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
SinceMi =supf(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) = supL(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 11 x
2.20 Example: Find
0
2x dx = lim f(xn,i)∆n,ix = lim f 2i2 = 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 =2i−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 =3i2−3i+1andso i=1 i=1 i=1
nnn
6i2 =2n3 +6i−21=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 =4i3−6i2+4i−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+21+ 2 i32 n→∞ n n n
i=1 n
lim 1+2 i+21+6 i+12 i2+ 8 i32 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
bbb (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
bcc 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
bcc
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 ab 2.26Definition: Wedefine f=0andfora
f(ti)∆ix − f < ε and |f(ti)|∆ix − |f| < ε . 22
i=1 ai=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 −
aa aa
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 axx 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
bn
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
bn
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 aa
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−13 −1√−1π
2=tanx =tan 3−tan0=3. 01+x0
12