程序代写代做代考 C Chapter 2. The Riemann Integral

Chapter 2. The Riemann Integral
The Riemann Integral
2.1 Definition: A partition of the closed interval [a, b] is a set X = {x0, x1, · · · , xn} with
a=x0 0 there exists δ > 0 such that for every partition X of [a, b] with |X| < δ we have |S − I| < ε for every Riemann sum for f on X, that is 􏰈􏰈􏰒n 􏰈􏰈 􏰈i=1 􏰈􏰈 f (ti )∆i x − I 􏰈􏰈 < ε . 􏰈 for every choice of ti ∈ [xi−1, xi] This number I is unique (as we prove below); it is called the (Riemann) integral of f on [a, b], and we write 􏰓b 􏰓b I = f , or I = f(x)dx. aa Proof: Suppose that I and J are two such numbers. Let ε > 0 be arbitrary. Choose δ1 so that for every partition X with |X| < δ1 we have |S − I| < ε for every Riemann 2 sumSonX,andchooseδ2 >0sothatforeverypartitionXwith|X|<δ2 wehave |S −J| < ε for every Riemann sum S on X. Let δ = min{δ1,δ2}. Let X be any partition 2 n of [a, b] with |X| < δ. Choose ti ∈ [xi−1, xi] and let S = 􏰁 f(ti)∆ix. Then we have i=1 |I−J|≤|I−S|+|S−J|<ε +ε =ε.Sinceεwasarbitrary,wemusthaveI=J. 22 1 􏰎1 if x∈Q 2.4 Example: Let f(x) = 0 if x ∈/ Q. Show that f is not integrable on [0,1]. Solution: Suppose, for a contradiction, that f is integrable on [0, 1], and write I = 􏰑 1 f . 0 Let ε = 1. Choose δ so that for every partition X with |X| < δ we have |S−I| < 1 for every 22 n Riemann sum S for f on X. Choose a partition X with |X| < δ. Let S1 = 􏰁 f(ti)∆ix i=1 n where each ti ∈ [xi−1,xi] is chosen with ti ∈ Q, and let S2 = 􏰁f(si)∆ix where each i=1 si ∈[xi−1,xi]ischosenwithsi ∈/Q. Notethatwehave|S1−I|<1 and|S2−I|<1. 22 Since each ti ∈ Q we have f(ti) = 1 and so S1 = 􏰁 f(ti)∆ix = 􏰁 ∆ix = 1 − 0 = 1, and i=1 i=1 n sinceeachsi∈/Qwehavef(si)=0andsoS2=􏰁f(si)∆ix=0.Since|S1−I|<1 we 2 nn i=1 have|1−I|<1 andso1 0. Choose δ = 2ε . Let X be any partition of [a,b] with |X| < δ. Let b−a nn ti ∈ [xi−1,xi] and set S = 􏰁 f(ti)∆ix = 􏰁 ti∆ix. We must show that |S−1(b2−a2)| < ε. Notice that i=1 i=1 nn 2 i=1 i=1 =(x12 −x02)+(x22 −x12)+···+(xn−12 −xn−22)+(xn2 −xn−12) =−x02 +(x12 −x12)+···+(xn−12 −xn−12)+xn2 = xn2 − x02 = b2 − a2 􏰁n 􏰒 􏰒22 (xi + xi−1)∆ix = (xi + xi−1)(xi − xi−1) = xi − xi−1 i=1 and that when ti ∈ [xi−1, xi] we have 􏰈􏰈ti − 1 (xi + xi−1)􏰈􏰈 ≤ 1 (xi − xi−1) = 1 ∆ix, and so 222 􏰈 1 2 2 􏰈 􏰈 􏰁n 􏰈S−2(b −a)􏰈=􏰈􏰈 i=1 1 􏰁n 􏰈 ti∆ix−2 􏰈􏰁n􏰂1 􏰃􏰈 i=1 (xi+xi−1)∆ix􏰈􏰈 i=1 ti−2(xi+xi+1) ∆ix􏰈􏰈 2 =􏰈􏰈 ≤ 􏰁n 􏰈􏰈ti − 1 (xi + xi+1)􏰈􏰈∆ix i=1 nn ≤ 􏰁 1∆ix∆ix ≤ 􏰁 1δ∆ix 22 i=1 i=1 = 1 δ(b − a) = ε . 2 2 Upper and Lower Riemann Sums 2.7 Definition: Let X be a partition for [a, b] and let f : [a, b] → R be bounded. The upper Riemann sum for f on X, denoted by U(f,X), is n U(f,X) = 􏰒Mi ∆ix where Mi = sup􏰆f(t)􏰈􏰈t ∈ [xi−1,xi]􏰇 i=1 and the lower Riemann sum for f on X, denoted by L(f,X) is n L(f,X) = 􏰒mi ∆ix where mi = inf 􏰆f(t)􏰈􏰈t ∈ [xi−1,xi]􏰇. i=1 2.8 Remark: The upper and lower Riemann sums U(f,X) and L(f,X) are not, in general, Riemann sums at all, since we do not always have Mi = f(ti) or mi = f(si) for any ti,si ∈ [xi−1,xi]. If f is increasing, then Mi = f(xi) and mi = f(xi−1), and so in this case U(f,X) and L(f,X) are indeed Riemann sums. Similarly, if f is decreasing then U(f,X) and L(f,X) are Riemann sums. Also, if f is continuous then, by the Extreme Value Theorem, we have Mi = f(ti) and mi = f(si) for some ti,si ∈ [xi−1,xi], and so in this case U(f,X) and L(f,X) are again Riemann sums. 2.9 Note: Let X be a partition of [a, b], and let f : [a, b] → R. be bounded. Then U(f,X) = sup􏰆S􏰈􏰈S is a Riemann sum for f on X􏰇 , and L(f,X) = inf 􏰆S􏰈􏰈S is a Riemann sum for f on X􏰇. In particular, for every Riemann sum S for f on X we have L(f,X) ≤ S ≤ U(f,X) Proof: We show that U(f,X) = sup􏰆S􏰈􏰈S is a Riemann sum for f on X􏰇 (the other state- ment is proved similarly). Let T = 􏰆S􏰈􏰈S is a Riemann sum for f on X􏰇. For S ∈ T , say n S = 􏰁 f(ti)∆ix where ti ∈ [xi−1,xi], we have i=1 nn S = 􏰁 f(ti)∆ix ≤ 􏰁 Mi∆ix = U(f,X). i=1 i=1 Thus U(f,X) is an upper bound for T so we have U(f,X) ≥ supT . It remains to show thatgivenanyε>0wecanfindS∈T withU(f,X)−S<ε. Letε>0bearbitrary.
SinceMi =sup􏰆f(t)􏰈􏰈t∈[xi−1,xi]􏰇,wecanchooseti ∈[xi−1,xi]withMi−f(ti)< ε . b−a Then we have nnnn U(f,X)−S=􏰒Mi∆ix−􏰒f(ti)∆ix=􏰒􏰂Mi−f(ti)􏰃∆ix<􏰒 ε ∆ix=ε b−a i=1 i=1 i=1 i=1 3 2.10 Lemma: Let f : [a,b] → R be bounded with upper and lower bounds M and m. LetX andY bepartitionsof[a,b]suchthatY =X∪{c}forsomec∈/X. Then 0 ≤ L(f, Y ) − L(f, X) ≤ (M − m)|X| , and 0 ≤ U(f, X) − U(f, Y ) ≤ (M − m)|X| . Proof: We shall prove that 0 ≤ L(f,Y)−L(f,X) ≤ (M −m)|X| (the proof that 0 ≤ U(f,X)−U(f,Y)≤(M−m)|X|issimilar). SayX={x0,x1,···,xn}andc∈[xi−1,xi] so Y = {x0,x1,···,xi−1,c,xi,···,xn}. Then L(f,Y)−L(f,X)=ki(c−xi−1)+li(xi −c)−mi(xi −xi−1) where ki = inf 􏰆f(t)􏰈􏰈t ∈ [xi−1,c]􏰇, li = inf 􏰆f(t)􏰈􏰈t ∈ [c,xi]􏰇 , mi = inf 􏰆f(t)􏰈􏰈t ∈ [xi−1,xi]􏰇. Since mi = min{ki,li} we have ki ≥ mi and li ≥ mi, so L(f,Y)−L(f,X)≥mi(c−xi−1)+mi(xi −c)−mi(xi −xi−1)=0. Sinceki ≤M andli ≤M andmi ≥mwehave L(f,Y)−L(f,X)≤M(c−xi−1)+M(xi −c)−m(xi −xi−1) = (M − m)(xi − xi−1) ≤ (M − m)|X| . 2.11 Note: Let X and Y be partitions of [a, b] with X ⊂ Y . Then L(f, X) ≤ L(f, Y ) ≤ U(f, Y ) ≤ U(f, X) . Proof: If Y is obtained by adding one point to X then this follows from the above lemma. In general, Y can be obtained by adding finitely many points to X, one point at a time. 2.12Note: LetXandY beanypartitionsof[a,b].ThenL(f,X)≤U(f,Y). Proof: LetZ=X∪Y. Thenbytheabovenote, L(f, X) ≤ L(f, Z) ≤ U(f, Z) ≤ U(f, Y ) . 2.13 Definition: Let f : [a,b] → R be bounded. The upper integral of f on [a,b], denoted by U(f), is given by U(f) = inf 􏰆U(f,X)􏰈􏰈X is a partition of [a,b]􏰇 and the lower integral of f on [a,b], denoted by L(f), is given by L(f) = sup􏰆L(f,X)􏰈􏰈X is a partition of [a,b]􏰇. 2.14 Note: The upper and lower integrals of f both exist even when f is not integrable. 2.15 Note: We always have L(f) ≤ U(f). Proof: Let ε > 0 be arbitrary. Choose a partition X1 so that L(f) − L(f, X1) < ε and chooseapartitionX2 sothatU(f,X2)−U(f)< ε. Then 2 2 U(f) − L(f) = 􏰂U(f) − U(f, X2)􏰃 + 􏰂U(f, X2) − L(f, X1)􏰃 + 􏰂L(f, X1) − L(f)􏰃 >−ε +0−ε =−ε.
22
Since ε was arbitrary, this implies that U (f ) − L(f ) ≥ 0. 4

2.16 Theorem: (Equivalent Definitions of Integrability) Let f : [a, b] → R be bounded. Then the following are equivalent.
(1) L(f) = U(f).
(2)Forallε>0thereexistsapartitionX suchthatU(f,X)−L(f,X)<ε. (3) f is integrable on [a, b]. Proof: (1) =⇒ (2). Suppose that L(f) = U(f). Let ε > 0. Choose a partition X1 so
that L(f)−L(f,X1) < ε and choose a partition X2 so that U(f,X2)−U(f) < ε. Let 22 X =X1∪X2. ThenL(f,X1)≤L(f,X)≤L(f)soL(f)−L(f,X)≤L(f)−L(f,X1)< ε, andU(f)≤U(f,X)≤U(f,X2)soU(f,X)−U(f)< ε. Thus 2 U(f, X) − L(f, X) = 􏰂U(f, X) − U(f)􏰃 + 􏰂U(f) − L(f)􏰃 + 􏰂L(f) − L(f, X)􏰃 <ε+0+ε=ε. 22 2 (2) =⇒ (1). Suppose that for all ε > 0 there is a partition X such that U(f,X)−L(f,X) < ε. Letε>0. ChooseX sothatU(f,X)−L(f,X)<ε. Then U(f) − L(f) = 􏰂U(f) − U(f, X)􏰃 + 􏰂U(f, X) − L(f, X)􏰃 + 􏰂L(f, X) − L(f)􏰃 <0+ε+0=ε. Since 0 ≤ U(f) − L(f) < ε for every ε > 0, we have U(f) = L(f).
(3)=⇒(2). Supposethatf isintegrableon[a,b]withI=􏰑bf. Letε>0. Chooseδ>0 aε
so that for every partition X with |X| < δ we have |S − I| < 4 for every Riemann sum S on X. Let X be a partition with |X| < δ. Let S1 be a Riemann sum for f on X with |U(f,X)−S1| < ε, and let S2 be a Riemann sum for f on X with |S2 −L(f,X)| < ε. 44 Then |U(f,X)−L(f,X)|≤|U(f,X)−S1|+|S1 −I|+|I−S2|+|S2 −L(f,X)| <ε+ε+ε+ε=ε. 4444 (1) =⇒ (3). Suppose that L(f) = U(f) and let I = L(f) = U(f). Let ε > 0. Choose a partition X0 of [a,b] so that L(f) − L(f,X0) < ε and U(f,X0) − U(f) < ε. Say 22 X0 = {x0,x1,···,xn} and set δ = ε , where M and m are upper and lower 2(n−1)(M −m) bounds for f on [a,b]. Let X be any partition of [a,b] with |X| < δ. Let Y = X0 ∪X. Note that Y is obtained from X by adding at most n − 1 points, and each time we add a point, the size of the new partition is at most |X| < δ. By lemma 1.10, applied n − 1 times, we have 0≤U(f,X)−U(f,Y)≤(n−1)(M−m)|X|<(n−1)(M−m)δ= ε ,and 2 0≤L(f,Y)−L(f,X)≤(n−1)(M−m)|X|<(n−1)(M−m)δ= ε . 2 Now let S be any Riemann sum for f on X. Note that L(f,X0) ≤ L(f,Y) ≤ L(f) = U(f) ≤ U(f, Y ) ≤ U(f, X0) and L(f, X) ≤ S ≤ U(f, X), so we have S−I ≤U(f,X)−I =U(f,X)−U(f)=􏰂U(f,X)−U(f,Y)􏰃+􏰂U(f,Y)−U(f)􏰃 ≤􏰂U(f,X)−U(f,Y)􏰃+􏰂U(f,X0)−U(f)􏰃<ε +ε =ε 22 and I − S = I − L(f, X) = L(f) − L(f, X) = 􏰂L(f) − L(f, Y )􏰃 + 􏰂L(f, Y ) − L(f, X)􏰃 ≤􏰂L(f)−L(f,X0)􏰃+􏰂L(f,Y)−L(f,X)􏰃<ε +ε =ε. 22 5 Evaluating Integrals of Continuous Functions 2.17 Theorem: (Continuous Functions are Integrable) Let f : [a, b] → R be continuous. Then f is integrable on [a, b]. Proof: Let ε > 0. Since f is uniformly continuous on [a,b], we can choose δ > 0 such
thatforallx,y∈[a,b]wehave|x−y|<δ=⇒|f(x)−f(y)|< ε . LetXbeany b−a partition of [a,b] with |X| < δ. By the Extreme Value Theorem we have Mi = f(ti) and mi = f(si) for some ti,si ∈ [xi−1,xi]. Since |ti − si| ≤ |xi − xi−1| ≤ |X| = δ, we have |Mi −mi|=|f(ti)−f(si)|< ε . Thus b−a nnn n U(f,X)−L(f,X)=􏰁Mi∆ix−􏰁mi∆ix=􏰁(Mi−mi)∆ix< ε 􏰁∆ix=ε. i=1 b−a 2.18 Note: Let f be integrable on [a,b]. Let Xn be any sequence of partitions of [a,b] i=1 i=1 i=1 with lim |Xn| = 0. Let Sn be any Riemann sum for f on Xn. Then {Sn} converges with n→∞ 􏰓b lim Sn = f(x)dx. n→∞ a Proof: Write I = 􏰑b f. Given ε > 0, choose δ > 0 so that for every partition X of [a,b]
a
with |X| < δ we have |S − I| < ε for every Riemann sum S for f on X, and then choose N so that n > N =⇒ |Xn| < δ. Then we have n > N =⇒ |Sn − I| < ε. 2.19 Note: Let f be integrable on [a,b]. If we let Xn be the partition of [a,b] into n equal-sized subintervals, and we let Sn be the Riemann sum on Xn using right-endpoints, then by the above note we obtain the formula 􏰓b 􏰒n f(x)dx= lim f(xn,i)∆n,ix,wherexn,i=a+b−aiand∆n,ix=b−a. an→∞ nn i=1 􏰓2 2x dx. Solution: Let f(x) = 2x. Note that f is continuous and hence integrable, so we have 􏰓2nnn n→∞ n n n→∞ n i=1 i=1 i=1 2·41/n 4−1 · 41/n − 1 , by the formula for the sum of a geometric sequence 􏰉􏰊􏰉1􏰊1 x 2.20 Example: Find 0 2x dx = lim 􏰒f(xn,i)∆n,ix = lim 􏰒f 􏰂2i􏰃􏰂2 􏰃 = lim 􏰒22i/n 􏰂2 􏰃 0 n→∞ = lim n→∞ n = lim6·41/n lim 􏰂 􏰃=6lim n =6lim x→04x−1 n→∞ = 6 lim 1 n→∞n 41/n−1 , by l’Hoˆpital’s Rule n→∞41/n−1 x→0 ln4·4x =6=3. ln4 ln2 6 2.21 Lemma: (Summation Formulas) We have nnnn 􏰒 1 = n , 􏰒 i = n(n + 1) , 􏰒 i2 = n(n + 1)(2n + 1) , 􏰒 i3 = n2(n + 1)2 i=1 i=1 2 i=1 6 i=1 4 Proof: These formulas could be proven by induction, but we give a more constructive nnn proof. It is obvious that 􏰁 1 = 1+1+···1 = n. To find 􏰁 i, consider 􏰁 􏰂i2 −(i−1)2􏰃. i=1 i=1 n=1 On the one hand, we have n 􏰁􏰂i2 −(i−1)2􏰃=(12 −02)+(22 −12)+···+((n−1)2 −(n−2)2)+(n2 −(n−1)2) i=1 =−02 +(12 −12)+(22 −22)+···+((n−1)2 −(n−1)2)+n2 = n2 and on the other hand, nnnnn 􏰁 􏰂i2 − (i − 1)2􏰃 = 􏰁 􏰂i2 − (i2 − 2i + 1)􏰃 = 􏰁(2i − 1) = 2 􏰁 i − 􏰁 1 i=1 i=1 i=1 i=1i=1 nn Equatingthesegivesn2 =2􏰁i−􏰁1andso i=1 i=1 nn 2 􏰁 i = n2 + 􏰁 1 = n2 + n = n(n + 1) , i=1 i=1 ∞ as required. Next, to find 􏰁 i2, consider 􏰁 􏰂i3 − (i − 1)3􏰃. On the one hand we have n=1 i=1 n 􏰁􏰂i3 −(i−1)3􏰃=(13 −03)+(23 −13)+(33 −23)+···+(n3 −(n−1)3) i=1 =−03 +(13 −13)+(23 −23)+···+((n−1)3 −(n−1)3)+n3 = n3 and on the other hand, nn 􏰁􏰂i3 −(i−1)3􏰃= 􏰁􏰂i3 −(i3 −3i2 +3i−1)􏰃 i=1 i=1 nnnn = 􏰁 (3i2 − 3i + 1) = 3 􏰁 i2 − 3 􏰁 i + 􏰁 1 . i=1 i=1 i=1 i=1 nnn Equatingthesegivesn3 =3􏰁i2−3􏰁i+􏰁1andso i=1 i=1 i=1 nnn 6􏰁i2 =2n3 +6􏰁i−2􏰁1=2n3 +3n(n+1)−2n=n(n+1)(2n+1) i=1 i=1 i=1 nn as required. Finally, to find 􏰁 i3, consider 􏰁 􏰂i4 − (i − 1)4􏰃. On the one hand we have i=1 i=1 􏰁􏰂i4 −(i−1)4􏰃=n4, n i=1 (as above) and on the other hand we have nnnnnn 􏰁 􏰂i4 − (i − 1)4 􏰃 = 􏰁 (4i3 − 6i2 + 4i − 1) = 4 􏰁 i3 − 6 􏰁 i2 + 4 􏰁 i − 􏰁 1 . i=1 i=1 i=1 i=1 i=1 i=1 7 nnnn Equatingthesegivesn4 =4􏰁i3−6􏰁i2+4􏰁i−􏰁1andso i=1 i=1 i=1 i=1 􏰓3 x+2x3 dx= 􏰒n lim f (xn,i )∆n,i x n→∞ i=1 n lim 􏰒 f 􏰂1 + 2 i􏰃 􏰂 2 􏰃 n→∞ nn i=1 n lim 􏰒􏰉􏰂1+ 2 i􏰃+2􏰂1+ 2 i􏰃3􏰊􏰂2􏰃 n→∞ n n n i=1 n lim 􏰒􏰂1+2 i+2􏰂1+6 i+12 i2+ 8 i3􏰃􏰃􏰂2􏰃 n→∞ n n n2 n3 n i=1 n lim 􏰒􏰂6 + 28 i+ 48 i2 + 32 i3􏰃 n→∞ n n2 n3 n4 i=1 􏰌nnnn􏰍 lim 6 􏰁1+28 􏰁i+48 􏰁i2+32 􏰁i3 1 nnnn 4 􏰁 i3 = n4 + 6 􏰁 i2 − 4 􏰁 i + 􏰁 1 = = = = = = = n→∞n n2 n3 n4 i=1 i=1 i=1 i=1 i=1 i=1 i=1 =n4 +n(n+1)(2n+1)−2n(n+1)+n =n4 +2n3 +n2 =n2(n+1)2, 􏰓3 x + 2x3 dx. Solution: Let f(x) = x + 2x3. Then as required. 2.22 Example: Find 1 i=1 lim 􏰉6 ·n+ 28 · n(n+1) + 48 · n(n+1)(2n+1) + 32 · n2(n+1)2 􏰊 n→∞n n2 2 n3 6 n4 4 6 + 28 + 48·2 + 32 = 44 . 264 8 Basic Properties of Integrals 2.23 Theorem: (Linearity) Let f and g be integrable on [a, b] and let c ∈ R. Then f + g and cf are both integrable on [a, b] and and 􏰓b􏰓b􏰓b (f + g) = f + g aaa 􏰓b 􏰓b cf = c f . aa Proof: The proof is left as an exercise. 2.24 Theorem: (Comparison) Let f and g be integrable on [a,b]. If f(x) ≤ g(x) for all x ∈ [a, b] then Proof: The proof is left as an exercise. 􏰓b 􏰓b f≤ g. aa 2.25 Theorem: (Additivity) Let a < b < c and let f : [a,c] → R be bounded. Then f is integrable on [a, c] if and only if f is integrable both on [a, b] and on [b, c], and in this case 􏰓b􏰓c􏰓c f+f=f. aba Proof: Suppose that f is integrable on [a,c]. Choose a partition X of [a,c] such that U(f,X) − L(f,X) < ε. Say that b ∈ [xi−1,xi] and let Y = {x0,x1,···,xi−1,b} and Z = {b,xi,xi+1,···,xn} so that Y and Z are partitions of [a,b] and of [b,c]. Then we have U(f, Y ) − L(f, Y ) ≤ U(f, X ∪ {b}) − L(f, X ∪ {b}) ≤ U(f, X) − L(f, X) < ε and also U(f,Z)−L(f,Z) ≤ U(f,X ∪{b})−L(f,X ∪{b}) ≤ U(f,X)−L(f,X) < ε and so f is integrable both on [a, b] and on [b, c]. Conversely, suppose that f is integrable both on [a, b] and on [b.c]. Choose a partition Y of [a,b] so that U(f,Y)−L(f,Y) < ε and choose a partition Z of [b,c] such that 2 U(f,Z)−L(f,Z) < ε. Let X = Y ∪Z. Then X is a partition of [a,c] and we have 􏰂2 􏰃􏰂 􏰃 U(f,X)−L(f,X)= U(f,Y)+U(f,Z) − L(f,Y)+L(f,Z) <ε. Now suppose that f is integrable on [a,c] (hence also on [a,b] and on [b,c]) with 􏰓b􏰓c􏰓c I1 = f, I2 = f and I = f. Let ε > 0. Choose δ > 0 so that for all partitions aba
X1, X2 and X of [a,b], [b,c] and [a,c] respectively with |X1| < δ, |X2| < δ and |X| < δ, wehave|S1−I1|<ε,|S2−I2|<ε and|S−I|<ε forallRiemannsumsS1,S2and 333 S for f on X1, X2 and X respectively. Choose partitions X1 and X2 of [a,b] and [b,c] with |X1| < δ and |X2| < δ. Choose Riemann sums S1 and S2 for f on X1 and X2. Let X=X1∪X2 andnotethat|X|<δandthatS=S1+S2 isaRiemannsumforfonX. Then we have 􏰈􏰈I−(I1+I2)􏰈􏰈=􏰈􏰈(I−S)+(S1−I1)+(S2−I2)􏰈􏰈≤􏰈􏰈I−S|+|S1−I1|+|S2−I2|≤ ε+ε+ε =ε. 333 9 􏰓a 􏰓a􏰓b 2.26Definition: Wedefine f=0andfora 0. Choose a partition X of [a,b] such that U(f,X) − L(f,X) < ε. Write Mi(f) = sup 􏰆f(t)􏰈􏰈t ∈ [xi−1, xi]􏰇 and Mi(|f|) = sup 􏰆|f(t)|􏰈􏰈t ∈ [xi−1, xi]􏰇, and similarly for mi(f) and mi(|f|). When 0 ≤ mi(f) ≤ Mi(f) we have Mi(|f|) = Mi(f) and mi(|f|) = mi(f). When mi(f) ≤ 0 ≤ Mi(f) we have Mi(|f|) = max{Mi(f),−mi(f)} and mi(|f|) ≥ 0, and so Mi(|f|)−mi(|f|) ≤ max 􏰆Mi(f), −mi(f)􏰇 ≤ Mi(f)−mi(f). When mi(f) ≤ Mi(f) ≤ 0 we have Mi(|f|) = −mi(f) and mi(|f|) = −Mi(f), and so Mi(|f|)−mi(|f|) = Mi(f)−mi(f). In all three cases we have Mi(|f|) − mi(|f|) ≤ Mi(f) − mi(f) and so U (|f |, X ) − L(|f |, X ) = 􏰒 􏰂Mi (|f |) − mi (|f |)􏰃∆i x ≤ 􏰒 􏰂Mi (f ) − mi (f )􏰃∆i x = U(f, X) − L(f, X) < ε . Thus |f| is integrable on [a,b]. that Again, let ε > 0. Choose a partition X on [a,b] and choose values ti ∈ [xi−1,xi] so 􏰈􏰈􏰒n 􏰓b􏰈􏰈 􏰈􏰈􏰒n 􏰓b 􏰈􏰈
􏰈􏰈 f(ti)∆ix − f􏰈􏰈 < ε and 􏰈􏰈 |f(ti)|∆ix − |f|􏰈􏰈 < ε . 22 􏰈i=1 a􏰈􏰈i=1 a􏰈 􏰈􏰁n 􏰈􏰁n nn i=1 i=1 Note that by the triangle inequality we have 􏰈􏰈 f(ti)∆ix􏰈􏰈 ≤ i=1 􏰈􏰈􏰓b 􏰈􏰈 􏰓b 􏰈􏰈 f􏰈􏰈 − 􏰈a 􏰈 a 􏰈􏰈􏰓b 􏰈􏰈 􏰓b Since 􏰈􏰈 f􏰈􏰈 − |f(ti)|∆ix, and so i=1 􏰏􏰈􏰈􏰓b 􏰈􏰈 􏰈􏰈n 􏰈􏰈􏰐 􏰏􏰈􏰈n 􏰒􏰒􏰒􏰈􏰈 |f| = 􏰈􏰈 f􏰈􏰈 − 􏰈􏰈 f(ti)∆ix􏰈􏰈 + 􏰈a 􏰈 􏰈i=1 􏰈 􏰏n 􏰓b􏰐 + 􏰒􏰈􏰈f(ti)􏰈􏰈∆ix− |f| 􏰈􏰈 n 􏰐 i=1 <ε+0+ε=ε a 􏰈􏰈 f(ti)∆ix􏰈􏰈 − 􏰈i=1 􏰈f(ti)􏰈∆ix |f| ≤ 0, as required. 􏰈 i=1 22 􏰈􏰈􏰓b 􏰈􏰈 􏰓b |f| < ε for every ε > 0, we have 􏰈􏰈 f􏰈􏰈 −
􏰈a􏰈a 􏰈a􏰈a
10

The Fundamental Theorem of Calculus
2.29 Notation: For a function F , defined on an interval containing [a, b], we write 􏰔 􏰕b
F(x) =F(b)−F(a). a
2.30 Theorem: (The Fundamental Theorem of Calculus) (1) Let f be integrable on [a,b]. Define F : [a,b] → R by
􏰓x 􏰓x
F(x) = f = f(t)dt.
aa
Then F is continuous on [a, b]. Moreover, if f is continuous at a point x ∈ [a, b] then F is differentiable at x and
F′(x) = f(x).
(2) Let f be integrable on [a, b]. Let F be differentiable on [a, b] with F ′ = f . Then
􏰓b 􏰔 􏰕b
f = F(x) =F(b)−F(a).
a
a
Proof: (1)LetM beanupperboundfor|f|on[a,b]. Fora≤x,y≤bwehave 􏰈􏰓y 􏰓x 􏰈 􏰈􏰓y 􏰈 􏰈􏰓y 􏰈 􏰈􏰓y 􏰈
􏰈􏰈 F ( y ) − F ( x ) 􏰈􏰈 = 􏰈􏰈 f − f 􏰈􏰈 = 􏰈􏰈 f 􏰈􏰈 ≤ 􏰈􏰈 | f | 􏰈􏰈 ≤ 􏰈􏰈 M 􏰈􏰈 = M | y − x | 􏰈a a􏰈􏰈x􏰈􏰈x 􏰈􏰈x 􏰈
sogivenε>0wecanchooseδ= ε toget M
| y − x | < δ =⇒ 􏰈􏰈 F ( y ) − F ( x ) 􏰈􏰈 ≤ M | y − x | < M δ = ε . Thus F is continuous (indeed uniformly continuous) on [a,b]. Now suppose that f is continuousatthepointx∈[a,b]. Notethatfora≤x,y≤bwithx̸=ywehave 􏰈F(y)−F(x) 􏰈 􏰈􏰈􏰑yf−􏰑xf 􏰈􏰈 􏰈􏰈aa 􏰈 − f ( x ) 􏰈 = 􏰈􏰈 − f ( x ) 􏰈􏰈 􏰈y−x 􏰈􏰈y−x 􏰈 􏰈􏰈 􏰑 y f 􏰑 y f ( x ) 􏰈􏰈 xx =􏰈􏰈 − 􏰈􏰈 􏰈y−x y−x 􏰈 and then for 0 < |y − x| < δ we have 􏰈F(y)−F(x) 􏰈 1 􏰈􏰓 y 1 􏰈􏰈􏰓y 􏰈􏰈 1 ≤ |y−x|􏰈 εdt􏰈= |y−x|ε|y−x|=ε. 􏰈x􏰈 and thus we have F′(x) = f(x) as required. 11 􏰈􏰈 − f (x)􏰈􏰈 ≤ 􏰈 y−x 􏰈 􏰈􏰈 |y−x|􏰈 x 􏰈 􏰈􏰈f (t) − f (x)􏰈􏰈 dt􏰈􏰈 􏰈 1 􏰈􏰓y 􏰈 = 􏰈􏰈 􏰂f(t)−f(x)􏰃dt􏰈􏰈 |y−x|􏰈 x 1 􏰈􏰓y ≤ 􏰈􏰈 |y−x|􏰈 x 􏰈 􏰈 􏰈􏰈f(t)−f(x)􏰈􏰈dt􏰈􏰈 . 􏰈 Givenε>0,sincef iscontinuousatxwecanchooseδ>0sothat | y − x | < δ = ⇒ 􏰈􏰈 f ( y ) − f ( x ) 􏰈􏰈 < ε (2) Let f be integrable on [a, b]. Suppose that F is differentiable on [a, b] with F ′ = f . Let ε > 0 be arbitrary. Choose δ > 0 so that for every partition X of [a,b] with |X| < δ we 􏰈􏰈􏰓b􏰒n 􏰈􏰈 have 􏰈􏰈 f − f(ti)∆ix􏰈􏰈 < ε for every choice of sample points ti ∈ [xi−1,xi]. Choose 􏰈a i=1 􏰈 sample points ti ∈ [xi−1, xi] as in the Mean Value Theorem so that F′(ti) = F(xi) − F(xi−1) , xi − xi−1 􏰈􏰈􏰓b􏰒n 􏰈􏰈 that is f(ti)∆ix = F(xi) − F(xi−1). Then 􏰈􏰈 f − f(ti)∆ix􏰈􏰈 < ε, and nn 􏰒 f (ti )∆i x = 􏰒 􏰂F (xi ) − F (xi−1 􏰃 i=1 i=1 = 􏰂F (x1 ) − F (x)􏰃 + 􏰂F (x2 ) − F (x1 )􏰃 + · · · + 􏰂F (n − 1) − F (xn )􏰃 = −F (x) + 􏰂F (x1 ) − F (x1 )􏰃 + · · · + 􏰂F (xn−1 ) − F (xn−1 )􏰃 + F (xn ) 􏰈a i=1 􏰈 = F (xn ) − F (x) = F (b) − F (a) . 􏰈􏰈􏰓 b 􏰈􏰈 􏰈􏰈􏰓 b 􏰈􏰈 􏰂􏰃 􏰂􏰃 f− F(b)−F(a) 􏰈􏰈<ε. Sinceεwasarbitrary,􏰈􏰈 f− F(b)−F(a) 􏰈􏰈=0. andso􏰈􏰈 􏰈a􏰈􏰈a􏰈 2.31 Definition: A function F such that F ′ = f on an interval is called an antiderivative of f on the interval. 2.32 Note: If G′ = F′ = f on an interval, then (G−F)′ = 0, and so G−F is constant on the interval, that is G = F + c for some constant c. 2.33 Notation: We write 􏰓􏰓 f = F + c , or f (x) dx = F (x) + c when F is an antiderivative of f on an interval, so that the antiderivatives of f on the interval are the functions of the form G = F + c for some constant c. √ 􏰓 3 dx 2.34 Example: Find 􏰓dx−1 d−11 1 + x2 . Solution: Wehave 1+x2 =tan x+c,since dx(tan x)= 1+x2,andsobyPart2 0 of the Fundamental Theorem of Calculus, we have √ √ 􏰓3dx􏰔−1􏰕3 −1√−1π 2=tanx =tan 3−tan0=3. 01+x0 12