AMA 1120 Week 4 Lecture note
Antiderivatives, definite integral, indefinite integral
Xindong Tang
February 8th, 2023
AMA1120 Week 4
Outline
Antiderivatives
Indefinite integral
Definite integral
AMA1120 Week 4
Antiderivatives: definition
Definition 1 (Antiderivative)
Suppose that f (x ) is a function defined in an interval I. A function
F (x ) is said to be an antiderivative of f (x ) if
F ′ (x ) = f (x )
for all x ∈ I.
Example 1
(i) Let f (x ) = 3x 2 , then F (x ) = x 3 is an antiderivative of f (x )
because
F ′ (x ) = 3x 2 = f (x ).
(ii) Let f (x ) = e x , then F (x ) = e x is an antiderivative of f (x )
because
F ′ (x ) = e x = f (x ).
AMA1120 Week 4
Antiderivatives: definition
Why do we say ‘an’ antiderivative? How many antiderivative can a
function have?
Theorem 1
Suppose that f (x ) is a function defined in an interval I, and F (x )
is an antiderivative of f (x ), then a function G(x ) is also an
antiderivative of f (x ) if and only if
G(x ) = F (x ) + c
for all x ∈ I,
where c is a constant.
So there are always infinitely many antiderivatives, as long as
they exist.
AMA1120 Week 4
Antiderivatives: definition
Theorem 1
Suppose that f (x ) is a function defined in an interval I, and F (x ) is an
antiderivative of f (x ), then a function G(x ) is also an antiderivative of
f (x ) if and only if
G(x ) = F (x ) + c
for all
x ∈ I,
where c is a constant.
Sketch of Proof.
If G(x ) = F (x ) + c, then G ′ (x ) = F ′ (x ) = f (x ), then G(x ) is an
antiderivative of f (x ). Conversely, if G(x ) is an antiderivative of f (x ),
then
(G(x ) − F (x ))′ = G ′ (x ) − F ′ (x ) = f (x ) − f (x ) = 0.
Hence G(x ) − F (x ) is a constant, namely c, in I. Thus
G(x ) − F (x ) = c
for all x ∈ I.
AMA1120 Week 4
Antiderivatives: existence
Theorem 2 (Existence of antiderivative)
If a function f (x ) is continuous in an interval I, then it has an
antiderivative.
▶ The converse of the theorem is not true: even if f (x ) has an
antiderivative, f (x ) could still be discontinuous. For example,
consider
2
x sin(1/x )
when x ̸= 0,
F (x ) =
0
when x = 0.
Then it is the antiderivative of
2x sin(1/x ) − cos(1/x )
f (x ) =
0
when x ̸= 0,
when x = 0.
▶ There exist functions who have no antiderivative. An example?
AMA1120 Week 4
Indefinite integral: definition
Definition 2 (Indefinite integral)
Suppose that f (x ) is a function on an interval I. If F (x ) is an
antiderivative of f (x ), then we will
also call F (x ) an indefinite
R
integral of f (x ) and denoted as f (x )dx
▶ ‘Antiderivative’ and ‘indefinite integral’ are synonyms. They
are also called ‘primitive function’.
R
▶ The is the integral symbol, f (x ) is called ‘integrand’, and x
is called the integration variable.
▶ If F (x ) is an antiderivative of f (x ), then
Z
f (x )dx = F (x ) + c,
where c is a constant, namely the ‘constant of integration’.
AMA1120 Week 4
Indefinite integral: definition
Example 3
Z
3x 2 dx = x 3 + c,
because x 3 is an antiderivative of 3x 2 .
Example 4
Z
e x dx = e x + c,
because e x is an antiderivative of e x .
AMA1120 Week 4
Indefinite integral vs derivatives
Integration is the inverse of differentiation. Let f (x ) be a
differentiable function in an interval I. Then according to the
definition of antiderivative (or definite integral), we have
( f (x )dx )′ = f (x ),
( R
R ′
f (x )dx = f (x ) + c.
Using the notation
d
dx f (x )
for f ′ (x ), we have

R
 d ( f (x )dx ) = f (x ),
Rdx d

f
(x
)
dx = f (x ) + c.
dx
In other words, the operators
R
and
d
dx
cancel each other.
AMA1120 Week 4
Indefinite integral: basic formulas
▶
R
adx = ax + c;
x a dx =
x a+1
a+1
▶
R
▶
R
▶
R
ax dx =
▶
R
sin(x )dx = − cos(x ) + c;
▶
R
cos(x )dx = sin(x ) + c;
+ c, if a ̸= −1;
x −1 dx = ln(|x |) + c;
ax
ln(a)
+ c, if a > 0;
▶
R
sec2 (x )dx = tan(x ) + c;
▶
R
csc2 (x )dx = − cot(x ) + c;
▶
R
sec(x ) tan(x )dx = sec(x ) + c;
▶
R
▶
R
csc(x ) cot(x )dx =
− csc(x ) + c;
R
▶
√ 1 dx = arcsin(x ) + c;
2
1−x
1
1+x 2 dx
= arctan(x ) + c.
AMA1120 Week 4
Indefinite integral: basic formulas
▶
R
af (x )dx = a f (x )dx ;
▶
R
(f (x ) + g(x ))dx =
▶
R
(af (x ) + bg(x ))dx = a f (x )dx + b g(x )dx .
R
R
R
f (x )dx + g(x )dx ;
R
R
Exercise
Show these formulas. (Hint: consider their analogs for derivatives)
AMA1120 Week 4
Indefinite integral
Example 5
2
3
+ 2 + e x )dx
x
x
Z
Z
Z
Z
Z
Z
3
2
2
= x dx − xdx + 1dx −
dx +
dx + e x dx
x
x2
x3 x2
x −1
= −
+ x − 2 ln(|x |) + 3
+ ex + c
3
2
−1
Z
(x 2 − x + 1 −
AMA1120 Week 4
Indefinite integral
Example 6
Z
Z
=3
3 tan2 (x )dx =
Z
2
sec (x )dx − 3
3(sec2 (x ) − 1)dx
Z
1dx
=3 tan(x ) − 3x + c.
AMA1120 Week 4
Definite integral
The definition of definite integral is based on three steps: partition,
summation and taking limit.
Definition 3 (Definite integral: part I: Partition)
Let f (x ) be a function defined on an interval [a, b]. Let ∆ be a
partition of the interval [a, b] given by
∆ := {a = x0 < x1 < x2 < · · · < xn−1 < xn = b},
in other words, we have the subintervals
[x0 , x1 ], [x1 , x2 ], . . . , [xn−1 , xn ].
AMA1120 Week 4
Definite integral
Definition 4 (Definite integral: part II: Riemann sum)
For each i = 1, 2, . . . , n, denote the length of the subinterval
[xi−1 , xi ] by
∆xi = xi − xi−1 ,
and let ξi be a number in the subinterval [xi−1 , xi ], then
n
X
f (ξi )∆xi
i=1
is called a Riemann sum of f (x ) for the partition ∆.
AMA1120 Week 4
Definite integral
Definition 5 (Definite integral: part III)
Let ∥∆∥ denote the largest value of ∆x1 , ∆x2 , . . . , ∆xn , i.e.,
∥∆∥ := max ∆xi . If there exists a finite number L such that
i=1,...,n
n
X
lim
∥∆∥→0
f (ξi )∆xi = L,
i=1
regardless of the choice of ξi , then we say that f (x ) is integrable
on [a, b], and L is the definite integral of f (x ) on [a, b]. In this
case, we use the notation
Z b
f (x )dx = L.
a
AMA1120 Week 4
Definite integral
AMA1120 Week 4
Definite integral: existence
Theorem 2
If f (x ) is continuous in [a, b], thenR f (x ) is integrable on [a, b],
meaning that the definite integral ab f (x )dx exists.
In the notation ab f (x )dx , we call
▶ f (x ) the integrand;
R
▶ x the integration variable;
▶ a the lower limit of integration;
▶ b the upper limit of integration.
AMA1120 Week 4
Definite integral: how to calculate?
Theorem 3
If
f (x ) is a non-negative continuous function in [a, b], then
Rb
curve of f (x ) and above the
a f (x )dx is the area below the
Rb
x -axis. We can also say that a f (x )dx is the area of the region
bounded by the curve of f (x ), the x -axis, and the vertical lines
x = a and x = b.
If f (x ) is negative, then the ab f (x )dx is the −1 times area above
the curve of f (x ) and below the x -axis.
R
Example 7
Use the theorem above, we have
Z 2
xdx = area of the triangle bounded by (0, 0), (0, 2), (2, 2) = 2.
0
AMA1120 Week 4
Definite integral: how to calculate?
Theorem 4
If f (x ) has the antiderivative
Z b
R
f (x )dx = F (x ) on [a, b], then
f (x )dx = F (b) − F (a).
a
Example 8
Calculate
Z 2
xdx .
0
Because the indefinite integral
Z 2
0
R
xdx = 21 x 2 ,
1
xdx = (22 − 02 ) = 2.
2
We will revisit this Theorem in the next lecture.
AMA1120 Week 4
Definite integral: additive interval property
Theorem 5
Let a, b and c be three constants and f (x ) be a function. We have
Z b
Z c
f (x )dx =
a
Z b
f (x )dx +
f (x )dx ,
a
c
as long as f (x ) is integrable on all the intervals concerned.
Example 9
By some elementary geometry knowledge,
So we still have
Z 2
Z 1
xdx =
0
Z 2
xdx +
0
R1
xdx =
1
0
xdx = 12 ,
R2
1
xdx = 32 .
1 3
+ = 2.
2 2
This theorem do not require f to be positive.
AMA1120 Week 4
Definite integral: additive interval property
If we define
Z b
f (x )dx = −
a
Z a
f (x )dx
b
when a > b, and define
Z a
f (x )dx = 0,
a
then the proposition does not require a ≤ c ≤ b. For example,
R
R
R
▶ 02π sin(x )dx = 0π sin(x )dx + π2π sin(x )dx ;
▶
R1 x
R5 x
R1 x
2 e dx = 2 e dx + 5 e dx .
AMA1120 Week 4
Definite integral: linearity
Proposition 1 (Linearity)
Suppose that f (x ) and g(x ) are integrable on [a, b], and α, β are
constants. Then
Z b
Z b
[αf (x ) + βg(x )]dx = α
a
Z b
f (x )dx + β
a
g(x )dx .
a
Exercise
Show Theorem 5 Using this proposition. (Hint: Consider
(
f (x )
0
when
(
0
f (x )
when 0 ∈ [a, c),
)
otherwise.
f1 (x ) =
f2 (x ) =
x ∈ [a, c),
otherwise.
AMA1120 Week 4
Definite integral: Preservation of inequality
Proposition 2 (Preservation of positiveness)
If a ≤ b, the function f (x ) is integrable on [a, b] and f (x ) ≥ 0 for
all x ∈ [a, b], then
Z b
f (x )dx ≥ 0.
a
Proposition 3 (Preservation of inequality)
If a ≤ b, the functions f (x ) and g(x ) are integrable on [a, b] and
f (x ) ≥ g(x ) for all x ∈ [a, b], then
Z b
a
f (x )dx ≥
Z b
g(x )dx .
a
AMA1120 Week 4