follows from the Extreme Vlue Theorem tht there exist c 1 [, ] nd c 2 [, ] such tht f(c 1 ) f(x) f(c 2 ) for ll x [, ]. It follows from Theorems 1, 9, nd 10 tht: (64) f(c 1 )( ) = f(c 1 )dx f(x)dx Since <, > 0. It follows from eqution (64) tht: (65) f(c 1 ) 1 f(x)dx f(c 2 ). f(c 2 )dx = f(c 2 )( ). Suppose, on the one hnd, tht c 1 = c 2. Then from eqution (65) we conclude tht: (66) f(c 1 ) = 1 Since c 1 [, ], the desired conclusion follows. Suppose, on the other hnd, tht c 1 c 2. Let: (67) L. = 1 It follows from equtions (65) nd (67) tht L is numer etween the vlues f(c 1 ) nd f(c 2 ) of f t the endpoints of closed intervl I contined in [, ]. Since f is continuous on [, ], f is continuous on I. It follows from the Intermedite Vlue Theorem tht there exists c I such tht: (68) f(c) = L = 1 Since c I [, ], the desired conclusion follows. Hence, in ny cse, the desired conclusion follows. Theorem 14. The Fundmentl Theorem of Clculus Prt 1 If f is continuous on [, ], then the rule: (69) F (x). = x f(t)dt defines continuous function F on [, ] which is differentile on (, ) such tht:
11 DEFINITE INTEGRALS 11 (70) F (x) = d dx x f(t)dt = f(x). Proof. Suppose tht x (, ). Let h e nonzero rel numer such tht x + h (, ). By Theorems 1 nd 8: (71) x+h f(t)dt = x By equtions (69) nd (71), it follows tht: (72) F (x + h) F (x) h f(t)dt + = 1 h x+h x x+h x f(t)dt. f(t)dt. Suppose, on the one hnd, tht h > 0. Since x, x + h [, ], [x, x + h] [, ]. Since f is continuous on [, ], it follows tht f is continuous on [x, x + h]. Thus, y Theorem 13, there exists c [x, x + h] such tht: (73) 1 h x+h x f(t)dt = f(c). Suppose, on the other hnd, tht h < 0. Since x, x + h [, ], [x + h, x] [, ]. Since f is continuous on [, ], it follows tht f is continuous on [x + h, x]. Thus, y Theorem 13, there exists c [x + h, x] such tht: (74) In other words, y Theorem 2: 1 x f(t)dt = f(c). ( h) x+h (75) Thus, in ny cse: 1 h x+h x f(t)dt = f(c). (76) 1 h x+h x f(t)dt = f(c) for some c etween x nd x + h. It follows from equtions (72) nd (76) tht: F (x + h) F (x) (77) = f(c) h for some c etween x nd x + h. Now suppose tht ɛ is positive rel numer. Since f is continuous on [, ] nd x (, ), f is continuous t x. Thus, there exists positive rel numer δ such tht if y x < δ, then f(y) f(x) < ɛ. Now suppose tht h 0 < δ. Since c is etween x nd x + h, it follows tht c x h = h 0 < δ. Hence, y the preceding prgrph, f(c) f(x) < ɛ. It follows from eqution (77) tht:
12 12 J. MCCARTHY F (x + h) F (x) (78) f(x) < ɛ. h Hence: F (x + h) F (x) (79) lim = f(x). h 0 h Tht is to sy, F (x) = f(x). This proves tht F is differentile on (, ) nd F (x) = f(x) for ll x (, ). A similr rgument shows tht the onesided derivtives of F t nd re given y the equtions F +() = f() nd F () = f(). Thus, F hs onesided derivtive t nd nd derivtive t x for ll x (, ). Since differentile functions re continuous, it follows tht F is continuous on [, ]. Theorem 15. The Fundmentl Theorem of Clculus Prt 2 If f is continuous on [, ] nd F is continuous function on [, ] such tht F is n ntiderivtive of f on (, ), then: (80) Proof. Let: F () F (). (81) G(x). = x f(t)dt for ll x [, ]. By Theorem 14, G is continuous on [, ] nd n ntiderivtive of f on (, ). Thus F nd G re ntiderivtives of f on (, ). It follows from Corollry 2 of the Men Vlue Theorem tht there exists constnt C such tht: (82) F (x) = G(x) + C for ll x (, ). Note tht F nd G re continuous on [, ] nd, hence, t nd. Thus: (83) lim F (x) = F (), x + (84) lim G(x) = G(), x + (85) lim F (x) = F (), x nd: (86) lim G(x) = G(). x
13 DEFINITE INTEGRALS 13 It follows from equtions (82), (83), (84), (85), nd (86), the sum rule for limits, nd the constnt rule for limits tht: (87) F () = lim x + F (x) = lim x + G(x) + C nd: = lim G(x) + lim C = G() + C, x + x + (88) F () = lim F (x) = lim G(x) + C x x = lim G(x) + lim C = G() + C, x x It follows from definition (2), equtions (81), (87), nd (88) tht: (89) F () F () = (G() + C) (G() + C) = G() G() = f(t)dt f(t)dt = f(t)dt 0 = f(t)dt. Since f(t)dt = f(x)dx, it follows from eqution (89) tht F () F (). Deprtment of Mthemtics, Michign Stte University, Est Lnsing, MI Emil ddress: URL:
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