introduction to integral calculus
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Integral Calculus Riemann Integral and other basics.TRANSCRIPT
Introduction to Integral Calculus
1 Preliminaries on Real Numbers
Recall the following definitions and properties of real numbers.
Definition 1 (Upper bound and lower bound) Let R be the set of realnumbers and X ⊆ R be any nonempty subset of R.
1. An upper bound for X is a number M with the property that
x ≤ M ∀ x ∈ X.
2. A lower bound for X is a number m with the property that
x ≥ m ∀ x ∈ X.
We say that X is bounded from above if X has some upper bound and Xis bounded from below if X has a lower bound.
Example 1 IfX = {x ∈ R : 0 ≤ x < 1},
then each number 1, 2, 3, . . . is an upper bound for X, whereas −1,−2,−3, . . .are all lower bounds for X. The number 1/2 is neither an upper bound for X(because 1/2 < 3/4 ∈ X) nor a lower bound for X (because 1/2 > 1/4 ∈ X).If X = Z, then X has no upper bound and has no lower bound.
Definition 2 (Least upper bound or Supremum) A numberM is calleda least upper bound of a nonempty set X ⊆ R, denoted by lubX or supX, if
(a) M is an upper bound for X.
(b) No number less than M is an upper bound for X.
Definition 3 (Greatest lower bound or Infimum) A number m is calleda greatest lower bound of a nonempty set X ⊆ R, denoted by glbX or infX,if
(a) m is a lower bound for X.
(b) No number greater than m is a lower bound for X.
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Example 2 If X = {x ∈ R : 0 ≤ x < 1}, then the number 1 is the leastupper bound for X and 0 is greatest lower bound for X. If
X = {x ∈ R : x is rational and x3 < 2},
then X has no greatest lower bound but 3√2 is the least upper bound for X.
Proposition 1 If a set X ⊆ R has a least upper bound, then it is unique. IfX ⊆ R has a greatest lower bound, then it is unique.
Proof: We prove the uniqueness of least upper bound for X; the proof foruniqueness of greatest lower bound for X is similar. Suppose that M1 andM2 are two least upper bounds for X. Then, since M2 is a least upper boundfor X and M1 is an upper bound for X, we get that M2 ≤ M1. Similarly,since M1 is a least upper bound for X and M2 is an upper bound for X, weget that M1 ≤ M2. Combining these two inequalities, we have M1 = M2.
Completeness axiom: Every nonempty subset X of real numbers R whichis bounded above has a least upper bound, that is there is a real number Msuch that M = lubX.
Remark 1 Using the completeness axiom, one can easily show that everynonempty subset X of real numbers which is bounded below has a greatestlower bound. Note that lubX need not be a member of X. However, if Xis nonempty, closed and bounded, then lubX is an element of X. For, ifM = lubX, then given any n > 0, n ∈ N, there exists an element, sayan ∈ X such that
M − 1
n≤ an ≤ M.
Hence, limn→∞
an = M. But X is closed so we get that M ∈ X. Similarly, if
m = glbX, then m ∈ X provided that X is nonempty, closed and bounded.
The following theorem gives a fundamental property of the supremum andinfimum for a nonempty subset X of real numbers:
Theorem 1 Let X be a nonempty subset of real numbers and ϵ > 0 be given.
(a) If X has a supremum M, then for some x ∈ X we have
x > M − ϵ.
(b) If X has an infimum m, then for some x ∈ X we have
x < m+ ϵ.
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Proof: We prove (a). The proof for (b) is similar. Suppose that x ≤ M − ϵfor every x ∈ X. Then M − ϵ is an upper bound for X. But M is the leastupper bound for X so that M < M − ϵ, a contradiction. Therefore, we musthave x > M − ϵ for at least one x ∈ X.
Theorem 2 Let S and T be two nonempty subsets of R such that s ≤ t forevery s ∈ S and t ∈ T. Then S has a supremum and T has an infimum, andthey satisfy the inequality
supS ≤ inf T.
Proof: Since s ≤ t for every s ∈ S and t ∈ T, each t ∈ T is an upper boundfor S. Therefore, S is bounded above and hence by the completeness axiom,S has a supremum which satisfies the inequality supS ≤ t for all t ∈ T.Hence, supS is a lower bound for T, so that again by completeness axiom,T has an infimum, which cannot be less than supS. In other words, we havesupS ≤ inf T.
Theorem 3 Let A and B be two nonempty subsets of R such that A ⊆ B.Then
(a) lubA ≤ lubB.
(b) glbA ≥ glbB.
Proof: First, we prove (a). Let lubB = M. Then, by definition, x ≤ M forall x ∈ B. Since A ⊆ B, we get that
x ≤ M ∀ x ∈ A.
Thus, M is an upper bound for A, and hence lubA ≤ M = lubB. Now, toprove (b), suppose glbB = m. Then, by definition, x ≥ m for all x ∈ B.Since A ⊆ B, we have x ≥ m for every x ∈ A. Thus, m is a lower bound forA, and hence glbA ≥ m = glbB.
Definition 4 (Partition) Let [a, b] be a closed and bounded interval of R.A partition of [a, b] is a finite, ordered set P of points in [a, b] defined by
P = {a = x0 < x1 < x2 < · · · < xn−1 < xn = b}.
Note that a partition P of [a, b] determines n closed subintervals of [a, b],namely
[x0, x1], [x1, x2], . . . , [xn−1, xn] such that [a, b] =n∪
i=1
[xi−1, xi].
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A typical closed subinterval is [xi−1, xi] and is referred to as the ith closedsubinterval of P. The corresponding open interval (xi−1, xi) is called the ith
open subinterval of P.
Definition 5 (Step function) A function s : [a, b] → R is called a stepfunction if there exists a partition
P = {a = x0 < x1 < x2 < · · · < xn−1 < xn = b}
of [a, b] such that s is constant on each open subinterval of P. Thus, for eachi = 1, 2, . . . , n, we have
s(x) = ci if xi−1 < x < xi,
where ci is an arbitrary real number.
Note that at each of the end points xi−1 and xi, smust have some well definedvalue, but it need not be the same as ci. Sometimes the step function s isalso called a piecewise constant function.
2 The Main Idea
The theory of integration is based on the following two principals of area:
I. If f and g are two continuous, nonnegative functions defined on [a, b]such that f(x) ≥ g(x) ∀ x ∈ [a, b], then the area under f is greaterthan the area under g. As we shall see later, it is possible to discard therestrictive assumptions that functions must be continuous, nonnegativeand that b > a.
II. There is an obvious way of defining the area under a step function: Lets : [a, b] → R be a step function and
P = {a = x0 < x1 < x2 < · · · < xn−1 < xn = b}
be a partition of [a, b] such that for each i = 1, 2, . . . , n,
s(x) = ci if xi−1 < x < xi, and s(x) = 0 if x = xi, xi−1
where ci is an arbitrary real number.
Definition 6 (Area under a step function) The area under a stepfunction s : [a, b] → R from a to b, denoted by A(s), is defined by theequation
A(s) =n∑
i=1
ci(xi − xi−1).
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Now, we extend the idea of the area under step functions to more generalfunctions f : [a, b] → R. The method is to approximate the function f frombelow and from above by step functions, that is we choose arbitrary stepfunctions s and t defined on [a, b] such that
s(x) ≤ f(x) ≤ t(x) ∀ x ∈ [a, b]. .......... (2)
If there exists one and only one number I such that
A(s) ≤ I ≤ A(t)
for every pair of step functions s and t satisfying (2), then we define thearea under f to be this number I. However, it is not possible to approximateevery function f satisfying (2). For example, the function f : R → R definedby the equation
f(x) =
{1/x x ̸= 0,0 x = 0
cannot be approximated by step functions satisfying (2) on any interval [a, b]containing the origin. This is due to the fact that f is unbounded in everyneighbourhood of the origin. Thus, unless stated otherwise, we shall assumethat the function f : [a, b] → R is bounded on [a, b], that is for which thereexists a number M > 0 such that
−M ≤ f(x) ≤ M ∀ x ∈ [a, b].
Geometrically, the graph of such a function lies between the graphs of twoconstant step functions s and t having the values −M and +M, respectively.
3 Upper and Lower Darboux sums
Let f : [a, b] → R be a bounded function and
P = {a = x0 < x1 < x2 < · · · < xn−1 < xn = b}
a partition of the interval of [a, b]. Define numbers Mi and mi, 1 ≤ i ≤ n, by
Mi = lub {f(x) : xi−1 ≤ x ≤ xi}; mi = glb {f(x) : xi−1 ≤ x ≤ xi}.
Notice that the numbers Mi and mi exist for each i = 1, 2, . . . , n because f isbounded on [a, b] and hence bounded on each closed subinterval [xi−1, xi] ofP. The upper Darboux sum of f on [a, b] relative to the partition P is thendefined by
U(P, f) =n∑
i=1
Mi(xi − xi−1),
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and the lower Darboux sum of f on [a, b] relative to the partition P is definedby
L(P, f) =n∑
i=1
mi(xi − xi−1).
Note that since mi ≤ Mi ∀ i = 1, 2, . . . , n, we get that
L(P, f) ≤ U(P, f).
To understand what these upper sums and lower sums mean, assume that fis a non-negative function, and define a function h : [a, b] → R by
h(x) =
{Mi if xi−1 ≤ x < xi,Mn if x = b.
Then h is a step function and its integral represents the area under its graph.In fact, the area under the graph of h is precisely U(P, f). Similarly, if wedefine a function g : [a, b] → R by
g(x) =
{mi if xi−1 ≤ x < xi,mn if x = b,
then g is a step function and the area under the graph of g is L(P, f). Now,since g(x) ≤ f(x) ≤ h(x) for every x ∈ [a, b], we have
A(g) ≤ A(f) ≤ A(h)
or equivalentlyL(P, f) ≤ A(f) ≤ U(P, f) ........(3)
Note that there are uncountably many partitions P of the interval [a, b], andequation (3) is true for every such partition. The main idea behind definingthese Darboux sums is to use a sequence of partitions Pn of [a, b] which make
U(Pn, f)− L(Pn, f)
arbitrary small; then to define the area under f as the common limit ofL(Pn, f) and U(Pn, f) as n → ∞. First, let us have some examples of upperand lower Darboux sums of a function f : [a, b] → R.
Example 3 Let f : [−1, 2] → R be defined by the equation f(x) = x2 − 2,and let P = {−1, 0, 1/2, 3/2, 2} be a partition of [−1, 2]. We have
M1 = lub {f(x) : −1 ≤ x < 0} = (−1)2 − 2 = −1,
M2 = lub {f(x) : 0 ≤ x <1
2} = (
1
2)2 − 2 =
−7
4,
M3 = lub {f(x) : 12≤ x <
3
2} = (
3
2)2 − 2 =
1
4,
M4 = lub {f(x) : 32≤ x < 2} = (2)2 − 2 = 2.
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Similarly, we have
m1 = glb {f(x) : −1 ≤ x < 0} = (0)2 − 2 = −2,
m2 = glb {f(x) : 0 ≤ x <1
2} = (0)2 − 2 = −2,
m3 = glb {f(x) : 12≤ x <
3
2} = (
1
2)2 − 2 =
−7
4,
m4 = glb {f(x) : 32≤ x < 2} = (
3
2)2 − 2 =
1
4.
Hence,
U(P, f) =n∑
i=1
Mi(xi − xi−1) = −1 + (−7
8) + (
1
4) + 1 =
−5
8,
and
L(P, f) =n∑
i=1
mi(xi − xi−1) = −2− 1 + (−7
4) + (
1
8) =
−37
8.
Example 4 Let f(x) = x be defined on [0, 1] and let
Pn = {0 <1
n<
2
n< . . . <
n− 1
n< 1}
be the partition of [0, 1]. Since f is strictly monotonic increasing, we have
Mi =i
nand mi =
i− 1
n∀ i = 1, 2, . . . , n.
Thus, we obtain
U(Pn, f) =n∑
i=1
(i
n)(1
n) = (
1
n2)
n∑i=1
(i) =n+ 1
2n
and
L(Pn, f) =n∑
i=1
(i− 1
n)(1
n) = (
1
n2)
n∑i=1
(i− 1) =n− 1
2n.
Note that
limn→∞
∆xi = limn→∞
1
n= 0 ∀ i = 1, 2, . . . , n,
and
limn→∞
U(Pn, f) =1
2= lim
n→∞L(Pn, f).
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Example 5 Let g : [0, b] → R be defined as g(x) = x2, and let
Pn = {0 <b
n<
2b
n< . . . <
(k − 1)b
n<
kb
n< . . . <
n− 1
n<
nb
n= 1}
be the partition of [0, b]. Since g is strictly monotonic increasing, we have
Mk =k2b2
n2and mk =
(k − 1)2b2
n2∀ k = 1, 2, . . . , n.
Also, the length of each subinterval [xk−1, xk] of [0, b] is given by
∆xk = xk − xk−1 =kb
n− (k − 1)b
n=
b
n.
Thus, we obtain
U(Pn, g) =n∑
k=1
(k2b2
n2)(b
n) = (
b3
n3)
n∑k=1
k2 = (b3
n3) · n(n+ 1)(2n+ 1)
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and
L(Pn, g) =n∑
k=1
((k − 1)2b2
n2)(b
n) = (
b3
n3)
n∑k=1
(k−1)2 = (b3
n3) · n(n− 1)(2n− 1)
6.
Note that
limn→∞
∆xk = limn→∞
b
n= 0 ∀ k = 1, 2, . . . , n,
and
limn→∞
U(Pn, g) =b3
3= lim
n→∞L(Pn, g).
4 Properties of Darboux sums
Definition 7 (Refinement) Let P1 and P2 be two partitions of the sameinterval [a, b]. We say that P2 is a refinement of P1 whenever P2 is obtainedfrom P1 by introducing some additional points, that is
P1 ⊂ P2.
Also, a partition P of [a, b] is said to be a common refinement of P1 andP2 if P is a refinement of both P1 and P2.
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Example 6 Let us consider the following two partitions of the interval [0, 2] :
P1 = {0 < 1/2 < 3/2 < 2} and P2 = {0 < 1/2 < 2/3 < 1 < 4/3 < 3/2 < 2}.
Since P1 ⊂ P2, P2 is a refinement of P1. Also,
P = P1 ∪ P2 = {0 < 1/2 < 2/3 < 1 < 4/3 < 3/2 < 2}
is a common refinement of both P1 and P2.
Theorem 4 Let f : [a, b] → R be a bounded function and P, P ∗ be any twopartitions of the interval [a, b]. If P ∗ is a refinement of P, then we have
U(P ∗, f) ≤ U(P, f) and L(P ∗, f) ≥ L(P, f).
Proof: We prove the case in which P ∗ is obtained from P by introducingone additional single point. The general case in which P ∗ is obtained fromP by introducing n additional points then follows by induction. So, let
P = {a = x0 < x1 < . . . < xi−1 < xi < . . . < xn = b}
be a partition of [a, b] and P ∗ be a refinement of P obtained by adding asingle point, say y ∈ (xi−1, xi), that is
P ∗ = {a = x0 < x1 < . . . < xi−1 < y < xi < . . . < xn}.
If, for each j = 1, 2, . . . , n, j ̸= i, Mj = lub {f(x) : xj−1 ≤ x ≤ xj}, and
Mi = lub {f(x) : xi−1 ≤ x ≤ xi},M
′
i = lub {f(x) : xi−1 ≤ x ≤ y},M
′′
i = lub {f(x) : y ≤ x ≤ xi},
then, by Theorem 3, we have M′i ≤ Mi and M
′′i ≤ Mi. Also, since
U(P, f) =n∑
i=1
Mi∆xi, U(P ∗, f) =∑j ̸=i
Mj∆xj +M′
i (y− xi−1)+M′′
i (xi − y),
we get that
U(P, f)− U(P ∗, f) = Mi(xi − xi−1)−M′
i (y − xi−1)−M′′
i (xi − y)
≥ Mi(xi − xi−1)−Mi(y − xi−1)−Mi(xi − y)
= 0
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which implies that U(P, f) ≥ U(P ∗, f). Similarly, if we let
mi = glb {f(x) : xi−1 ≤ x ≤ xi},m
′
i = glb {f(x) : xi−1 ≤ x ≤ y},m
′′
i = glb {f(x) : y ≤ x ≤ xi},
then, by Theorem 3 again, we get m′i ≥ mi and m
′′i ≥ mi. Thus, we obtain
L(P ∗, f)− L(P, f) = m′
i(y − xi−1) +m′′
i (xi − y)−mi(xi − xi−1)
≥ mi(y − xi−1) +mi(xi − y)−mi(xi − xi−1)
= 0.
Hence, L(P, f) ≤ L(P ∗, f).
Corollary 1 Let f : [a, b] → R be a bounded function and P1, P2 be any twopartitions of [a, b]. Then
L(P1, f) ≤ U(P2, f).
Proof: Let P ∗ be a common refinement of P1 and P2. Then, by Theorem 4,we get
L(P1, f) ≤ L(P ∗, f) and U(P ∗, f) ≤ U(P2, f).
Since, by definition, L(P ∗, f) ≤ U(P ∗, f), we have
L(P1, f) ≤ L(P ∗, f) ≤ U(P ∗, f) ≤ U(P2, f).
5 Darboux Integral
Let f : [a, b] → R be a bounded function, defined on a closed and boundedinterval [a, b] of R. Suppose
P = {a = x0 < x1 < x2 < . . . < xn = b}
is a partition of [a, b] which subdivides it into n closed subintervals
[x0, x1], [x1, x2], . . . , [xi−1, xi], . . . , [xn−1, xn].
Let △xi = xi − xi−1 (i = 1, 2, . . . , n) denote the length of the ith closedsubinterval [xi−1, xi] of P, and let Mi and mi be the supremum and infimumof f in [xi−1, xi], respectively. Form the upper Darboux sum
U(P, f) =n∑
i=1
Mi△xi
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and the lower Darboux sum
L(P, f) =n∑
i=1
mi△xi
corresponding to the partition P. If M and m are bounds of f on [a, b], wehave m ≤ mi ≤ Mi ≤ M which implies that
m△xi ≤ mi△xi ≤ Mi△xi ≤ M△xi .......(4)
Putting i = 1, 2, . . . , n in (4) above and adding all the inequalities, we get
m(b− a) ≤ L(P, f) ≤ U(P, f) ≤ M(b− a).
This is true for every partition P of [a, b]. Thus, the two sets of real numbers
S = {L(P, f) : P is a partition of [a, b]}
andT = {U(P, f) : P is a partition of [a, b]}
are bounded subsets of R. Therefore, by completeness axiom, each set S andT has the supremum as well as the infimum. In particular, inf T and supSexists, which are respectively called the upper Darboux integral of f and thelower Darboux integral of f . That is the upper Darboux integral of fon [a, b], denoted by U(f), is defined by
U(f) = inf{U(P, f) : P is a partition of [a, b]};
the lower Darboux integral of f on [a, b], denoted by L(f), is defined by
L(f) = sup{L(P, f) : P is a partition of [a, b]}.
Corollary 2 If f : [a, b] → R is a bounded function, then
L(f) ≤ U(f).
Proof: By Corollary 1, L(P1, f) ≤ U(P2, f) for any two partitions P1 andP2 of [a, b]. The proof now follows by applying Theorem 2.
Definition 8 A bounded function f : [a, b] → R is said to be Darboux inte-grable or simply integrable on [a, b] if
L(f) = U(f).
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In this case, the common value, denoted by the symbol∫ b
af(x)dx or
∫ b
afdx or
simply∫ b
af is called the Darboux integral of f on [a, b], or the integral
of f on [a, b]. Note that geometrically the integral∫ b
af represents the area
under f . Thus, we have
A(f) =
∫ b
a
f =
∫ b
a
f(x)dx = L(f) = U(f).
Remark 2 The existence of integral∫ b
af(x)dx of a function f : [a, b] → R
requires two very important assumptions, namely
(i) the function f is bounded on [a, b], and
(ii) the interval [a, b] is finite.
If any one of the above two assumptions or both are not satisfied, then wecall the integral
∫ b
af(x)dx as the improper integral of f on [a, b]. For this
reason, we sometimes call∫ b
afdx a proper integral of f on [a, b]..
Example 7 Consider a constant function f : [a, b] → R defined as f(x) = k,where k ∈ R. For any partition
P = {a = x0 < x1 < x2 < . . . < xn−1 < xn = b},
we have Mi = k = mi ∀ i = 1, 2, . . . , n. Thus
U(P, f) =n∑
i=1
Mi△xi =n∑
i=1
k · (xi − xi−1) = k ·n∑
i=1
(xi − xi−1) = k · (b− a),
and
L(P, f) =n∑
i=1
mi△xi =n∑
i=1
k · (xi − xi−1) = k ·n∑
i=1
(xi − xi−1) = k · (b− a).
It follows that L(f) = U(f), and hence f is integrable on [a, b] and∫ b
a
f(x)dx = k · (b− a).
Example 8 Let f : [a, b] → R be defined as
f(x) =
{1, if x ∈ Q0, if x ∈ Q′.
12
Here, Q denotes the set of rational numbers and Q′ denotes the set of ir-rational numbers. Since there are infinitely many rational numbers and in-finitely many irrational numbers in any interval [p, q] of R, we get that forany partition
P = {a = x0 < x1 < x2 < . . . < xn−1 < xn = b}
of [a, b], and for each i = 1, 2, . . . , n,
Mi = sup {f(x) : xi−1 ≤ x ≤ xi} = 1,
mi = inf {f(x) : xi−1 ≤ x ≤ xi} = 0.
Thus, we have
U(P, f) =n∑
i=1
Mi△xi =n∑
i=1
1 · (xi − xi−1) = (b− a),
L(P, f) =n∑
i=1
mi△xi =n∑
i=1
0 · (xi − xi−1) = 0.
It follows that U(f) = b− a and L(f) = 0, and hence U(f) ̸= L(f). So f isnot integrable.
Remark 3 In Example 7 and 8 above, both U(P, f) and L(P, f) had a con-stant value which do not depend on the choice of the partition P of the in-terval [a, b]. Therefore, we could easily find the values of the upper and lowerDarboux integrals of f , and hence decide whether the given functions are in-tegrable or not. In general, however, this does not happen as both U(P, f)and L(P, f) do depend on the partition of [a, b], and the number of partitionsof [a, b] that one has to consider to compute U(f) and L(f) is uncount-able. So our next goal shall be to prove certain conditions under which theintegrability of functions is easily settled.
6 Conditions of Integrability
First, we begin with the definition of an important term in the theory ofintegration.
Definition 9 (Norm or mesh) Let
P = {a = x0 < x1 < x2 < . . . < xn−1 < xn = b}
be a partition of a closed and bounded interval [a, b]. We define the norm ormesh of P, denoted by µ(P ), to be the number
µ(P ) = max {x1 − x0, x2 − x1, . . . , xn − xn−1}.
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Thus, the norm µ(P ) of a partition P of [a, b] is merely the length of thelargest subinterval into which the partition P divides [a, b].
Corollary 3 Let f : [a, b] → R be a bounded function such that |f(x)| ≤ kfor every x ∈ [a, b]. If P, P ∗ are two partitions of [a, b] such that P ∗ is arefinement P containing n additional points, then
(i) L(P, f) ≤ L(P ∗, f) ≤ L(P, f) + 2nkµ(P ).
(ii) U(P, f) ≥ U(P ∗, f) ≥ U(P, f)− 2nkµ(P ).
Proof: We prove (i); the proof for (ii) is similar. Since P ∗ is a refinement ofP, Theorem 4 implies that L(P, f) ≤ L(P ∗, f). To prove the next half of theinequality in (i), suppose that
P ∗ = P ∪ P1 ∪ P2 ∪ . . . ∪ Pn−1 ∪ Pn, where P ⊆ P1 ⊆ P2 ⊆ · · · ⊆ Pn,
and each Pi is a partition of [a, b] such that P1 contains one point more thanP, partition P2 contains one point more than P1, and so on. Thus, Pi containsone point more than Pi−1, i = 2, 3, . . . , n. Now, proceeding as in Theorem 4,since P1 is a partition of [a, b] which contains one point, say y ∈ (xi−1, xi),more than P, we have
L(P1, f)− L(P, f) = m′
i(y − xi−1) +m′′
i (xi − y)−mi(xi − xi−1)
= (m′
i −mi)(y − xi−1) + (m′′
i −mi)(xi − y).
Since |f(x)| ≤ k for every x ∈ [a, b], we have −k ≤ mi ≤ m′i ≤ k, which
implies that0 ≤ m
′
i −mi ≤ 2k.
Similarly, we obtain0 ≤ m
′′
i −mi ≤ 2k.
Therefore, we have
L(P1, f)− L(P, f) ≤ 2k(y − xi−1) + 2k(xi − y)
= 2k(xi − xi−1)
≤ 2kµ(P ).
Similarly, since P2 is a partition of [a, b] which contains one more point thanP1, we obtain
L(P2, f)− L(P1, f) ≤ 2kµ(P ).
Repeating the above process, we get that
L(Pi, f)− L(Pi−1, f) ≤ 2kµ(P ) ∀ i = 3, 4, . . . , n.
14
Therefore, adding all the above inequalities for i = 1, 2, . . . , n, we obtain(L(P1, f)− L(P, f)
)+( n∑
i=2
L(Pi, f)− L(Pi−1, f))≤ 2nkµ(P ),
or
L(Pn, f)− L(P, f) ≤ 2nkµ(P ) ⇒ L(Pn, f) < L(P, f) + 2nkµ(P ).
Thus, if P ∗ has n points more than P, then we get
L(P ∗, f) ≤ L(P, f) + 2nkµ(P ).
This completes the proof.
Theorem 5 (Darboux’s Theorem) If f : [a, b] → R is a bounded func-tion, then for every ϵ > 0, there exists a δ > 0 such that for all partitions Pof [a, b] with µ(P ) < δ,
(i) U(P, f) < U(f) + ϵ.
(ii) L(P, f) > L(f)− ϵ.
Proof: We prove (i); the proof for (ii) is similar. Since f is bounded, thereexists k > 0 such that
|f(x)| ≤ k ∀ x ∈ [a, b].
By definition, U(f) is the infimum of the set of upper Darboux sums, there-fore, for every ϵ > 0, there exists a partition
P1 = {a = x0 < x1 < x2 < . . . < xn = b}
of [a, b] such that
U(P1, f) < U(f) +ϵ
2. ...(1)
Let P be any partition of [a, b] and P ∗ = P ∪ P1. Then P ∗ is a commonrefinement of P and P1. Also, P
∗ has at most (n − 1) more points than Pbecause P1 has (n − 1) points other than x0 = a and xn = b. Thus, byCorollary 3, we get
U(P, f) ≥ U(P ∗, f) ≥ U(P, f)− 2(n− 1)kµ(P )
which implies using (1) that
U(P, f)− 2(n− 1)kµ(P ) ≤ U(P ∗, f) ≤ U(P1, f) < U(f) +ϵ
2.
15
Thus, we have
U(P, f) < U(f) + 2(n− 1)kµ(P ) +ϵ
2.
Choose δ = ϵ4k(n−1)
> 0 so that if µ(P ) < δ, then
[2(n− 1)k]µ(P ) < [2(n− 1)k]δ =2(n− 1)kϵ
4k(n− 1)=
ϵ
2.
Hence, for every partition P with µ(P ) < δ, we obtain
U(P, f) < U(f) +ϵ
2+
ϵ
2= U(f) + ϵ.
Now, we prove important theorems on the necessary and sufficient condi-tions of integrability, which are easily applied in establishing whether a givenfunction f : [a, b] → R is integrable or not.
Theorem 6 A bounded function f on [a, b] is integrable if and only if foreach ϵ > 0 there exists a partition P of [a, b] such that
U(P, f)− L(P, f) < ϵ.
Proof: Let ϵ > 0 be a given positive real number. Suppose first that f isintegrable on [a, b] so that∫ b
a
fdx = L(f) = U(f).
Since upper and lower Darboux integrals are infimum and supremum of theset of upper and lower Darboux sums, respectively, there exist partitions P1
and P2 of [a, b] satisfying
U(P1, f) < U(f) +ϵ
2=
∫ b
a
fdx+ϵ
2
and
L(P2, f) > L(f)− ϵ
2=
∫ b
a
fdx− ϵ
2.
Let P = P1 ∪P2 so that P is a common refinement of both P1 and P2. UsingTheorem 4 along with the above two inequalities, we get that
U(P, f) ≤ U(P1, f) <
∫ b
a
fdx+ϵ
2< L(P2, f) +
ϵ
2+
ϵ
2≤ L(P, f) + ϵ
16
which implies that U(P, f)− L(P, f) < ϵ.
Conversely, suppose that for every ϵ > 0, there exists a partition P0 of[a, b] such that
U(P0, f)− L(P0, f) < ϵ. ...(1)
Now, for any partition P of [a, b], we have by definition
L(P, f) ≤ L(f) ≤ U(f) ≤ U(P, f).
In particular, for the partition P0, we have the inequality
L(P0, f) ≤ L(f) ≤ U(f) ≤ U(P0, f).
This implies that
U(f)− L(f) ≤ U(P0, f)− L(P0, f).
By using (1), we get that
U(f)− L(f) < ϵ.
Since ϵ is arbitrary, we conclude that U(f) = L(f), and hence f is integrableon [a, b].
Theorem 7 A necessary and sufficient condition for the integrability of abounded function f on [a, b] is that for each ϵ > 0 there exists a δ > 0 suchthat for every partition P of [a, b] with µ(P ) < δ,
U(P, f)− L(P, f) < ϵ
or equivalentlylim
µ(P )→ 0{U(P, f)− L(P, f)} = 0.
Proof: First, suppose that the given condition holds. Then, given ϵ > 0, wehave for any partition P of [a, b] with µ(P ) < δ,
U(P, f)− L(P, f) < ϵ. ...(1)
This implies that given ϵ > 0, there is not one but infinitely many partitionsP of [a, b] (of course, all such partitions must have their norm µ(P ) < δ) suchthat (1) holds. Applying Theorem 6, we get that f is integrable on [a, b].
Conversely, suppose that f is integrable on [a, b] so that∫ b
a
fdx = L(f) = U(f).
17
Let ϵ > 0 be given. By Darboux’s Theorem, there exists a δ > 0 such thatfor every partition P of [a, b] with µ(P ) < δ, we have
U(P, f) < U(f) +ϵ
2=
∫ b
a
fdx+ϵ
2, ...(2)
and
L(P, f) > L(f)− ϵ
2=
∫ b
a
fdx− ϵ
2. ...(3)
Hence, by using (2) and (3), we obtain U(P, f) − L(P, f) < ϵ for everypartition P of [a, b] with µ(P ) < δ.
Theorem 8 Let < Pn > be a sequence of partitions of [a, b] such that
limn→∞
µ(Pn) = 0.
Then a function f : [a, b] → R is integrable on [a, b] if and only if
limn→∞
{U(Pn, f)− L(Pn, f)} = 0 ⇔ limn→∞
U(Pn, f) = limn→∞
L(Pn, f).
Also, in this case,∫ b
afdx = lim
n→∞U(Pn, f) = lim
n→∞L(Pn, f).
Proof: f is integrable on [a, b] is equivalent to ϵ − δ condition in Theorem7, namely
limµ(P )→ 0
{U(P, f)− L(P, f)} = 0 ⇔ limµ(Pn)→ 0
{U(Pn, f)− L(Pn, f)} = 0.
Since limn→∞
µ(Pn) = 0, this preceding condition is equivalent to
limn→∞
{U(Pn, f)− L(Pn, f)} = 0.
The value of the integral follows using the inequality
L(Pn, f) ≤ L(f) ≤ U(f) ≤ U(Pn, f) ∀ n
by taking the limit as n → ∞ so that
limn→∞
L(Pn, f) = L(f) = U(f) = limn→∞
U(Pn, f).
18
Example 9 Recall from Example 4 the function f : [0, 1] → R defined asf(x) = x and the given sequence of partitions
Pn = {0 <1
n<
2
n< . . . <
n− 1
n< 1}
of the interval [0, 1]. Since
limn→∞
µ(Pn) = limn→∞
1
n= 0,
and
limn→∞
U(Pn, f) =1
2= lim
n→∞L(Pn, f),
Theorem 8 implies that f is integrable on [0, 1] and∫ 1
0
fdx =
∫ 1
0
xdx =1
2.
Example 10 Recall from Example 5 the function g : [0, b] → R defined asg(x) = x2, and the given sequence of partitions
Pn = {0 <b
n<
2b
n< . . . <
(k − 1)b
n<
kb
n< . . . <
n− 1
n<
nb
n= 1}
of [0, b]. Since
limn→∞
µ(Pn) = limn→∞
b
n= 0,
and
limn→∞
U(Pn, g) =b3
3= lim
n→∞L(Pn, g),
Theorem 8 implies that g is integrable on [0, b] and∫ b
0
gdx =
∫ b
0
x2dx =b3
3.
Example 11 Let f : [0, 1] → R be defined as
f(x) =
{−x, if x ∈ Qx, if x ∈ Q′.
Suppose that f is integrable on [0, 1]. Then, by Theorem 7, for each ϵ > 0there exists a δ > 0 such that for every partition P of [a, b] with µ(P ) < δ,
U(P, f)− L(P, f) < ϵ.
19
Now let 0 < ϵ < 1 and choose the partition
P0 = {0 <1
n<
2
n< . . . <
n− 1
n< 1},
where n is a positive integer such that 1n< δ or nδ > 1. Since
△xi =i
n− i− 1
n=
1
n
for all i = 1, 2, . . . , n, we have µ(P0) =1n< δ. Now, by definition,
Mi = sup {f(x)|xi−1 < x < xi} =i
n∀ i = 1, 2, . . . , n,
mi = inf {f(x)|xi−1 < x < xi} =−i
n∀ i = 1, 2, . . . , n.
Hence, we obtain
U(P0, f)− L(P0, f) =n∑
i=1
{ in− (
−i
n)}( 1
n) =
n∑i=1
(2i
n)1
n= (
2
n2)
n∑i=1
i
which implies that
U(P0, f)− L(P0, f) = (2
n2)n(n+ 1)
2= 1 +
1
n> 1 > ϵ,
a contradiction to Theorem 7. Hence f is not integrable on [0, 1].
7 Riemann Integration
We now discuss the concept of Riemann integrability of real-valued functionsdefined on a closed and bounded interval of R, given by Bernhard Riemannin 1850’s. He associated certain sums, formally called the Riemann sums,with each real-valued function f : [a, b] → R, and used the limit process todefine the integral of f.
Definition 10 (Riemann sum) Let I := [a, b] be a closed and boundedinterval in R and P = {a = x0 < x1 < x2 < · · · < xn−1 < xn = b}be a partition of I. Choose a point ti from each subinterval Ii = [xi−1, xi],for i = 1, 2, . . . , n, and let f : [a, b] → R be a bounded function. Then theRiemann sum f corresponding to the partition P is a finite sum of the form
S(P, f) =n∑
i=1
f(ti)(xi − xi−1).
20
Remark 4
(i) A partition P = {a = x0 < x1 < x2 < · · · < xn−1 < xn = b},together with the choice of points ti, called tags, from each subintervalIi = [xi−1, xi], for i = 1, 2, . . . , n, is usually called a tagged partitionof I. Also, note that the Riemann sum S(P, f) is not a function of thenorm µ(P ) of the partition P.
(ii) If f ≥ 0 on [a, b] then the Riemann sum S(P, f) associated with thetagged partition P is the sum of the areas of n rectangles, each ofwhich has width equal to the length (xi − xi−1) of the subinterval Iiand height f(ti). Since the choice of ti’s is arbitrary, there are infinitelymany Riemann sums associated with a given partition P of [a, b] andthe function f. Also, if m, M is the infimum and supremum of f on[a, b], respectively, then
m(b− a) ≤ L(P, f) ≤ S(P, f) ≤ U(P, f) ≤ M(b− a).
Definition 11 (Riemann Integral) A function f : [a, b] → R is said to beRiemann integrable on [a, b] if
limµ(P )→ 0
S(P, f) exists.
That is, f is Riemann integrable on [a, b] if there exists a number L ∈ R withthe following property: “for each ϵ > 0 there exists a δ > 0 such that if P isa tagged partition of [a, b] with its norm µ(P ) < δ, then
| S(P, f)− L |< ϵ.”
The number L is called the Riemann integral of f on [a, b] and is denoted bythe symbol
R∫ b
a
f.
Theorem 9 If f : [a, b] → R is Riemann integrable, then the value of theRiemann integral of f is uniquely determined.
Proof: Assume that L1 and L2 both satisfy the definition, so that
R∫ b
a
f = L1 and R∫ b
a
f = L2.
21
Let ϵ > 0. Then there exists δ1 > 0 such that if P1 is any tagged partition of[a, b] with norm µ(P ) < δ1, then
| S(P1, f)− L1 |<ϵ
2. (1)
Also, there exists δ2 > 0 such that if P2 is any tagged partition of [a, b] withnorm µ(P ) < δ2, then
| S(P2, f)− L2 |<ϵ
2. (2)
Let δ := min {δ1, δ2} > 0 and let P be any tagged partition of [a, b] withnorm µ(P ) < δ. Then µ(P ) < δ1 and µ(P ) < δ2 so that both the inequalities(1) and (2) hold for µ(P ) < δ. Thus, we obtain
| S(P, f)− L1 |<ϵ
2and | S(P, f)− L2 |<
ϵ
2. (3)
By using the triangle inequality, we get
| L1−L2 |=| L1−S(P, f)+S(P, f)−L2 |≤| L1−S(P, f) | + | S(P, f)−L2 |
which, on using (3), gives | L1 − L2 |< ϵ2+ ϵ
2< ϵ. Since ϵ > 0 is arbitrary, it
follows that L1 = L2.
Remark 5 The preceding theorem implies that the limµ(P )→ 0
S(P, f) is inde-
pendent of the choice of the points ti from each subinterval Ii = [xi−1, xi], fori = 1, 2, . . . , n.
Theorem 10 A function f : [a, b] → R is Riemann integrable on [a, b] ifand only if it is [Darboux] integrable on [a, b].
Proof: Suppose first that f is [Darboux] integrable on [a, b] and let ϵ > 0.Then, by Theorem 7, there exists a δ > 0 such that for every partition Pwith norm µ(P ) < δ,
U(P, f)− L(P, f) < ϵ. (1)
Let P = {a = x0 < x1 < x2 < · · · < xn−1 < xn = b} be a partition of [a, b],
mi = inf {f(x) : x ∈ [xi−1, xi]} and Mi = sup {f(x) : x ∈ [xi−1, xi]}.
Then, for any ti ∈ [xi−1, xi], i = 1, 2, . . . , n, we have
mi ≤ f(ti) ≤ Mi
⇒n∑
i=1
mi∆xi ≤n∑
i=1
f(ti)∆xi ≤n∑
i=1
Mi∆xi
22
⇒ L(P, f) ≤ S(P, f) ≤ U(P, f). (2)
Also, for any partition P of [a, b], we have
L(P, f) ≤∫ b
a
f ≤ U(P, f)
or equivalently
−U(P, f) ≤ −∫ b
a
f ≤ −L(P, f). (3)
Adding (2) and (3), we get
−[U(P, f)− L(P, f)] ≤ S(P, f)−∫ b
a
f ≤ U(P, f)− L(P, f)
which implies that for any partition P of [a, b],∣∣∣S(P, f)− ∫ b
a
f∣∣∣ ≤ U(P, f)− L(P, f). (4)
Now, by using (1) together with (4), we get that for any partition P of [a, b]with norm µ(P ) < δ,∣∣∣S(P, f)− ∫ b
a
f∣∣∣ < ϵ ⇔ lim
µ(P )→ 0S(P, f) =
∫ b
a
f.
Thus, limµ(P )→ 0
S(P, f) exists and equals∫ b
af, so f is Riemann integrable.
Conversely, suppose that limµ(P )→ 0
S(P, f) exists and equals L. Then there
exists a δ > 0 such that for every tagged partition P with norm µ(P ) < δ,we have ∣∣∣S(P, f)− L
∣∣∣ < ϵ
4⇒ L− ϵ
4< S(P, f) < L+
ϵ
4. (5)
For each i = 1, 2, . . . , n, allow the tag points ti to range over the entiresubinterval [xi−1, xi], and then choose ti ∈ [xi−1, xi] such that
f(ti) < mi +ϵ
4(b− a).
Such a choice of ti is possible by the definition of mi for each i = 1, 2, . . . , n.Multiplying the preceding inequality by ∆xi and adding, we obtain
S(P, f) < L(P, f) +ϵ
4(b− a)· (b− a) = L(P, f) +
ϵ
4.
23
⇒ S(P, f)− ϵ
4< L(P, f).
Using (5), we get
(L− ϵ
4)− ϵ
4< S(P, f)− ϵ
4< L(P, f) ⇒ L− ϵ
2< L(P, f). (6)
Similarly, for each i = 1, 2, . . . , n, choose ti ∈ [xi−1, xi] such that
f(ti) > Mi −ϵ
4(b− a).
Such a choice of ti is possible by the definition of Mi for each i = 1, 2, . . . , n.Again, multiplying the preceding inequality by ∆xi and adding, we obtain
S(P, f) > U(P, f)− ϵ
4(b− a)· (b− a) = U(P, f)− ϵ
4.
⇒ U(P, f) < S(P, f) +ϵ
4.
Using (5), we get
U(P, f) < S(P, f) +ϵ
4< (L+
ϵ
4) +
ϵ
4⇒ U(P, f) < L+
ϵ
2. (7)
Hence, from (6) and (7), we obtain
L− ϵ
2< L(P, f) ≤ U(P, f) < L+
ϵ
2
⇒ U(P, f)− L(P, f) <(L+
ϵ
2
)−(L− ϵ
2
)= ϵ
for all partitions P of [a, b] with norm µ(P ) < δ. By Theorem 7, it followsthat f is [Darboux] integrable on [a, b].
Remark 6
(i) Riemann’s definition of the integral of f on [a, b] is a little differentfrom the definition given by Darboux. However, Theorem 10 tells usthat the two definitions are equivalent. For this reason, the Darbouxintegral is also called the Riemann integral. So, henceforth we denotethe value of the Riemann integral, if it exists, by the symbol∫ b
a
f.
(ii) The set of all Riemann integrable functions on [a, b] is denoted by thesymbol R[a, b], and therefore f ∈ R[a, b] means that f is Riemannintegrable on [a, b].
24
8 Properties of the Riemann Integral
In this section, we prove theorems that enable us to form certain combinationsof Riemann integrable functions.
Theorem 11 Suppose that f and g are in R[a, b]. Then:
(i) If c ∈ R, the function cf is in R[a, b] and∫ b
a
cf = c
∫ b
a
f.
(ii) The function f ± g is in R[a, b] and∫ b
a
(f ± g) =
∫ b
a
f ±∫ b
a
g.
(iii) If f(x) ≤ g(x) for all x ∈ [a, b], then∫ b
a
f ≤∫ b
a
g.
Proof: We prove (i) by applying the direct limiting process and (ii) and (iii)by the equivalent ϵ − δ definition of the limiting process, just to illustratetwo different arguments to prove the integrability.
(i) Since f ∈ R[a, b], we have that for any partition P of [a, b],
limµ(P )→ 0
S(P, f) exists and equals
∫ b
a
f.
Let P = {a = x0 < x1 < x2 < · · · < xn−1 < xn = b} be any partition of [a, b]with tag points ti ∈ [xi−1, xi], for i = 1, 2, . . . , n, chosen arbitrarily. Then
S(P, cf) =n∑
i=1
(cf)(ti)∆xi =n∑
i=1
c(f(ti))∆xi = cn∑
i=1
f(ti)∆xi = c · S(P, f).
Thus, proceeding to the limits as µ(P ) → 0, we get
limµ(P )→ 0
S(P, cf) = c · limµ(P )→ 0
S(P, f) = c ·∫ b
a
f.
25
Hence cf is in R[a, b] and ∫ b
a
cf = c ·∫ b
a
f.
(ii) We prove f + g is in R[a, b]. The proof for f − g is similar. Let ϵ > 0.Since f and g are in R[a, b], there exist δ1 > 0, δ2 > 0 such that if P1 and P2
are any two partitions of [a, b] with µ(P1) < δ1 and µ(P2) < δ2, then∣∣∣S(P1, f)−∫ b
a
f∣∣∣ < ϵ
2and
∣∣∣S(P2, g)−∫ b
a
g∣∣∣ < ϵ
2.
Let δ = min {δ1, δ2} > 0 so that if P is any partition of [a, b] with µ(P ) < δ,then both ∣∣∣S(P, f)− ∫ b
a
f∣∣∣ < ϵ
2and
∣∣∣S(P, g)− ∫ b
a
g∣∣∣ < ϵ
2(1)
hold. Now, using the fact that S(P, f + g) = S(P, f) + S(P, g) (prove it !!),we have∣∣∣S(P, f + g)−
(∫ b
a
f +
∫ b
a
g)∣∣∣ = ∣∣∣S(P, f)− ∫ b
a
f + S(P, g)−∫ b
a
g∣∣∣.
Thus, by Triangle inequality together with (1), we get that if P is any par-tition of [a, b] with µ(P ) < δ,∣∣∣S(P, f+g)−
(∫ b
a
f+
∫ b
a
g)∣∣∣ ≤ ∣∣∣S(P, f)−∫ b
a
f∣∣∣+∣∣∣S(P, g)−∫ b
a
g∣∣∣ < ϵ
2+ϵ
2= ϵ.
Since ϵ is arbitrary, we conclude that f + g ∈ R[a, b] and that its integral isthe sum of the integrals of f and g.
(iii) Given ϵ > 0, we construct a δ > 0 as in step (ii), so that if P is anypartition of [a, b] with µ(P ) < δ, then (1) implies that∫ b
a
f − ϵ
2< S(P, f) and S(P, g) <
∫ b
a
g +ϵ
2. (2)
We are given that f(x) ≤ g(x) for all x ∈ [a, b], so we obtain (prove it !!)
S(P, f) ≤ S(P, g).
Hence, from (2), we have ∫ b
a
f <
∫ b
a
g + ϵ.
But, since ϵ is arbitrary, we conclude that∫ b
af ≤
∫ b
ag.
26
Theorem 12 Let f and g be two integrable functions on [a, b]. Then
(i) their product f · g is integrable on [a, b].
(ii) the quotient f/g is integrable on [a, b] if |g(x)| ≥ λ > 0 ∀ x ∈ [a, b].
Proof: Let ϵ > 0 be given.(i) Since f and g are bounded on [a, b], there exists k > 0 such that
|f(x)| ≤ k and |g(x)| ≤ k ∀ x ∈ [a, b].
Therefore, |(f · g)(x)| = |f(x) · g(x)| = |f(x)| · |g(x)| ≤ k2 ∀x ∈ [a, b], whichimplies that f · g is bounded on [a, b]. Also, f and g are integrable, therefore,by Theorem 7, there exists a δ > 0 such that for any partition
P = {a = x0 < x1 < x2 < . . . < xn = b}
of [a, b] with norm µ(P ) < δ,
U(P, f)− L(P, f) <ϵ
2k, U(P, g)− L(P, g) <
ϵ
2k. (1)
Now, for each i = 1, 2, . . . , n, define
mi = inf {f(x) : x ∈ [xi−1, xi]}, Mi = sup {f(x) : x ∈ [xi−1, xi]};
m′
i = inf {g(x) : x ∈ [xi−1, xi]}, M′
i = sup {g(x) : x ∈ [xi−1, xi]};
m′′
i = inf {(f ·g)(x) : x ∈ [xi−1, xi]}, M′′
i = sup {(f ·g)(x) : x ∈ [xi−1, xi]}.
Then, we have for all x, y ∈ [xi−1, xi],
(f · g)(x)− (f · g)(y) = g(x)[f(x)− f(y)] + f(y)[g(x)− g(y)]
⇒∣∣∣(f · g)(x)− (f · g)(y)
∣∣∣ ≤ k(Mi −mi) + k(M′
i −m′
i)
⇒ M′′
i −m′′
i ≤ k(Mi −mi) + k(M′
i −m′
i). (2)
Multiplying (2) by ∆xi and adding all such inequalities, we get
U(P, f · g)− L(P, f · g) ≤ k[U(P, f)− L(P, f)] + k[U(P, g)− L(P, g)].
Thus, using (1), we obtain that for every partition P with norm µ(P ) < δ,
U(P, f · g)− L(P, f · g) < ϵ
2+
ϵ
2= ϵ,
which implies that f · g is integrable on [a, b].
27
(ii) Since f and g are bounded on [a, b], there exists k > 0 such that
|f(x)| ≤ k and λ ≤ |g(x)| ≤ k ∀ x ∈ [a, b].
Therefore, |(f/g)(x)| = |f(x)/g(x)| = |f(x)|/|g(x)| ≤ k/λ ∀ x ∈ [a, b], whichimplies that f/g is bounded on [a, b]. Also, f and g are integrable, therefore,by Theorem 7, there exists a δ > 0 such that for any partition
P = {a = x0 < x1 < x2 < . . . < xn = b}
of [a, b] with norm µ(P ) < δ,
U(P, f)− L(P, f) <ϵλ2
2k, U(P, g)− L(P, g) <
ϵλ2
2k. (3)
Now, for each i = 1, 2, . . . , n, define
mi = inf {f(x) : x ∈ [xi−1, xi]}, Mi = sup {f(x) : x ∈ [xi−1, xi]};
m′
i = inf {g(x) : x ∈ [xi−1, xi]}, M′
i = sup {g(x) : x ∈ [xi−1, xi]};
m′′
i = inf {(f/g)(x) : x ∈ [xi−1, xi]}, M′′
i = sup {(f/g)(x) : x ∈ [xi−1, xi]}.
Then, we have for all x, y ∈ [xi−1, xi],
(f/g)(x)− (f/g)(y) =∣∣∣g(y)[f(x)− f(y)]− f(y)[g(x)− g(y)]
g(x)g(y)
∣∣∣⇒
∣∣∣(f/g)(x)− (f/g)(y)∣∣∣ ≤ (k/λ2)|f(x)− f(y)|+ (k/λ2)|g(x)− g(y)|
⇒∣∣∣(f/g)(x)− (f/g)(y)
∣∣∣ ≤ (k/λ2)(Mi −mi) + (k/λ2)(M′
i −m′
i)
⇒ M′′
i −m′′
i ≤ (k/λ2)(Mi −mi) + (k/λ2)(M′
i −m′
i). (4)
Multiplying (4) by ∆xi and adding all such inequalities, we get
U(P, f/g)−L(P, f/g) ≤ (k/λ2)[U(P, f)−L(P, f)]+(k/λ2)[U(P, g)−L(P, g)].
Thus, using (3), we obtain that for every partition P with norm µ(P ) < δ,
U(P, f/g)− L(P, f/g) <ϵ
2+
ϵ
2= ϵ,
which implies that f/g is integrable on [a, b].
Theorem 13 Let f be a function defined on [a, b] and a < c < b.
28
(i) If f integrable on [a, b], then f is integrable on [a, c] and [c, b].
(ii) If f is integrable on [a, c] and [c, b], then f is integrable on [a, b].
(iii) If either (i) or (ii) above holds, then∫ b
a
f =
∫ c
a
f +
∫ b
c
f.
Proof: Let ϵ > 0 be given.(i) Since f is integrable on [a, b], Theorem 6 asserts that there exists a par-tition P of [a, b] such that
U(P, f)− L(P, f) < ϵ. (1)
Let P ∗ = P ∪ {c} so that P ∗ is a refinement of P. Thus, we have
L(P, f) ≤ L(P ∗, f) ≤ U(P ∗, f) ≤ U(P, f),
which, on using (1), implies that
⇒ U(P ∗, f)− L(P ∗, f) ≤ U(P, f)− L(P, f) < ϵ. (2)
Let P ∗ = P1 ∪ P2, where P1, and P2 denote the set of points of P ∗ between[a, c], [c, b], respectively. Then P1 is a partition of [a, c] and P2 is a partitionof [c, b]. Also, by definition, we have
U(P ∗, f) = U(P1, f) + U(P2, f), and L(P ∗, f) = L(P1, f) + L(P2, f).
Therefore, using (2), we get
{U(P1, f)− L(P1, f)}+ {U(P2, f)− L(P2, f)} = U(P ∗, f)− L(P ∗, f) < ϵ.
Since each bracket on the left is non-negative, it follows that
U(P1, f)− L(P1, f) < ϵ/2 and U(P2, f)− L(P2, f) < ϵ/2,
which shows that f is integrable on [a, c] and [c, b].
(ii) Since f is bounded on both [a, c] and [c, b], f is bounded on [a, b]. Further,f is integrable on [a, c] and [c, b] implies, by using Theorem 6, that there existpartitions P1 and P2 of [a, c] and [c, b], respectively, such that
U(P1, f)− L(P1, f) <ϵ
2and U(P2, f)− L(P2, f) <
ϵ
2. (3)
29
Thus, the set P = P1 ∪ P2 is a partition of [a, b] such that
U(P, f)− L(P, f) = {U(P1, f) + U(P2, f)} − {L(P1, f) + L(P2, f)}= {U(P1, f)− L(P1, f)}+ {U(P2, f)− L(P2, f)}<
ϵ
2+
ϵ
2(by (3))
= ϵ.
Hence, f is integrable on [a, b].
(iii) Using the sets P, P1 and P2 as defined in the proof of (ii) above, we have∫ b
a
f ≤ U(P, f) = U(P1, f) + U(P2, f)
< L(P1, f) + L(P2, f) + ϵ (by (3))
≤∫ c
a
f +
∫ b
c
f + ϵ
so that∫ b
af <
∫ c
af +
∫ b
cf + ϵ. Similarly, we have
∫ b
af >
∫ c
a+∫ b
cf − ϵ. Thus,
we obtain ∣∣∣ ∫ b
a
f −(∫ c
a
f +
∫ b
c
f)∣∣∣ < ϵ.
Since ϵ is arbitrary, we conclude that∫ b
af =
∫ c
af +
∫ b
cf.
Corollary 4 If f is integrable on [a, b], then f is integrable on every interval[c, d] ⊆ [a, b].
Proof: By Theorem 13 (i), since f is integrable on [a, b], f is integrable on[a, c] and [c, b], where a < c < b. Applying Theorem 13 (i) again on [c, b],we get that f is integrable on [c, d] and [d, b], where c < d < b. Thus, f isintegrable on every interval [c, d] ⊆ [a, b].
Theorem 14 If f is integrable on [a, b], then |f | is integrable on [a, b] and∣∣∣ ∫ b
a
f∣∣∣ ≤ ∫ b
a
|f |.
Proof: Since f is integrable on [a, b], it is bounded on [a, b]. So, there existsk > 0 such that
|f(x)| ≤ k ∀ x ∈ [a, b],
which implies that the function |f | is bounded on [a, b]. Let ϵ > 0. Since f isintegrable on [a, b], Theorem 6 shows that there exists a partition
P ∗ = {a = x0 < x1 < x2 < . . . < xn = b}
30
of [a, b] such thatU(P ∗, f)− L(P ∗, f) < ϵ. (1)
For each i = 1, 2, . . . , n, let
mi = inf {f(x) : x ∈ [xi−1, xi]}, Mi = sup {f(x) : x ∈ [xi−1, xi]
and
m′
i = inf {|f |(x) : x ∈ [xi−1, xi]}, M′
i = sup {|f |(x) : x ∈ [xi−1, xi].
Now, we have for all x, y ∈ [xi−1, xi], i = 1, 2, . . . , n,∣∣∣|f |(x)− |f |(y)∣∣∣ = ∣∣∣|f(x)| − |f(y)|
∣∣∣ ≤ ∣∣∣f(x)− f(y)∣∣∣ ≤ Mi −mi
which implies that
M′
i −m′
i ≤ Mi −mi ∀ i = 1, 2, . . . , n.
Hence, for any partition P of [a, b], we have
U(P, |f |)− L(P, |f |) ≤ U(P, f)− L(P, f).
In particular, using (1), we have a partition P ∗ of [a, b] such that
U(P ∗, |f |)− L(P ∗, |f |) ≤ U(P ∗, f)− L(P ∗, f) < ϵ.
Theorem 6 then shows that |f | is integrable on [a, b]. Now, since
−|f | ≤ f ≤ |f |,
using Theorem 11 (iii), we get that
−∫ b
a
|f | ≤∫ b
a
f ≤∫ b
a
|f |
which implies that∣∣∣ ∫ b
af∣∣∣ ≤ ∫ b
a|f |.
Remark 7 The converse of Theorem 15 is not true, that is, |f | integrable on[a, b] does not imply that f is integrable on [a, b]. For Example, the functionf : [a, b] → R defined as
f(x) =
{1, if x ∈ Q−1, if x ∈ Q′.
31
is not integrable on [a, b] (see Example 8). However,
|f |(x) = 1 ∀ x ∈ [a, b],
which is a constant function and hence integrable on [a, b] (see Example 7).
As another example, consider the function f : [0, 1] → R defined as
f(x) =
{−x, if x ∈ Qx, if x ∈ Q′.
This function is not integrable on [0, 1] (see Example 11). However,
|f |(x) = x ∀ x ∈ [0, 1],
being the identity function, is integrable on [0, 1] (see Example 9).
Example 12 Let f : [1, 2] → R be defined by f(x) = 3x+ 1. Let
P = {1 = x0 < x1 < x2 < . . . < xn = 2}
be a partition of [1, 2] which divides [1, 2] into n subintervals, each of length1n. Then
xi = 1 +i
nand ∆xi =
1
n∀ i = 1, 2, . . . , n.
Hence, µ(P ) = 1n→ 0 as n → ∞ and
∑ni=1∆xi = 1. Now, choose the tag
point ti = xi for each i = 1, 2, . . . , n. Thus, we obtain
S(P, f) =n∑
i=1
f(ti)∆xi =n∑
i=1
f(xi)∆xi =n∑
i=1
(3xi + 1)∆xi
which on putting the values of xi gives
S(P, f) =n∑
i=1
{3(1+ i
n)+1}∆xi = 4
n∑i=1
∆xi+(3
n2)
n∑i=1
i = 4+(3
n2){n(n+ 1)
2}.
Hence, we get
limµ(P )→ 0
S(P, f) = limn→∞
{4 + (3
n2){n(n+ 1)
2}} = lim
n→∞(11
2+
3
2n) =
11
2.
Since f is bounded, this proves that f ∈ R[a, b] and∫ 2
1
f =11
2.
32
Example 13 Let f : [−1, 1] → R be defined by f(x) = |x|. It is integrableby using Example 9 together with Theorem 14. Alternatively, we prove it byusing the definition of Riemann integral, and also calculate the value of theintegral
∫ 1
−1f. Since 0 ≤ f(x) ≤ 1 for all x ∈ [−1, 1], f is bounded on [−1, 1].
Let
P = {−1 = x0 < x1 < x2 < . . . < xn = 0 = y0 < y1 < y2 < . . . < yn = 1}
be a partition of [−1, 1] which divides [−1, 1] into 2n subintervals, each oflength 1
n. Hence, µ(P ) = 1
n→ 0 as n → ∞ and
xi = −1 +i
n, yi =
i
n∀ i = 1, 2, . . . , n.
Also, we have
∆xi = ∆yi =1
n∀ i = 1, 2, . . . , n, and
n∑i=1
∆xi = 1 =n∑
i=1
∆yi.
Let us choose the tag points ti = xi ∈ [xi−1, xi] and t′i = yi ∈ [yi−1, yi] for
each i = 1, 2, . . . , n. Thus, we get
S(P, f) =n∑
i=1
f(ti)∆xi +n∑
i=1
f(t′
i)∆yi =n∑
i=1
(−xi)∆xi +n∑
i=1
yi∆yi
which on putting the values of xi and yi gives
S(P, f) =n∑
i=1
(1− i
n
)∆xi+
n∑i=1
( i
n
)∆yi =
n∑i=1
∆xi−( 1n
) n∑i=1
i∆xi+( 1
n2
) n∑i=1
i
or
S(P, f) = 1−( 1n
)2n∑
i=1
i+( 1n
)2n∑
i=1
i = 1.
Hence, limµ(P )→ 0
S(P, f) = limn→∞
S(P, f) exists and equals 1. This proves that
f ∈ R[−1, 1] and ∫ 1
−1
f = 1.
Remark 8 In the preceding Example 12 and Example 13, we have used thedefinition of Riemann integral to prove the existence of the integrals. Also,it gave us the value of the integral as the limit of a certain Riemann sum.However, this method works only for such simple examples and does not
33
carry us too far in proving the existence of the integrals and their evaluation.So, now we prove important theorems about the existence of integrals forcertain classes of functions, for example, class of continuous functions, classof monotonic functions, etc. which will immediately imply the existence of theintegrals in these examples as special cases. Also, we establish a fundamentaltheorem of integral calculus which will help us to compute the value of theseintegrals and many more integrals quite easily.
Theorem 15 Every continuous function f on [a, b] is integrable.
Proof: Let ϵ > 0. Since f is continuous on a closed and bounded interval[a, b], it is uniformly continuous on [a, b]. Therefore, there exists a δ > 0 suchthat
|x− y| < δ ⇒ |f(x)− f(y)| < ϵ
b− a∀ x, y ∈ [a, b]. (1)
Choose a partition P = {a = x0 < x1 < x2 < . . . , < xn = b} of [a, b] whosenorm µ(P ) < δ. Since the restriction of a continuous function is continuous,we get that f is continuous on each subinterval [xi−1, xi], i = 1, 2, . . . , n.Therefore, f is bounded and assumes its bounds, that is, f attains its supre-mum and infimum on [xi−1, xi] for each i = 1, 2, . . . , n. Thus, there existsi, ti ∈ [xi−1, xi] such that
Mi = sup {f(x) : xi−1 ≤ x ≤ xi} = f(si)
andmi = inf {f(x) : xi−1 ≤ x ≤ xi} = f(ti)
for each i = 1, 2, . . . , n. Hence, we obtain
U(P, f)− L(P, f) =n∑
i=1
(Mi −mi)∆xi =n∑
i=1
|f(si)− f(ti)|∆xi.
Since µ(P ) < δ, we have |si − ti| < δ for each i = 1, 2, . . . , n. So, using (1),we have
U(P, f)− L(P, f) <( ϵ
b− a
) n∑i=1
∆xi =( ϵ
b− a
)(b− a) = ϵ
and Theorem 6 shows that f is integrable on [a, b].
Since the definition of integral given by Darboux and Riemann are equivalent,the preceding theorem gives
34
Corollary 5 Let f : [a, b] → R be a continuous function. Then to everyϵ > 0, there exists a δ > 0 such that∣∣∣S(P, f)− ∫ b
a
f∣∣∣ < ϵ
for every tagged partition P of [a, b] with norm µ(P ) < δ.
Theorem 16 If f is bounded and monotonic on [a, b], then f is integrableon [a, b].
Proof: Assume that f is monotonically increasing on [a, b]. The proof whenf is monotonically decreasing on [a, b] is analogous. Let ϵ > 0, and
P = {a = x0 < x1 < x2 < . . . < xn = b}
be a partition of [a, b] such that ∆xi =b−an
for each i = 1, 2, . . . , n, where
n >(b− a)(f(b)− f(a))
ϵ. (1)
Since f is monotonically increasing on [a, b], it is monotonically increasingon each subinterval [xi−1, xi], i = 1, 2, . . . , n. Therefore, we have
f(xi−1) ≤ f(x) ≤ f(xi) ∀ x ∈ [xi−1, xi].
Hence, we get that for each i = 1, 2, . . . , n,
mi = inf {f(x) : x ∈ [xi−1, xi]} = f(xi−1)
andMi = sup {f(x) : x ∈ [xi−1, xi]} = f(xi).
Thus, we obtain
U(P, f)− L(P, f) =n∑
i=1
f(xi)∆xi −n∑
i=1
f(xi−1)∆xi
or
U(P, f)− L(P, f) = ∆xi ·n∑
i=1
[f(xi)− f(xi−1)] =(b− a
n
)(f(b)− f(a))
which, on using (1), gives that
U(P, f)− L(P, f) <ϵ · (f(b)− f(a))
f(b)− f(a)= ϵ.
Theorem 6 then implies that f is integrable on [a, b].
35
Theorem 17 If f is bounded and has finite number of points of discontinuityon [a, b], then f is integrable on [a, b].
Proof: Suppose f has p ̸= 0 points of discontinuity on [a, b]. Since f isbounded, there exist real numbers m and M such that
m ≤ f(x) ≤ M ∀ x ∈ [a, b].
Note that m ̸= M as otherwise f is a constant function on [a, b] having nopoints of discontinuity, contrary to our assumption. Let ϵ > 0. To show thatf is integrable, we use Theorem 6, so it suffices to find a partition P ∗ of [a, b]such that
U(P ∗, f)− L(P ∗, f) < ϵ.
Now, consider a partition
P = {a = x0 < x1 < x2 < . . . < xp < xp+1 < . . . < x2p < x2p+1 = b}
of [a, b] such that the ith point of discontinuity of f lies in the open subinterval(x2i−1, x2i) for each i = 1, 2, . . . , p, and the points of the partition are sochosen that
p∑i=1
(x2i − x2i−1) <ϵ
2(M −m). (1)
IfM2i,m2i denote, respectively, the supremum and infimum of f on the subin-terval [x2i−1, x2i], i = 1, 2, . . . , p, then the total contribution to the difference{U(P, f)− L(P, f)} from the p intervals {[x2i−1, x2i]}pi=1 is C, say, where
C =
p∑i=1
(M2i −m2i)(x2i − x2i−1) ≤ (M −m)
p∑i=1
(x2i − x2i−1) <ϵ
2, (2)
by using (1). On the remaining (p+ 1) subintervals, namely [x2i, x2i+1], i =0, 1, 2, . . . , p, f is continuous and hence integrable, by Theorem 15. On eachof these subintervals, f is integrable so by Theorem 6, there exists a partition,say P2i+1 corresponding to each subinterval [x2i, x2i+1], i = 0, 1, 2, . . . , p, suchthat
U(P2i+1, f)− L(P2i+1, f) <ϵ
2(p+ 1)∀ i = 0, 1, 2, . . . , p. (3)
LetP ∗ = P ∪ P1 ∪ P3 ∪ · · · ∪ P2p+1.
Note that all the open subintervals of [a, b] corresponding to the partition P ∗
are mutually disjoint. Thus, combining the contributions from each subin-terval of [a, b] corresponding to the partition P ∗, we obtain
U(P ∗, f)− L(P ∗, f) = C +
p∑i=0
{U(P2i+1, f)− L(P2i+1, f)},
36
which on using (2) and (3), gives that
U(P ∗, f)− L(P ∗, f) <ϵ
2+ (p+ 1)
( ϵ
2(p+ 1)
)= ϵ.
Example 14 Let f be integrable on [a, b], and suppose that g is a functionon [a, b] such that f(x) = g(x) except for finitely many points x ∈ [a, b]. Weclaim that
(a) g is integrable on [a, b].
(b)∫ b
af =
∫ b
ag.
To prove (a), define
F (x) = (f − g)(x) ∀x ∈ [a, b].
Then F (x) = 0 except for finitely many points, say c1, c2, . . . , cm ∈ [a, b].Thus, F is discontinuous at only finite number of points c1, c2, . . . , cm, so thatusing Theorem 17, we get that F is integrable on [a, b]. But then Theorem11 (ii) implies that g = f − F is integrable on [a, b].
For (b), since F is integrable on [a, b],
limµ(P )→ 0
S(P, F ) =
∫ b
a
F
for every tagged partition P of [a, b]. Let P = {a = x0 < x1 < . . . < xn = b}be any partition of [a, b] and choose the tags ti ∈ [xi−1, xi], i = 1, 2, . . . , nsuch that ti ̸= cj for any j = 1, 2, . . . ,m. Then, we have F (ti) = 0 for eachi = 1, 2, . . . , n, so that
limµ(P )→ 0
S(P, F ) = limµ(P )→ 0
( n∑i=1
F (ti)∆xi
)= 0.
Thus,∫ b
aF = 0 which implies that
∫ b
af =
∫ b
ag, by Theorem 11 (ii).
Theorem 18 (Intermediate Value Theorem for Integrals) If f is a con-tinuous function on [a, b], then for at least one x in [a, b], we have
f(x) =1
b− a
∫ b
a
f.
37
Proof: Since f is continuous on a closed and bounded interval [a, b], f isbounded and assumes its maximum value, say M and its minimum value,say m on [a, b]. Thus, there exist x0, x1 ∈ [a, b] such that f(x0) = m and
f(x1) = M. By using the definition of∫ b
af, we get that
f(x0) · (b− a) = m · (b− a) ≤∫ b
a
f ≤ M · (b− a) = f(x1) · (b− a),
or equivalently,
f(x0) ≤1
b− a
∫ b
a
f ≤ f(x1).
By intermediate value theorem, f assumes every value between f(x0) andf(x1). Hence, there exists x in [a, b] such that
f(x) =1
b− a
∫ b
a
f.
Example 15 Suppose f and g are continuous functions on [a, b] such that∫ b
a
f =
∫ b
a
g. (1)
We claim that there exists x0 in [a, b] such that f(x0) = g(x0). Consider thefunction
F (x) = (f − g)(x) ∀ x ∈ [a, b].
Since f and g are continuous on [a, b], the function F is continuous on [a, b].Theorem 15, therefore, implies that F is integrable on [a, b]. Also, usingTheorem 11 (ii) and equation (1) above, we obtain that∫ b
a
F =
∫ b
a
f −∫ b
a
g = 0. (2)
By Intermediate value theorem of Integrals, there exists x0 ∈ [a, b] such that
F (x0) =( 1
b− a
)∫ b
a
F.
Using (2)above, we get F (x0) = 0, which implies that f(x0) = g(x0).
Theorem 19 (Fundamental Theorem of Calculus I) Suppose there arefunctions f, F : [a, b] → R and a finite set E ⊆ [a, b] such that
38
(a) f is integrable on [a, b].
(b) F is continuous on [a, b].
(c) F ′(x) = f(x) for all x ∈ [a, b] ∼ E.
Then we have ∫ b
a
f = F (b)− F (a).
Proof: We prove the case when E = {a, b}. The general case can be obtainedby dividing [a, b] into finite number of subintervals using all the points of Eas some of the points of a partition of [a, b].
Let ϵ > 0. By (a), since f is integrable, Theorem 6 implies that thereexists a δ > 0 such that if
P = {a = x0 < x1 < x2 < . . . , xn = b}
is any partition of [a, b] with norm µ(P ) < δ, then∣∣∣S(P, f)− ∫ b
a
f∣∣∣ < ϵ ⇔
∣∣∣ n∑i=1
f(ti)(xi − xi−1)−∫ b
a
f∣∣∣ < ϵ (1)
for every choice of tag points ti ∈ [xi−1, xi], i = 1, 2, . . . , n. Now, considerthe subintervals [xi−1, xi] of [a, b] for each i = 1, 2, . . . , n. Using (b), we getthat F is continuous on each of these subintervals [xi−1, xi]. Also, (c) impliesthat F is differentiable on each of these subintervals [xi−1, xi]. Hence, F sat-isfies the hypothesis of Lagrange’s Mean value Theorem on each subinterval[xi−1, xi], i = 1, 2, . . . , n, of [a, b]. So, for each i = 1, 2, . . . , n, there exists ati ∈ (xi−1, xi) such that
F ′(ti) · (xi − xi−1) = F (xi)− F (xi−1) ∀ i = 1, 2, . . . , n. (2)
Putting i = 1, 2, . . . , n, in (2) above and adding, we get
n∑i=1
F ′(ti) · (xi − xi−1) =n∑
i=1
(F (xi)− F (xi−1)
)= F (b)− F (a).
Hence, by using (c), we obtain that
n∑i=1
f(ti) · (xi − xi−1) = F (b)− F (a). (3)
39
But the inequality (1) holds for every choice of ti ∈ [xi−1, xi], i = 1, 2, . . . , n.Thus, using (3) in (1), we have∣∣∣F (b)− F (a)−
∫ b
a
f∣∣∣ < ϵ.
Since ϵ > 0 is arbitrary, we have∫ b
a
f = F (b)− F (a).
Remark 9 We call the function F the primitive of the function f. Also,
(1) If F is differentiable on [a, b], the condition (b) holds automatically.
(2) If f is not defined for some c ∈ E, we let f(c) = 0.
(3) Even if F is differentiable on [a, b], condition (c) is not automaticallysatisfied, since there exist functions F such that F ′ is not integrable.For example, the function F : [0, 1] → R defined by
F (x) =
{x2 · cos( 1
x2 ), 0 < x ≤ 10, x = 0.
is continuous and differentiable at every point of [0, 1], but F ′ given by
F ′(x) =
{2x cos( 1
x2 ) + ( 1x) sin( 1
x2 ), 0 < x ≤ 10, x = 0.
is not integrable on [0, 1]. For, the first term 2x cos( 1x2 ) in F ′ is con-
tinuous on [0, 1], so it is integrable on [0, 1]. However, the second term( 1x) sin( 1
x2 ) in F ′ is not bounded (∵ limx→ 0
( 1x) sin( 1
x2 ) = ∞) and hence it
is not integrable on [0, 1]. Thus, F ′ is not integrable on [0, 1] and hencethe Fundamental Theorem of Calculus I does not apply to this F.
Corollary 6 If g is a continuous function on [a, b] that is differentiable on(a, b), and if g ′ is integrable on [a, b], then∫ b
a
g ′ = g(b)− g(a).
Proof: In the preceding theorem, consider the case E = {a, b} and replacef by g ′ and F by g.
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Example 16 If g(x) = xn+1
n+1, then g ′(x) = xn, so that Corollary 7 implies∫ b
a
xndx =bn+1
n+ 1− an+1
n+ 1=
bn+1 − an+1
n+ 1. (1)
In particular, ∫ b
a
x2dx =b3 − a3
3.
In fact, Formula (1) is valid for any powers n for which g(x) = xn+1
n+1is defined
on [a, b]. For example,∫ b
a
√xdx =
2
3(b3/2 − a3/2) ∀ 0 ≤ a < b.
Theorem 20 (Integration by Parts) If u and v are continuous functionson [a, b] that are differentiable on (a, b), and if u ′ and v ′ are integrable on[a, b], then∫ b
a
u(x)v ′(x)dx+
∫ b
a
u ′(x)v(x)dx = u(b)v(b)− u(a)v(a). (1)
Proof: Define g(x) = u(x)v(x) ∀x ∈ [a, b]. Then, by differentiating g, weobtain
g ′(x) = u(x)v ′(x) + u ′(x)v(x) ∀x ∈ [a, b].
Since u is continuous on [a, b], it is integrable on [a, b], by Theorem 15. Sinceit is given that v ′ is integrable on [a, b], so by using Theorem 12 (i), we getthat the product uv ′ is integrable on [a, b]. Similarly, u ′v is integrable on[a, b]. Hence, by Theorem 11 (ii), uv ′ + u ′v is integrable on [a, b], that is, g ′
is integrable on [a, b]. Now, by Fundamental Theorem of Calculus I, we have∫ b
a
g ′(x)dx = g(b)− g(a) = u(b)v(b)− u(a)v(a).
Using the equation in Theorem 11 (ii), we obtain∫ b
a
u(x)v ′(x)dx+
∫ b
a
u ′(x)v(x)dx = u(b)v(b)− u(a)v(a),
so (1) holds.
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Remark 10 By choosing f(x) = u(x) and g(x) = v ′(x) ∀ x ∈ [a, b], theformula (1) can be rewritten in the following form:∫ b
a
f(x) · g(x)dx =∣∣∣f(x) · ∫ g(x)dx
∣∣∣ba−∫ b
a
{f ′(x) ·∫
g(x)dx}dx, (2)
where∫g(x)dx denotes a primitive of g. Equation (2) is often used to inte-
grate the product of two functions, and this method of integrating theproduct of two functions, using equation (2), is called integrationby parts. Note that, in general, the derivative of a function is not necessarilycontinuous. For example, if
f(x) =
{2x sin( 1
x)− cos( 1
x), x ̸= 0
0, x = 0;
and
F (x) =
{x2 sin( 1
x), x ̸= 0
0, x = 0,
then F ′(x) = f(x) ∀x, but f(x) is not continuous at x = 0, becauselimx→ 0
cos( 1x) does not exist. However, the use of Theorem 12 (i) in the proof
of Theorem 21 can be avoided if u ′ and v ′ are continuous, which is normallythe case.
Example 17 To calculate∫ π
0x cosxdx, let f(x) = x and g(x) = cosx for all
x ∈ [0, π]. By the method of integration by parts, using equation (2), we get∫ π
0
x cosxdx =∣∣∣x · sinx
∣∣∣π0−
∫ π
0
1 · sinxdx = −∫ π
0
sinxdx = −2.
Definition 12 If f is integrable on [a, b], then the function F : [a, b] → Rdefined by
F (z) =
∫ z
a
f (1)
is called the indefinite integral of f with base point a.
Lemma 1 The indefinite integral F of f defined by equation (1) in Defini-tion 14 is [uniformly] continuous on [a, b].
Proof: Let w, z ∈ [a, b] and w ≤ z. Using Theorem 13 (i), since f is in-tegrable on [a, b], it is integrable on [a, z] and, therefore, f is integrable on[a, w] and [w, z]. Also, ∫ z
a
f =
∫ w
a
f +
∫ z
w
f.
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Thus, we obtain
F (z) = F (w) +
∫ z
w
f or F (z)− F (w) =
∫ z
w
f. (2)
Now, since f is integrable, it is bounded so there exists a positive number Msuch that
−M ≤ f(x) ≤ M ∀x ∈ [a, b]
⇒∫ z
w
(−M) ≤∫ z
w
f ≤∫ z
w
M
⇒ (−M)(z − w) ≤∫ z
w
f ≤ M(z − w)
⇒∣∣∣ ∫ z
w
f∣∣∣ ≤ M |z − w|,
which, on using (2), gives that
|F (z)− F (w)| ≤ M |z − w|. (3)
Let ϵ > 0 and δ = ϵM. Thus, if |z − w| < δ, then using (3), we get
|F (z)− F (w)| ≤ M |z − w| < M · δ = M · ϵ
M= ϵ,
or |F (z)− F (w)| < ϵ, and hence F is [uniformly] continuous on [a, b].
Remark 11 Equation (3) in the proof of the preceding theorem is calledthe Lipshitz condition. Thus, it follows from the proof of the precedingtheorem that if f is integrable on [a, b], then the indefinite integral F satisfiesthe Lipshitz condition, and hence it is [uniformly] continuous on [a, b].
Theorem 21 (Fundamental Theorem of Calculus II) Let f be an in-tegrable function on [a, b]. If f is continuous at c ∈ [a, b], then the functionF : [a, b] → R defined by
F (z) =
∫ z
a
f
is differentiable at c and F ′(c) = f(c).
Proof: Suppose first that c ∈ [a, b). We claim that at x = c,
limh→ 0+
F (c+ h)− F (c)
h= f(c).
43
Let ϵ > 0. Since f is continuous at c, there exists a δ > 0 such that
c ≤ x < c+ δ ⇒ f(c)− ϵ < f(x) < f(c) + ϵ. (3)
By Corollary 5, since f is integrable on [a, b], it is integrable on [c, c+ h] forevery h satisfying 0 < h < δ. Also, by definition of F, we have
F (c+ h)− F (c) =
∫ c+h
c
f. (4)
On [c, c+ h], the function f satisfies (3) so that∫ c+h
c
(f(c)− ϵ) <
∫ c+h
c
f <
∫ c+h
c
(f(c) + ϵ).
Hence, using (4), we obtain
(f(c)− ϵ) · h < F (c+ h)− F (c) < (f(c) + ϵ) · h
⇒ −ϵ <F (c+ h)− F (c)
h− f(c) < ϵ
⇒∣∣∣F (c+ h)− F (c)
h− f(c)
∣∣∣ < ϵ.
But ϵ is arbitrary, so we conclude that
limh→ 0+
F (c+ h)− F (c)
h= f(c), c ∈ [a, b). (5)
Similarly, for left hand derivative at c ∈ (a, b], we replace (3) by the inequality
c− δ < x ≤ c ⇒ f(c)− ϵ < f(x) < f(c) + ϵ,
and consider the interval [c− h, c], where 0 < h < δ, so that
(f(c)− ϵ) · h <
∫ c
c−h
f = F (c)− F (c− h) < (f(c) + ϵ) · h
⇒∣∣∣F (c+ h)− F (c)
h− f(c)
∣∣∣ < ϵ,
which is equivalent to
limh→ 0−
F (c)− F (c− h)
h= f(c), c ∈ (a, b]. (6)
Thus, using (5) and (6), we get that the function F is differentiable at everyc ∈ [a, b], and F ′(c) = f(c).
As a consequence, we have
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Corollary 7 If f is continuous on [a, b], then its indefinite integral F isdifferentiable on [a, b] and
F ′(x) = f(x) ∀ x ∈ [a, b].
Remark 12 The previous corollary implies that if f is a continuous functionon [a, b], then its indefinite integral is a primitive of f. However, in general,the indefinite integral need not be a primitive. For example, the functionf : [−1, 1] → R defined by
f(x) =
+1 x > 00, x = 0−1 x < 0,
is integrable on [−1, 1] since it has only two points of discontinuity. Also, fhas the indefinite integral
F (x) = |x| − 1
with the base point −1. But F ′(0) does not exist so that F is not a primitiveof f on [−1, 1].
Theorem 22 (Change of Variable or Substitution) Let u be a differ-entiable function on an open interval J such that u ′ is continuous, and let Ibe an open interval such that u(x) ∈ I for all x ∈ J. If f is continuous on I,then f ◦ u is continuous on J and∫ b
a
f ◦ u(x)u ′(x)dx =
∫ u(b)
u(a)
f(u)du ∀ a, b ∈ J. (1)
Proof: First, we show that f ◦u is continuous on J. Note that the composite
f ◦ u : J → I → R
is well defined, since u(x) ∈ I for all x ∈ J, and f : I → R. Let x0 ∈ J beany arbitrary point, and < xn > be a sequence in J which converges to x0.Since u is differentiable and hence continuous on J, we have
limn→∞
u(xn) = u(x0).
This implies that the sequence < u(xn) > converges to u(x0) ∈ I. Since f iscontinuous on I, therefore, f is continuous at u(x0). Thus, f(u(xn)) convergesto f(u(x0)), that is,
limn→∞
f ◦ u(xn) = f ◦ u(x0).
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Hence, f ◦ u is continuous at x0. Since x0 ∈ J is arbitrary, we conclude thatf ◦ u is continuous on J.
Now, f ◦ u is continuous on J and u ′ is continuous on J , so the productf ◦u(x) ·u ′(x) is continuous on J. By Theorem 15, f ◦u(x)·u ′(x) is integrableon J. Therefore, by Corollary 5, it is integrable on every subinterval [a, b]of J. Also, since f is continuous on I, f is integrable on the subinterval[u(a), u(b)] ⊆ I, again by using Corollary 5. Thus, both left and right handside terms in equation (1) hold. Now, fix c ∈ I, and let
F (u) =
∫ u
c
f(t)dt ∀u ∈ I.
By Fundamental Theorem of Calculus II, F ′(u) = f(u) for all u ∈ I. Letg = F ◦ u. By Chain rule, we have
g ′(x) = F ′(u(x)) · u ′(x) = f(u(x)) · u ′(x) ∀x ∈ J.
Hence, by Fundamental Theorem of Calculus I, we get∫ b
a
f(u(x)) · u ′(x)dx =
∫ b
a
g ′(x)dx = g(b)− g(a) = F (u(b))− F (u(a)),
⇒∫ b
a
f(u(x)) · u ′(x)dx =
∫ u(b)
c
f(t)dt−∫ u(a)
c
f(t)dt =
∫ u(b)
u(a)
f(u)du.
Example 18 Suppose that f is a continuous function on [a, b] and that
f(x) ≥ 0 for all x ∈ [a, b]. We claim that if∫ b
af = 0, then f(x) = 0 for all
x ∈ [a, b]. On the contrary, let f(c) > 0 for some c ∈ (a, b). Let ϵ > 0. Sincef is continuous at x = c, there exists a δ > 0 such that
|x− c| < δ ⇒ |f(x)− f(c)| < ϵ.
This implies that
f(c)− ϵ < f(x) < f(c) + ϵ ∀ x ∈ (c− δ, c+ δ). (1)
If ϵ = f(c)2
> 0, then using (1), we obtain that
f(x) >f(c)
2> 0 ∀ x ∈ (c− δ, c+ δ). (2)
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Now, since f is continuous on [a, b], it is integrable on [a, b] and therefore fis integrable on each interval [a, c − δ], [c − δ, c + δ] and [c + δ, b]. Also, wehave ∫ b
a
f =
∫ c−δ
a
f +
∫ c+δ
c−δ
f +
∫ b
c+δ
f ≥∫ c+δ
c−δ
f
because f ≥ 0. Thus, using (2), we get∫ b
a
fdx ≥∫ c+δ
c−δ
fdx >f(c)
2
∫ c+δ
c−δ
dx = δ · f(c) > 0,
a contradiction. Thus, f(x) = 0 for every x ∈ (a, b). Similarly, we obtain acontradiction when f(a) > 0 and f(b) > 0. This proves our claim.
Example 19 Suppose that f is a continuous real valued function on [a, b].
We claim that if∫ b
af · g = 0 for every continuous function g on [a, b], then
f(x) = 0 for all x ∈ [a, b]. Suppose first that f(c) > 0 for some c ∈ (a, b). Letϵ > 0. Since f is continuous at x = c, there exists a δ > 0 such that
|x− c| < δ ⇒ |f(x)− f(c)| < ϵ.
This implies that
f(c)− ϵ < f(x) < f(c) + ϵ ∀ x ∈ (c− δ, c+ δ). (1)
If ϵ = f(c)2
> 0, then using (1), we obtain that
f(x) >f(c)
2> 0 ∀ x ∈ (c− δ, c+ δ). (2)
Now, since f and g are continuous on [a, b], their product f · g is continuouson [a, b]. Thus, f · g integrable on [a, b] and therefore it is integrable on each
interval [a, c − δ], [c − δ, c + δ] and [c + δ, b]. We are given∫ b
af · g = 0 for
every continuous function g on [a, b]. But for f = g, we have∫ b
a
f · g dx =
∫ c−δ
a
f 2 dx+
∫ c+δ
c−δ
f 2 dx+
∫ b
c+δ
f 2 dx ≥∫ c+δ
c−δ
f 2 dx
because f 2 ≥ 0, so that, by using (2),∫ b
a
f · g dx ≥∫ c+δ
c−δ
f 2dx >f 2(c)
4
∫ c+δ
c−δ
dx =f 2(c)
2δ > 0,
a contradiction. Similarly, we obtain a contradiction when f(a) > 0 andf(b) > 0.
47
Now, suppose that f(c) < 0 for some c ∈ (a, b). Let ϵ > 0. Since f iscontinuous at x = c, there exists a δ > 0 such that
|x− c| < δ ⇒ |f(x)− f(c)| < ϵ.
This implies that
f(c)− ϵ < f(x) < f(c) + ϵ ∀ x ∈ (c− δ, c+ δ). (1)
If ϵ = −f(c)2
> 0, then using (1), we obtain that
f 2(x) >f 2(c)
4> 0 ∀ x ∈ (c− δ, c+ δ). (2)
Now, since f and g are continuous on [a, b], their product f · g is continuouson [a, b]. Thus, f · g integrable on [a, b] and therefore it is integrable on each
interval [a, c − δ], [c − δ, c + δ] and [c + δ, b]. We are given∫ b
af · g = 0 for
every continuous function g on [a, b]. But for f = g, we have∫ b
a
f · g dx =
∫ c−δ
a
f 2 dx+
∫ c+δ
c−δ
f 2 dx+
∫ b
c+δ
f 2 dx ≥∫ c+δ
c−δ
f 2 dx
because f 2 ≥ 0, so that by using (2),∫ b
a
f · g dx ≥∫ c+δ
c−δ
f 2dx >1
4· f 2(c)
∫ c+δ
c−δ
dx =f 2(c)
2δ > 0,
a contradiction. Similarly, we obtain a contradiction when f(a) < 0 andf(b) < 0. This proves our claim.
48