LECTURE 12: APPROXIMATING AREAS UNDER CURVES AND DEFINITE INTEGRALS
MINGFENG ZHAO
January 30, 2015
Example 1. For a regular partition x0 = a < x1 < · · · < xn = b of [a, b], then ∆x =
b−a
and the Riemann sum can
n
be written:
f (x∗1 )∆x + f (x∗2 )∆x + · · · + f (x∗n )∆x =
n
X
f (x∗k )∆x.
k=1
For the choices
x∗k ’s,
we have
a. For the left Riemann sum, we have x∗k = xk−1 = a + (k − 1)∆x.
b. For the right Riemann sum, we have x∗k = xk = a + k∆x.
a + (k − 1)∆x + a + k∆x
1
xk−1 + xk
=
=a+ k−
∆x.
c. For the midpoint Riemann sum, we have x∗k =
2
2
2
Example 2. Express the left, right, and midpoint Riemann sums with sigma notations for f (x) = x3 + 1 between a = 0
and b = 2 using n = 50 subintervals.
2−0
2
Since n = 50, then ∆x =
=
= 0.04. So we have
50
50
for all k = 0, 1, · · · , 50.
xk = 0 + k∆x = 0.04k,
Therefore, we have
a. For the left Riemann sum: x∗k = xk−1 = 0.04(k − 1). Then
50
X
f (x∗k )∆x =
k=1
b. For the right Riemann sum:
x∗k
50
X
[(0.04(k − 1))3 + 1] · 0.04
k=1
= xk = 0.04k. Then
50
X
f (x∗k )∆x =
k=1
50
X
[(0.04k)3 + 1] · 0.04
k=1
0.04(k − 1) + 0.04k
1
xk−1 + xk
=
= 0.04 k −
. Then
c. For the left Riemann sum:
=
2
2
2
"
#
3
50
50
X
X
1
f (x∗k )∆x =
0.04 k −
+ 1 · 0.04
2
x∗k
k=1
k=1
1
2
MINGFENG ZHAO
Example 3. Express the sum 5 + 7 + 9 + 11 + · · · + 23 + 25 using sigma notation.
Let x1 = 5, x2 = 7, x3 = 9,· · · , we see the pattern:
xk = 3 + 2k,
for all k = 1, 2, · · · .
Set xn = 25, then 3 + 2n = 25, which implies that n = 11. Hence we get
5 + 7 + 9 + 11 + · · · + 23 + 25 = x1 + x2 + x3 + · · · + x11 =
11
X
(3 + 2k).
k=1
Theorem 1. Let n be a positive integer, c be a real number and two sequences ak and bk , then
n
X
cak
= c
k=1
n
X
n
X
ak
k=1
(ak + bk )
=
k=1
n
X
ak +
k=1
n
X
bk .
k=1
Theorem 2 (Sums of Powers of Integers). Let n be a positive integer and c be a real number, then
n
X
c = cn
k=1
n
X
k
=
n(n + 1)
2
k2
=
n(n + 1)(2n + 1)
6
k3
=
n2 (n + 1)2
.
4
k=1
n
X
k=1
n
X
k=1
Example 4. For the sum in Example 3, we have
5 + 7 + 9 + 11 + · · · + 23 + 25
=
11
X
(3 + 2k) =
k=1
=
11
X
11
X
3+
k=1
3+2
k=1
11
X
k
11
X
2k
By Theorem 1
k=1
By Theorem 1
k=1
11(11 + 1)
2
=
3 · 11 + 2 ·
=
33 + 11 ∗ 12
=
33 + 132
=
165.
By Theorem 2
LECTURE 12: APPROXIMATING AREAS UNDER CURVES AND DEFINITE INTEGRALS
3
Net area
Definition 1. Consider the region R bounded by the graph of a continuous function f (x) and the x-axis between x = a
and x = b, the net area of R is the sum of the areas of the parts of R lies above the x-axis minus the sum of the areas
of the parts of R that lie below the x-axis on [a, b].
Figure 1. Yellow=positive net area, Pink=negative net area
Example 5. Let f (x) = −5 and R be the region bounded by the graph of f and the x-axis on [1, 5], then
Area of R = 5 · 4 = 20,
and
Net Area of R = −20.
Example 6. Let f (x) = sin(x), the net area of the region bounded by the graph of f and the x-axis on [−π, π] is 0. In
future, we will see the area of this region is 4.
The definite integral
For a partition a = x0 < x1 < x2 < · · · < xn = b, let
∆ = max {∆x1 , ∆x2 , · · · , ∆xn }.
Then
∆ → 0 =⇒ ∆xk → 0, for all k = 1, · · · , n.
4
MINGFENG ZHAO
Definition 2. A function f (x) defined on [a, b] is integrable if lim
∆→0
n
X
f (x∗k )∆xk exists and is unique over all partitions
k=1
of [a, b] and all choices of x∗k on this partition. This limit is called the definite integral of f from a to b, and denoted by:
Z b
n
X
f (x) dx = lim
f (x∗k )∆xk .
∆→0
a
Z
In the notation
k=1
b
f (x) dx, the function f (x) is called the integrand of the integral, a is called the lower limit of the
a
integral, and b is called the upper limit of the integral.
b
Z
f (x) dx, the variable x is not essential, one can change x into any another letter, that
Remark 1. For the notation
a
is,
Z
b
Z
b
a
Z
Remark 2. The integral
Z
a
b
f (s) ds.
f (w) dw =
f (x) dx =
a
b
f (x) dx is just the net area of the region bounded by the graph of f and the x-axis between
a
x = a and x = b.
Theorem 3. If f is continuous on [a, b] or bounded on [a, b] with a finite number of discontinuities, then f is integrable
on [a, b].

 0
Example 7. Let f (x) =
 1
if x is a rational number
, then f is not integrable on [1, 2].
if x is an irrational number
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca