Courses

# Riemann Sums and Riemann Integral - Integral Calculus, CSIR-NET Mathematical Sciences Mathematics Notes | EduRev

## Mathematics : Riemann Sums and Riemann Integral - Integral Calculus, CSIR-NET Mathematical Sciences Mathematics Notes | EduRev

``` Page 1

1. Review
Theorem 1.1 (Least Upper Bound Property). Let E be a nonempty subset ofR. If E
has some upper bound, then E has exactly one least upper bound.
Theorem 1.2. Let a < b be real numbers, and let f : [a;b]! R be a function which is
continuous on [a;b]. Then f is also uniformly continuous on [a;b].
Proposition 1.3. Let X be a subset ofR, let x
0
be a limit point of X, let f : X!R be a
function, and let L be a real number. Then the following two statements are equivalent.
 f is dierentiable at x
0
on X with derivative L.
 For every"> 0, there exists a =(")> 0 such that, ifx2X satisesjxx
0
j<,
then
jf(x) [f(x
0
) +L(xx
0
)]j"jxx
0
j:
Corollary 1.4 (Mean Value Theorem). Let a<b be real numbers, and let f : [a;b]!R
be a continuous function which is dierentiable on (a;b). Then there exists x2 (a;b) such
that
f
0
(x) =
f(b)f(a)
ba
:
Proposition 1.5. Let X be a subset ofR, let x
0
be a limit point of X, and let f : X!R
be a function. If f is dierentiable at x
0
, then f is also continuous at x
0
.
Theorem 1.6 (Properties of Derivatives). LetX be a subset ofR, letx
0
be a limit point
of X, and let f : X!R and g : X!R be functions.
(iv) If f;g are dierentiable at x
0
, then fg is dierentiable at x
0
, and (fg)
0
(x
0
) =
f
0
(x
0
)g(x
0
) +g
0
(x
0
)f(x
0
). (Product Rule)
Page 2

1. Review
Theorem 1.1 (Least Upper Bound Property). Let E be a nonempty subset ofR. If E
has some upper bound, then E has exactly one least upper bound.
Theorem 1.2. Let a < b be real numbers, and let f : [a;b]! R be a function which is
continuous on [a;b]. Then f is also uniformly continuous on [a;b].
Proposition 1.3. Let X be a subset ofR, let x
0
be a limit point of X, let f : X!R be a
function, and let L be a real number. Then the following two statements are equivalent.
 f is dierentiable at x
0
on X with derivative L.
 For every"> 0, there exists a =(")> 0 such that, ifx2X satisesjxx
0
j<,
then
jf(x) [f(x
0
) +L(xx
0
)]j"jxx
0
j:
Corollary 1.4 (Mean Value Theorem). Let a<b be real numbers, and let f : [a;b]!R
be a continuous function which is dierentiable on (a;b). Then there exists x2 (a;b) such
that
f
0
(x) =
f(b)f(a)
ba
:
Proposition 1.5. Let X be a subset ofR, let x
0
be a limit point of X, and let f : X!R
be a function. If f is dierentiable at x
0
, then f is also continuous at x
0
.
Theorem 1.6 (Properties of Derivatives). LetX be a subset ofR, letx
0
be a limit point
of X, and let f : X!R and g : X!R be functions.
(iv) If f;g are dierentiable at x
0
, then fg is dierentiable at x
0
, and (fg)
0
(x
0
) =
f
0
(x
0
)g(x
0
) +g
0
(x
0
)f(x
0
). (Product Rule)
2. Riemann Sums
Within calculus, the two most fundamental concepts are dierentiation and integration.
We have covered dierentiation already, and we now move on to integration. Dening an
integral is fairly delicate. In the case of the derivative, we created one limit, and the existence
of this limit dictated whether or not the function in question was dierentiable. In the case
of the Riemann integral, there is also a limit to discuss, but it is much more complicated
than in the case of dierentiation.
We should mention that there is more than one way to construct an integral, and the
Riemann integral is only one such example. Within this course, we will only be discussing
the Riemann integral. The Riemann integral has some deciencies which are improved upon
by other integration theories. However, those other integration theories are more involved,
so we focus for now only on the Riemann integral.
Our starting point will be partitions of intervals into smaller intervals, which will form the
backbone of the Riemann sum. The Riemann sum will then be used to create the Riemann
integral through a limiting constructing.
Denition 2.1 (Partition). Let a < b be real numbers. A partition P of the interval
[a;b] is a nite subset of real numbers x
0
;:::;x
n
such that
a =x
0
<x
1
<<x
n1
<x
n
=b:
We write P =fx
0
;x
1
;:::;x
n
g.
Remark 2.2. Let P;P
0
be partitions of [a;b]. Then the union P[P
0
of P and P
0
is also a
partition of [a;b].
Denition 2.3 (Upper and Lower Riemann Sums). Let a < b be real numbers, let
f : [a;b]!R be a bounded function, and letP =fx
0
;:::;x
n
g be a partition of [a;b]. For ev-
ery integer 1in, the functionfj
[x
i1
;x
i
]
is also a bounded function. So, sup
x2[x
i1
;x
i
]
f(x)
and inf
x2[x
i1
;x
i
]
f(x) exist by the Least Upper Bound property (Theorem 1.1). We therefore
dene the upper Riemann sum U(f;P ) by
U(f;P ) :=
n
X
i=1

sup
x2[x
i1
;x
i
]
f(x)

(x
i
x
i1
):
We also dene the lower Riemann sum L(f;P ) by
L(f;P ) :=
n
X
i=1

inf
x2[x
i1
;x
i
]
f(x)

(x
i
x
i1
):
Remark 2.4. For each integer 1 i n, we dene a function g : [a;b]! R such that
g(x) := sup
y2[x
i1
;x
i
]
f(y) for all x
i1
 x < x
i
, with g(b) := f(b). Then g is constant on
[x
i1
;x
i
) for all 1 i n, and f(x) g(x) for all x2 [a;b]. The upper Riemann sum
U(f;P ) then represents the area under the function g, which is meant to upper bound the
area under the function f. Similarly, for each integer 1  i n, we dene a function
h: [a;b]!R such that h(x) := inf
y2[x
i1
;x
i
]
f(y) for all x
i1
 x < x
i
, with h(b) := f(b).
Then h is constant on [x
i1
;x
i
) for all 1 i n, and h(x) g(x) for all x2 [a;b]. The
lower Riemann sum L(f;P ) then represents the area under the function g, which is meant
to lower bound the area under the function f.
Page 3

1. Review
Theorem 1.1 (Least Upper Bound Property). Let E be a nonempty subset ofR. If E
has some upper bound, then E has exactly one least upper bound.
Theorem 1.2. Let a < b be real numbers, and let f : [a;b]! R be a function which is
continuous on [a;b]. Then f is also uniformly continuous on [a;b].
Proposition 1.3. Let X be a subset ofR, let x
0
be a limit point of X, let f : X!R be a
function, and let L be a real number. Then the following two statements are equivalent.
 f is dierentiable at x
0
on X with derivative L.
 For every"> 0, there exists a =(")> 0 such that, ifx2X satisesjxx
0
j<,
then
jf(x) [f(x
0
) +L(xx
0
)]j"jxx
0
j:
Corollary 1.4 (Mean Value Theorem). Let a<b be real numbers, and let f : [a;b]!R
be a continuous function which is dierentiable on (a;b). Then there exists x2 (a;b) such
that
f
0
(x) =
f(b)f(a)
ba
:
Proposition 1.5. Let X be a subset ofR, let x
0
be a limit point of X, and let f : X!R
be a function. If f is dierentiable at x
0
, then f is also continuous at x
0
.
Theorem 1.6 (Properties of Derivatives). LetX be a subset ofR, letx
0
be a limit point
of X, and let f : X!R and g : X!R be functions.
(iv) If f;g are dierentiable at x
0
, then fg is dierentiable at x
0
, and (fg)
0
(x
0
) =
f
0
(x
0
)g(x
0
) +g
0
(x
0
)f(x
0
). (Product Rule)
2. Riemann Sums
Within calculus, the two most fundamental concepts are dierentiation and integration.
We have covered dierentiation already, and we now move on to integration. Dening an
integral is fairly delicate. In the case of the derivative, we created one limit, and the existence
of this limit dictated whether or not the function in question was dierentiable. In the case
of the Riemann integral, there is also a limit to discuss, but it is much more complicated
than in the case of dierentiation.
We should mention that there is more than one way to construct an integral, and the
Riemann integral is only one such example. Within this course, we will only be discussing
the Riemann integral. The Riemann integral has some deciencies which are improved upon
by other integration theories. However, those other integration theories are more involved,
so we focus for now only on the Riemann integral.
Our starting point will be partitions of intervals into smaller intervals, which will form the
backbone of the Riemann sum. The Riemann sum will then be used to create the Riemann
integral through a limiting constructing.
Denition 2.1 (Partition). Let a < b be real numbers. A partition P of the interval
[a;b] is a nite subset of real numbers x
0
;:::;x
n
such that
a =x
0
<x
1
<<x
n1
<x
n
=b:
We write P =fx
0
;x
1
;:::;x
n
g.
Remark 2.2. Let P;P
0
be partitions of [a;b]. Then the union P[P
0
of P and P
0
is also a
partition of [a;b].
Denition 2.3 (Upper and Lower Riemann Sums). Let a < b be real numbers, let
f : [a;b]!R be a bounded function, and letP =fx
0
;:::;x
n
g be a partition of [a;b]. For ev-
ery integer 1in, the functionfj
[x
i1
;x
i
]
is also a bounded function. So, sup
x2[x
i1
;x
i
]
f(x)
and inf
x2[x
i1
;x
i
]
f(x) exist by the Least Upper Bound property (Theorem 1.1). We therefore
dene the upper Riemann sum U(f;P ) by
U(f;P ) :=
n
X
i=1

sup
x2[x
i1
;x
i
]
f(x)

(x
i
x
i1
):
We also dene the lower Riemann sum L(f;P ) by
L(f;P ) :=
n
X
i=1

inf
x2[x
i1
;x
i
]
f(x)

(x
i
x
i1
):
Remark 2.4. For each integer 1 i n, we dene a function g : [a;b]! R such that
g(x) := sup
y2[x
i1
;x
i
]
f(y) for all x
i1
 x < x
i
, with g(b) := f(b). Then g is constant on
[x
i1
;x
i
) for all 1 i n, and f(x) g(x) for all x2 [a;b]. The upper Riemann sum
U(f;P ) then represents the area under the function g, which is meant to upper bound the
area under the function f. Similarly, for each integer 1  i n, we dene a function
h: [a;b]!R such that h(x) := inf
y2[x
i1
;x
i
]
f(y) for all x
i1
 x < x
i
, with h(b) := f(b).
Then h is constant on [x
i1
;x
i
) for all 1 i n, and h(x) g(x) for all x2 [a;b]. The
lower Riemann sum L(f;P ) then represents the area under the function g, which is meant
to lower bound the area under the function f.
Denition 2.5 (Upper and Lower Integrals). Leta<b be real numbers, letf : [a;b]!
R be a bounded function. We dene the upper Riemann integral
R
b
a
f of f on [a;b] by
Z
b
a
f := inffU(f;P ): P is a partition of [a;b]g:
We also dene the lower Riemann integral
R
b
a
f of f on [a;b] by
Z
b
a
f := supfL(f;P ): P is a partition of [a;b]g:
Lemma 2.6. Let f : [a;b]!R be a bounded function, so that there exists a real number M
such thatMf(x)M for all x2 [a;b]. Then
M(ba)
Z
b
a
f
Z
b
a
fM(ba)
In particular,
R
b
a
and
R
b
a
exist as real numbers.
Proof. If we choose P to be the partition P =fa;bg, then U(f;P ) = (ba) sup
x2[a;b]
f(x)
and L(f;P ) = (ba) inf
x2[a;b]
f(x). So, U(f;P ) (ba)M and L(f;P ) (ba)(M).
So,M(ba)
R
b
a
f and
R
b
a
f  M(ba) by the denition of supremum and inmum,
respectively.
We now show that
R
b
a
f 
R
b
a
f. Let P be any partition of [a;b]. By the denition of
L(f;P ) and U(f;P ), we have1 < L(f;P ) U(f;P ) < +1. So, we know that the
setfU(f;P ): P is a partition of [a;b]g is nonempty and bounded from below. Similarly, the
setfL(f;P ): P is a partition of [a;b]g is nonempty and bounded from above. Then, by the
least upper bound property (Theorem 1.1),
R
b
a
f and
R
b
a
f exist as real numbers. So, given
any "> 0, choose a partition P such that L(f;P )
R
b
a
f". (Such a partition P exists by
the denition of the supremum.) We then have
R
b
a
f L(f;P ) +" U(f;P ) +". Taking
the inmum over partitions P of [a;b] of both sides of this inequality, we get
R
b
a
f
R
b
a
f +".
Since "> 0 is arbitrary, we conclude that
R
b
a
f
R
b
a
f, as desired. 
3. Riemann Integral
Denition 3.1 (Riemann Integral). Let a < b be real numbers, let f : [a;b]!R be a
bounded function. If
R
b
a
f =
R
b
a
f we say that f is Riemann integrable on [a;b], and we
dene
Z
b
a
f :=
Z
b
a
f =
Z
b
a
f:
Remark 3.2. Dening the Riemann integral of an unbounded function takes more care,
and we defer this issue to later courses.
Page 4

1. Review
Theorem 1.1 (Least Upper Bound Property). Let E be a nonempty subset ofR. If E
has some upper bound, then E has exactly one least upper bound.
Theorem 1.2. Let a < b be real numbers, and let f : [a;b]! R be a function which is
continuous on [a;b]. Then f is also uniformly continuous on [a;b].
Proposition 1.3. Let X be a subset ofR, let x
0
be a limit point of X, let f : X!R be a
function, and let L be a real number. Then the following two statements are equivalent.
 f is dierentiable at x
0
on X with derivative L.
 For every"> 0, there exists a =(")> 0 such that, ifx2X satisesjxx
0
j<,
then
jf(x) [f(x
0
) +L(xx
0
)]j"jxx
0
j:
Corollary 1.4 (Mean Value Theorem). Let a<b be real numbers, and let f : [a;b]!R
be a continuous function which is dierentiable on (a;b). Then there exists x2 (a;b) such
that
f
0
(x) =
f(b)f(a)
ba
:
Proposition 1.5. Let X be a subset ofR, let x
0
be a limit point of X, and let f : X!R
be a function. If f is dierentiable at x
0
, then f is also continuous at x
0
.
Theorem 1.6 (Properties of Derivatives). LetX be a subset ofR, letx
0
be a limit point
of X, and let f : X!R and g : X!R be functions.
(iv) If f;g are dierentiable at x
0
, then fg is dierentiable at x
0
, and (fg)
0
(x
0
) =
f
0
(x
0
)g(x
0
) +g
0
(x
0
)f(x
0
). (Product Rule)
2. Riemann Sums
Within calculus, the two most fundamental concepts are dierentiation and integration.
We have covered dierentiation already, and we now move on to integration. Dening an
integral is fairly delicate. In the case of the derivative, we created one limit, and the existence
of this limit dictated whether or not the function in question was dierentiable. In the case
of the Riemann integral, there is also a limit to discuss, but it is much more complicated
than in the case of dierentiation.
We should mention that there is more than one way to construct an integral, and the
Riemann integral is only one such example. Within this course, we will only be discussing
the Riemann integral. The Riemann integral has some deciencies which are improved upon
by other integration theories. However, those other integration theories are more involved,
so we focus for now only on the Riemann integral.
Our starting point will be partitions of intervals into smaller intervals, which will form the
backbone of the Riemann sum. The Riemann sum will then be used to create the Riemann
integral through a limiting constructing.
Denition 2.1 (Partition). Let a < b be real numbers. A partition P of the interval
[a;b] is a nite subset of real numbers x
0
;:::;x
n
such that
a =x
0
<x
1
<<x
n1
<x
n
=b:
We write P =fx
0
;x
1
;:::;x
n
g.
Remark 2.2. Let P;P
0
be partitions of [a;b]. Then the union P[P
0
of P and P
0
is also a
partition of [a;b].
Denition 2.3 (Upper and Lower Riemann Sums). Let a < b be real numbers, let
f : [a;b]!R be a bounded function, and letP =fx
0
;:::;x
n
g be a partition of [a;b]. For ev-
ery integer 1in, the functionfj
[x
i1
;x
i
]
is also a bounded function. So, sup
x2[x
i1
;x
i
]
f(x)
and inf
x2[x
i1
;x
i
]
f(x) exist by the Least Upper Bound property (Theorem 1.1). We therefore
dene the upper Riemann sum U(f;P ) by
U(f;P ) :=
n
X
i=1

sup
x2[x
i1
;x
i
]
f(x)

(x
i
x
i1
):
We also dene the lower Riemann sum L(f;P ) by
L(f;P ) :=
n
X
i=1

inf
x2[x
i1
;x
i
]
f(x)

(x
i
x
i1
):
Remark 2.4. For each integer 1 i n, we dene a function g : [a;b]! R such that
g(x) := sup
y2[x
i1
;x
i
]
f(y) for all x
i1
 x < x
i
, with g(b) := f(b). Then g is constant on
[x
i1
;x
i
) for all 1 i n, and f(x) g(x) for all x2 [a;b]. The upper Riemann sum
U(f;P ) then represents the area under the function g, which is meant to upper bound the
area under the function f. Similarly, for each integer 1  i n, we dene a function
h: [a;b]!R such that h(x) := inf
y2[x
i1
;x
i
]
f(y) for all x
i1
 x < x
i
, with h(b) := f(b).
Then h is constant on [x
i1
;x
i
) for all 1 i n, and h(x) g(x) for all x2 [a;b]. The
lower Riemann sum L(f;P ) then represents the area under the function g, which is meant
to lower bound the area under the function f.
Denition 2.5 (Upper and Lower Integrals). Leta<b be real numbers, letf : [a;b]!
R be a bounded function. We dene the upper Riemann integral
R
b
a
f of f on [a;b] by
Z
b
a
f := inffU(f;P ): P is a partition of [a;b]g:
We also dene the lower Riemann integral
R
b
a
f of f on [a;b] by
Z
b
a
f := supfL(f;P ): P is a partition of [a;b]g:
Lemma 2.6. Let f : [a;b]!R be a bounded function, so that there exists a real number M
such thatMf(x)M for all x2 [a;b]. Then
M(ba)
Z
b
a
f
Z
b
a
fM(ba)
In particular,
R
b
a
and
R
b
a
exist as real numbers.
Proof. If we choose P to be the partition P =fa;bg, then U(f;P ) = (ba) sup
x2[a;b]
f(x)
and L(f;P ) = (ba) inf
x2[a;b]
f(x). So, U(f;P ) (ba)M and L(f;P ) (ba)(M).
So,M(ba)
R
b
a
f and
R
b
a
f  M(ba) by the denition of supremum and inmum,
respectively.
We now show that
R
b
a
f 
R
b
a
f. Let P be any partition of [a;b]. By the denition of
L(f;P ) and U(f;P ), we have1 < L(f;P ) U(f;P ) < +1. So, we know that the
setfU(f;P ): P is a partition of [a;b]g is nonempty and bounded from below. Similarly, the
setfL(f;P ): P is a partition of [a;b]g is nonempty and bounded from above. Then, by the
least upper bound property (Theorem 1.1),
R
b
a
f and
R
b
a
f exist as real numbers. So, given
any "> 0, choose a partition P such that L(f;P )
R
b
a
f". (Such a partition P exists by
the denition of the supremum.) We then have
R
b
a
f L(f;P ) +" U(f;P ) +". Taking
the inmum over partitions P of [a;b] of both sides of this inequality, we get
R
b
a
f
R
b
a
f +".
Since "> 0 is arbitrary, we conclude that
R
b
a
f
R
b
a
f, as desired. 
3. Riemann Integral
Denition 3.1 (Riemann Integral). Let a < b be real numbers, let f : [a;b]!R be a
bounded function. If
R
b
a
f =
R
b
a
f we say that f is Riemann integrable on [a;b], and we
dene
Z
b
a
f :=
Z
b
a
f =
Z
b
a
f:
Remark 3.2. Dening the Riemann integral of an unbounded function takes more care,
and we defer this issue to later courses.
Theorem 3.3 (Laws of integration). Let a<b be real numbers, and let f;g : [a;b]!R
be Riemann integrable functions on [a;b]. Then
(i) The function f +g is Riemann integrable on [a;b], and
R
b
a
(f +g) = (
R
b
a
f) + (
R
b
a
g).
(ii) For any real number c, cf is Riemann integrable on [a;b], and
R
b
a
(cf) =c(
R
b
a
f).
(iii) The function fg is Riemann integrable on [a;b], and
R
b
a
(fg) = (
R
b
a
f) (
R
b
a
g).
(iv) If f(x) 0 for all x2 [a;b], then
R
b
a
f 0.
(v) If f(x)g(x) for all x2 [a;b], then
R
b
a
f
R
b
a
g.
(vi) If there exists a real number c such that f(x) =c for x2 [a;b], then
R
b
a
f =c(ba).
(vii) Let c;d be real numbers such that c a < b d. Then [c;d] contains [a;b]. Dene
F (x) := f(x) for all x 2 [a;b] and F (x) := 0 otherwise. Then F is Riemann
integrable on [c;d], and
R
d
c
F =
R
b
a
f.
(viii) Let c be a real number such that a < c < b. Then fj
[a;c]
and fj
[c;b]
are Riemann
integrable on [a;c] and [c;b] respectively, and
Z
b
a
f =
Z
c
a
fj
[a;c]
+
Z
b
c
fj
[c;b]
:
Exercise 3.4. Prove Theorem 3.3.
Remark 3.5. Concerning Theorem 3.3(viii), we often write
R
c
a
R
c
a
fj
[a;c]
.
3.1. Riemann integrability of continuous functions. So far we have discussed some
properties of Riemann integrable functions, but we have not shown many functions that are
actually Riemann integrable. In this section, we show that a continuous function on a closed
interval is Riemann integrable.
Theorem 3.6. Let a<b be real numbers, and let f : [a;b]!R be a continuous function on
[a;b]. Then f is Riemann integrable.
Proof. We will produce a family of partitions of the interval [a;b] such that the upper and
lower Riemann integrals of f are arbitrarily close to each other.
From Theorem 1.2, f is uniformly continuous on [a;b]. Let " > 0. Then, by uniform
continuity of f, there exists  = (") > 0 such that, if x;y2 [a;b] satisfyjxyj < , then
jf(x)f(y)j<". By the Archimedean property, there exists a positive integer N such that
(ba)=N <.
Consider the partition P of the interval [a;b] of the form
P =fx
0
;:::;x
N
g =fa;a+(ba)=N;a+2(ba)=N;a+3(ba)=N;:::;a+(N1)(ba)=N;bg:
Note that x
i
x
i1
= (ba)=N for all 1iN. Since f is continuous on [a;b], f is also
continuous on [x
i1
;x
i
] for each 1 i N. In particular, fj
[x
i1
;x
i
]
achieves its maximum
and minimum for all 1iN. So, for each 1iN, there exist m
i
;M
i
2 [x
i1
;x
i
] such
that
inf
x2[x
i1
;x
i
]
f(x) =f(m
i
); and sup
x2[x
i1
;x
i
]
f(x) =f(M
i
):
Since x
i
x
i1
= (ba)=N < , we havejm
i
M
i
j <  for each 1 i n. Since f is
uniformly continuous, we conclude that
inf
x2[x
i1
;x
i
]
f(x) =f(m
i
)>f(M
i
)" = ( sup
x2[x
i1
;x
i
]
f(x))"; 8 1in: ()
Page 5

1. Review
Theorem 1.1 (Least Upper Bound Property). Let E be a nonempty subset ofR. If E
has some upper bound, then E has exactly one least upper bound.
Theorem 1.2. Let a < b be real numbers, and let f : [a;b]! R be a function which is
continuous on [a;b]. Then f is also uniformly continuous on [a;b].
Proposition 1.3. Let X be a subset ofR, let x
0
be a limit point of X, let f : X!R be a
function, and let L be a real number. Then the following two statements are equivalent.
 f is dierentiable at x
0
on X with derivative L.
 For every"> 0, there exists a =(")> 0 such that, ifx2X satisesjxx
0
j<,
then
jf(x) [f(x
0
) +L(xx
0
)]j"jxx
0
j:
Corollary 1.4 (Mean Value Theorem). Let a<b be real numbers, and let f : [a;b]!R
be a continuous function which is dierentiable on (a;b). Then there exists x2 (a;b) such
that
f
0
(x) =
f(b)f(a)
ba
:
Proposition 1.5. Let X be a subset ofR, let x
0
be a limit point of X, and let f : X!R
be a function. If f is dierentiable at x
0
, then f is also continuous at x
0
.
Theorem 1.6 (Properties of Derivatives). LetX be a subset ofR, letx
0
be a limit point
of X, and let f : X!R and g : X!R be functions.
(iv) If f;g are dierentiable at x
0
, then fg is dierentiable at x
0
, and (fg)
0
(x
0
) =
f
0
(x
0
)g(x
0
) +g
0
(x
0
)f(x
0
). (Product Rule)
2. Riemann Sums
Within calculus, the two most fundamental concepts are dierentiation and integration.
We have covered dierentiation already, and we now move on to integration. Dening an
integral is fairly delicate. In the case of the derivative, we created one limit, and the existence
of this limit dictated whether or not the function in question was dierentiable. In the case
of the Riemann integral, there is also a limit to discuss, but it is much more complicated
than in the case of dierentiation.
We should mention that there is more than one way to construct an integral, and the
Riemann integral is only one such example. Within this course, we will only be discussing
the Riemann integral. The Riemann integral has some deciencies which are improved upon
by other integration theories. However, those other integration theories are more involved,
so we focus for now only on the Riemann integral.
Our starting point will be partitions of intervals into smaller intervals, which will form the
backbone of the Riemann sum. The Riemann sum will then be used to create the Riemann
integral through a limiting constructing.
Denition 2.1 (Partition). Let a < b be real numbers. A partition P of the interval
[a;b] is a nite subset of real numbers x
0
;:::;x
n
such that
a =x
0
<x
1
<<x
n1
<x
n
=b:
We write P =fx
0
;x
1
;:::;x
n
g.
Remark 2.2. Let P;P
0
be partitions of [a;b]. Then the union P[P
0
of P and P
0
is also a
partition of [a;b].
Denition 2.3 (Upper and Lower Riemann Sums). Let a < b be real numbers, let
f : [a;b]!R be a bounded function, and letP =fx
0
;:::;x
n
g be a partition of [a;b]. For ev-
ery integer 1in, the functionfj
[x
i1
;x
i
]
is also a bounded function. So, sup
x2[x
i1
;x
i
]
f(x)
and inf
x2[x
i1
;x
i
]
f(x) exist by the Least Upper Bound property (Theorem 1.1). We therefore
dene the upper Riemann sum U(f;P ) by
U(f;P ) :=
n
X
i=1

sup
x2[x
i1
;x
i
]
f(x)

(x
i
x
i1
):
We also dene the lower Riemann sum L(f;P ) by
L(f;P ) :=
n
X
i=1

inf
x2[x
i1
;x
i
]
f(x)

(x
i
x
i1
):
Remark 2.4. For each integer 1 i n, we dene a function g : [a;b]! R such that
g(x) := sup
y2[x
i1
;x
i
]
f(y) for all x
i1
 x < x
i
, with g(b) := f(b). Then g is constant on
[x
i1
;x
i
) for all 1 i n, and f(x) g(x) for all x2 [a;b]. The upper Riemann sum
U(f;P ) then represents the area under the function g, which is meant to upper bound the
area under the function f. Similarly, for each integer 1  i n, we dene a function
h: [a;b]!R such that h(x) := inf
y2[x
i1
;x
i
]
f(y) for all x
i1
 x < x
i
, with h(b) := f(b).
Then h is constant on [x
i1
;x
i
) for all 1 i n, and h(x) g(x) for all x2 [a;b]. The
lower Riemann sum L(f;P ) then represents the area under the function g, which is meant
to lower bound the area under the function f.
Denition 2.5 (Upper and Lower Integrals). Leta<b be real numbers, letf : [a;b]!
R be a bounded function. We dene the upper Riemann integral
R
b
a
f of f on [a;b] by
Z
b
a
f := inffU(f;P ): P is a partition of [a;b]g:
We also dene the lower Riemann integral
R
b
a
f of f on [a;b] by
Z
b
a
f := supfL(f;P ): P is a partition of [a;b]g:
Lemma 2.6. Let f : [a;b]!R be a bounded function, so that there exists a real number M
such thatMf(x)M for all x2 [a;b]. Then
M(ba)
Z
b
a
f
Z
b
a
fM(ba)
In particular,
R
b
a
and
R
b
a
exist as real numbers.
Proof. If we choose P to be the partition P =fa;bg, then U(f;P ) = (ba) sup
x2[a;b]
f(x)
and L(f;P ) = (ba) inf
x2[a;b]
f(x). So, U(f;P ) (ba)M and L(f;P ) (ba)(M).
So,M(ba)
R
b
a
f and
R
b
a
f  M(ba) by the denition of supremum and inmum,
respectively.
We now show that
R
b
a
f 
R
b
a
f. Let P be any partition of [a;b]. By the denition of
L(f;P ) and U(f;P ), we have1 < L(f;P ) U(f;P ) < +1. So, we know that the
setfU(f;P ): P is a partition of [a;b]g is nonempty and bounded from below. Similarly, the
setfL(f;P ): P is a partition of [a;b]g is nonempty and bounded from above. Then, by the
least upper bound property (Theorem 1.1),
R
b
a
f and
R
b
a
f exist as real numbers. So, given
any "> 0, choose a partition P such that L(f;P )
R
b
a
f". (Such a partition P exists by
the denition of the supremum.) We then have
R
b
a
f L(f;P ) +" U(f;P ) +". Taking
the inmum over partitions P of [a;b] of both sides of this inequality, we get
R
b
a
f
R
b
a
f +".
Since "> 0 is arbitrary, we conclude that
R
b
a
f
R
b
a
f, as desired. 
3. Riemann Integral
Denition 3.1 (Riemann Integral). Let a < b be real numbers, let f : [a;b]!R be a
bounded function. If
R
b
a
f =
R
b
a
f we say that f is Riemann integrable on [a;b], and we
dene
Z
b
a
f :=
Z
b
a
f =
Z
b
a
f:
Remark 3.2. Dening the Riemann integral of an unbounded function takes more care,
and we defer this issue to later courses.
Theorem 3.3 (Laws of integration). Let a<b be real numbers, and let f;g : [a;b]!R
be Riemann integrable functions on [a;b]. Then
(i) The function f +g is Riemann integrable on [a;b], and
R
b
a
(f +g) = (
R
b
a
f) + (
R
b
a
g).
(ii) For any real number c, cf is Riemann integrable on [a;b], and
R
b
a
(cf) =c(
R
b
a
f).
(iii) The function fg is Riemann integrable on [a;b], and
R
b
a
(fg) = (
R
b
a
f) (
R
b
a
g).
(iv) If f(x) 0 for all x2 [a;b], then
R
b
a
f 0.
(v) If f(x)g(x) for all x2 [a;b], then
R
b
a
f
R
b
a
g.
(vi) If there exists a real number c such that f(x) =c for x2 [a;b], then
R
b
a
f =c(ba).
(vii) Let c;d be real numbers such that c a < b d. Then [c;d] contains [a;b]. Dene
F (x) := f(x) for all x 2 [a;b] and F (x) := 0 otherwise. Then F is Riemann
integrable on [c;d], and
R
d
c
F =
R
b
a
f.
(viii) Let c be a real number such that a < c < b. Then fj
[a;c]
and fj
[c;b]
are Riemann
integrable on [a;c] and [c;b] respectively, and
Z
b
a
f =
Z
c
a
fj
[a;c]
+
Z
b
c
fj
[c;b]
:
Exercise 3.4. Prove Theorem 3.3.
Remark 3.5. Concerning Theorem 3.3(viii), we often write
R
c
a
R
c
a
fj
[a;c]
.
3.1. Riemann integrability of continuous functions. So far we have discussed some
properties of Riemann integrable functions, but we have not shown many functions that are
actually Riemann integrable. In this section, we show that a continuous function on a closed
interval is Riemann integrable.
Theorem 3.6. Let a<b be real numbers, and let f : [a;b]!R be a continuous function on
[a;b]. Then f is Riemann integrable.
Proof. We will produce a family of partitions of the interval [a;b] such that the upper and
lower Riemann integrals of f are arbitrarily close to each other.
From Theorem 1.2, f is uniformly continuous on [a;b]. Let " > 0. Then, by uniform
continuity of f, there exists  = (") > 0 such that, if x;y2 [a;b] satisfyjxyj < , then
jf(x)f(y)j<". By the Archimedean property, there exists a positive integer N such that
(ba)=N <.
Consider the partition P of the interval [a;b] of the form
P =fx
0
;:::;x
N
g =fa;a+(ba)=N;a+2(ba)=N;a+3(ba)=N;:::;a+(N1)(ba)=N;bg:
Note that x
i
x
i1
= (ba)=N for all 1iN. Since f is continuous on [a;b], f is also
continuous on [x
i1
;x
i
] for each 1 i N. In particular, fj
[x
i1
;x
i
]
achieves its maximum
and minimum for all 1iN. So, for each 1iN, there exist m
i
;M
i
2 [x
i1
;x
i
] such
that
inf
x2[x
i1
;x
i
]
f(x) =f(m
i
); and sup
x2[x
i1
;x
i
]
f(x) =f(M
i
):
Since x
i
x
i1
= (ba)=N < , we havejm
i
M
i
j <  for each 1 i n. Since f is
uniformly continuous, we conclude that
inf
x2[x
i1
;x
i
]
f(x) =f(m
i
)>f(M
i
)" = ( sup
x2[x
i1
;x
i
]
f(x))"; 8 1in: ()
We now estimate U(f;P ) andL(f;P ). By the denition of U(f;P ) andL(f;P ), we have
L(f;P )U(f;P ): ()
However, L(f;P ) is also close to U(f;P ) by ():
L(f;P ) =
ba
N
N
X
i=1
( inf
x2[x
i1
;x
i
]
f(x))>
ba
N
N
X
i=1
[( sup
x2[x
i1
;x
i
]
f(x))"] =(ba)" +U(f;P ):
By the denition of
R
b
a
f, we conclude that
Z
b
a
f >(ba)" +U(f;P ):
By the denition of
R
b
a
f, we conclude that
Z
b
a
f >(ba)" +
Z
b
a
f:
Since "> 0 is arbitrary, we get
Z
b
a
f
Z
b
a
f:
Combining this inequality with Lemma 2.6, we conclude that
R
b
a
f =
R
b
a
f. That is, f is
Riemann integrable. 
Exercise 3.7. Let a < b be real numbers. Let f : [a;b]!R be a bounded function. Let
c2 [a;b]. Assume that, for each  > 0, we know that f is Riemann integrable on the set
fx2 [a;b]: jxcjg. Then f is Riemann integrable on [a;b].
3.2. Piecewise Continuous Functions. We can now expand a bit more the family of
functions that are Riemann integrable.
Proposition 3.8. Let a < b be real numbers. Assume that f : [a;b]!R is continuous at
every point of [a;b], except for a nite number of points. Then f is Riemann integrable.
Proof. By Theorem 3.3(viii) and an inductive argument, it suces to consider the case that
f is discontinuous at a single point c2 [a;b]. Let  > 0. Then f is continuous on the set
E :=fx2 [a;b]: jxcj g. Note that E consists of either one or two closed intervals.
Since fj
E
is continuous, we then conclude that fj
E
Riemann integrable by Theorem 3.6.
Then Exercise 3.7 says that f is Riemann integrable on [a;b], as desired. 
3.3. Monotone Functions. It turns out that monotone functions are Riemann integrable
as well. There exist monotone functions that are not piecewise continuous, so the current
section is not subsumed by the previous one.
Proposition 3.9. Let a<b be real numbers, and let f : [a;b]!R be a monotone function.
Then f is Riemann integrable.
```
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

197 docs

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;