1.1_The_Definite_Integral.pdf odjoqwddoio

Partitions and the Definite Integral
MAT070 - Calculus with Analytic Geometry 2
2nd Semester, A.Y. 2021-2022
MARIBETH B. MONTERO, PhD
Mathematics Department
Mindanao State University Main Campus
Marawi City
maribeth.montero@msumain.edu.ph

Partitions
The concept of partitions or divisions and the formation of sums are the initial steps in the
development of the theory of integration.
Let f be a continuous function dened on the closed interval [a, b].
The partition D of [a, b] is a collection of points x0, x1, x2, · · · , xn such that
a = x0 x1 x2 · · · xn = b;
that is,
D = {x0, x1, x2, · · · , xn}.

The norm of D, denoted by kDk, is the largest of the dierences
∆i x = xi − xi−1, where i = 1, 2, 3, · · · , n;
that is, kDk is the length of the longest interval in D.
The sum
s(f : D) =
n
X
i=1
f (ξi )(xi − xi−1), where ξi ∈ [xi−1, xi ],
is called the Riemann sum of f with respect to a partition D of [a, b].

Example.
(a) A partition D1 on [−1, 1] with n = 4 and
x0 = a = −1, x1 = −0.5, x2 = 0, x3 = 0.5, x4 = b = 1
is the set
D1 = {x0, x1, x2, x3, x4} = {−1, −0.5, 0, 0.5, 1}.
Observe that for each i = 1, 2, 3, 4,
∆i x = xi − xi−1 =
1
2
.
Hence, kD1k = 1
2.

(b) Another partition D2 on [−1, 1] with n = 8 can be made by setting
kD2k =
b − a
n
=
1 − (−1)
8
= 0.25.
In this case, for i = 1, 2, . . . , 8,
xi = a + ikD2k = a + 0.25i.
That is,
D2 = {x0, x1, x2, x3, x4, x5, x6, x7, x8}
= {−1, −0.75, −0.5, −0.25, 0, 0.25, 0.5, 0.75, 1}.

(c) D3 = {−1, 0, 2
5, 1} is another partition of [−1, 1]. Here, kD3k = 1.
A Riemann sum s(f : D3) of f (x) = (x + 1)2 with respect to the partition
D3 = {−1, 0, 2
5, 1} on [−1, 1] is
s(f : D3) = f (ξ1)(0 − (−1)) + f (ξ2)(2
5 − 0) + f (ξ3)(1 − 2
5).
If we choose ξ1 = −0.5, ξ2 = 0.1 and ξ3 = 0.5, then
s(f : D3) = f (−0.5)(0 − (−1)) + f (0.1)(0.4 − 0) + f (0.5)(1 − 0.4)
= 0.25(1) + 1.21(0.4) + 2.25(0.6)
= 0.25 + 0.484 + 1.35
= 2.084.

Geometrically, the Riemann sum s(f : D3) is the sum of the areas of the shaded
rectangles below.
y
1
2
3
x
−1 0 1
y = (x + 1)2

The Definite Integral
Let f be a continuous function dened on the closed interval [a, b].
The denite integral of f from a to b is given by
lim
kDk→0
n
X
i=1
f (ξi )(xi − xi−1),
where D = {x0, x1, . . . , xn} is a partition of [a, b].
The symbol
Z b
a
f (x)dx is used to denote this denite integral.
The numbers a and b are then called the lower and the upper limits of the
denite integral, respectively.

Properties of the Definite Integrals
Theorem.
If f is a continuous function on [a, b], then
Z c
c
f (x)dx = 0
for all c ∈ [a, b].
Theorem.
If f is a continuous function on [a, b], and c ∈ (a, b), then
Z b
a
f (x)dx =
Z c
a
f (x)dx +
Z b
c
f (x)dx.

Theorem.
If f is a continuous function on [a, b] and k is any real number, then
Z b
a
kf (x)dx = k
Z b
a
f (x)dx.
Theorem.
If f and g are continuous functions on [a, b], then f + g is continuous on
[a, b] and
Z b
a
[f (x) + g(x)] dx =
Z b
a
f (x)dx +
Z b
a
g(x)dx.

Theorem. (First Fundamental Theorem of Calculus)
Let f be a continuous function function on [a, b]. If F is the function
dened by
F(x) =
Z x
a
f (t)dt
for every x in [a, b], then F0
(x) = f (x) for all x ∈ [a, b], that is, F is an
antiderivative of f .

Theorem. (Second Fundamental Theorem of Calculus)
Let f be a continuous function on [a, b] and let F be an antiderivative of f ,
that is, F0
(x) = f (x) for all x ∈ [a, b]. Then
Z b
a
f (x)dx = F(x)|b
a = F(b) − F(a).

Examples. Evaluate the following denite integrals.
1.
Z 2
−1
2 − 3x2
+ 3x5

dx
2.
Z 2
0
2x
p
1 + 3x2dx
3.
Z π
4
0
sin(cos(2x)) sin(2x)dx
4.
Z 4
−4
|x + 2| dx

1.
Z 2
−1
2 − 3x2
+ 3x5

dx
Solution: Using the Fundamental Theorems of Calculus, we have
Z 2
−1
2 − 3x2
+ 3x5

dx =

2x − 3
x3
3
+ 3
x6
6
2
−1
=

2x − x3
+
x6
2
2
−1
=

4 − 8 +
64
2

−

−2 + 1 +
1
2

=
56
2
+
1
2
=
57
2
.

2.
Z 2
0
2x
p
1 + 3x2dx
Solution: Let u = 1 + 3x2. Then du = 6xdx; that is,
du
3
= 2xdx. Note that
when x = 0, u = 1 and when x = 2, u = 13. Thus,
Z 2
0
2x
p
1 + 3x2 dx =
Z 13
1
√
u ·
du
3
=
1
3
Z 13
1
u
1
2 du
=
1
3
·
u
3
2
3
2
13
1
=
2
9
u
√
u
13
1
=
2
9

13
√
13 − 1

.

3.
Z π
4
0
sin(cos(2x)) sin(2x)dx
Solution: Let u = cos 2x. Then du = −2 sin 2xdx; that is,
du
−2
= sin 2xdx.
Note that when x = 0, u = 1 and when x =
π
4
, u = 0. Thus,
Z π
4
0
sin(cos(2x)) sin(2x)dx =
Z 0
1
sin u ·
du
−2
=
−1
2
Z 0
1
sin udu =
−1
2
(− cos u)
0
1
=
1
2
cos u
0
1
=
1
2
(1 − cos 1) .

4.
Z 4
−4
|x + 2| dx
Solution: Note that f (x) = |x + 2| =
(
−(x + 2), if − 4 ≤ x −2
x + 2, if − 2 ≤ x ≤ 4
.
Thus,
Z 4
−4
|x + 2|dx =
Z −2
−4
|x + 2|dx +
Z 4
−2
|x + 2|dx
=
Z −2
−4
−(x + 2)dx +
Z 4
−2
(x + 2)dx
= −

x2
2
+ 2x
−2
−4
+

x2
2
+ 2x
4
−2
= (2 − 0) + (16 + 2) = 20 .

Exercises. Evaluate each of the given denite integral.
1.
Z 3
1
(x2
− 2x + 5) dx 6.
Z 1
−1
(x2 + 6x − 1)2
9
dx
2.
Z 0
−1
(5x4
− 5x2
− 7) dx 7.
Z x
0
t2
dt
3.
Z 2
1
(x3/2
+ 2x1/2
) dx 8.
Z π/6
0
cos(2x) dx
4.
Z 4
1
x2 + x + 1
√
x
dx 9.
Z π/3
π/6
sin(3x) dx
5.
Z 2
0
(x + 1)(2x − 3) dx 10.
Z π/2
0
sec 2
x dx

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