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MAT 1100 Integral Calculus I - 2020
Introduction to mathematics (1100)
University of Zambia
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MAT
LECTURE NOTES
6 Integral Calculus I
6 The Indefinite Integral
6.1 Integration as the reverse process of differentiation
Recall that when a function 𝑓(𝑥) is differentiated 𝑓′(𝑥) is obtained. In indefinite
integral when a function 𝑓′(𝑥) is given, we find a function which was differentiated to
give 𝑓′(𝑥).
The symbol for integration is For example, for 𝑓′
(
𝑥
)
= 𝑛𝑥
𝑛−
, to find 𝑓
(
𝑥
)
we
write ∫(𝑛𝑥
𝑛−
)𝑑𝑥 = 𝑓(𝑥). In general, if
𝑑𝑦
𝑑𝑥
= 𝑓′(𝑥) then ∫ 𝑓
′
(𝑥)𝑑𝑥 = 𝑦.
Now, the process of finding the function which was differentiated is called
integration , and clearly it is the reverse operation of differentiation.
Example 6.1 Find what was differentiated to give us.
Here. clearly, the majority of people will say it was 𝑥
2
. Others would say it was
𝑥
2
- 1, May be you may say. All these are correct answers. Thus, we can
receive an infinite number of correct answers. Even , where 𝑐 is a constant, is
also a correct answer. 𝑓
(
𝑥
)
= 𝑥
2
- 𝑐 is a general answer for the integral of.
Therefore,
,
and c is called the constant of integration.
Similarly,
∫ 𝑓
′
(
𝑥
)
𝑑𝑥 = 𝑓
(
𝑥
)
+ 𝑐.
Definition 6.1 Suppose that. Then a function
𝐹(𝑥)=
∫
𝑓(𝑥)𝑑𝑥 is called the indefinite integral of.
Note: If is an integral of with respect to then where c is any
constant is also such an integral. Therefore all indefinite integrals of are
contained in the formula. This is the more reason why the indefinite
integral of is written in the form
.
.
2 x
29
2
x
x c
2
2 x
xdx x c
2
2
xf a x b
dx
dy
( ),
xf )(
Fx )( xf )(
x F x )( c
xf )(
y Fx )( c
xf )(
xf )( dx F x )( c
6.1 Fundamental integration formulae
A number of fundamental integration formulae below follow immediately from the
standard differentiation formulae, while other may be checked by differentiation.
These formulae are important in the evaluation of some integrals.
- If k is any constant then
∫𝑘𝑑𝑥 =𝑘𝑥 + 𝑐.
2. ∫ 𝑥
𝑛
𝑑𝑥 =
1
𝑛+
𝑥
𝑛+
+ 𝑐, 𝑛 ≠ −1.
Example: ∫
𝑥
14
𝑑𝑥 =
1
15
𝑥
15
+ 𝑐
- c, since
𝑑
𝑑𝑥
(ln𝑥)=
1
𝑥
. (This will be proved when we do Further
Differential Calculus.)
4.
∫
1
𝑥+𝑘
𝑑𝑥 =ln|𝑥 + 𝑘|+ 𝑐. (This follows from 3.)
- where k is any constant.
Example: ∫
4𝑥
3
𝑑𝑥 = 4
∫
𝑥
3
𝑑𝑥 = 4 (
1
4
) 𝑥
4
+ 𝑐 = 𝑥
4
+ 𝑐.
6.
Examples: (a) ∫
(
𝑥
2
− 5𝑥
3
)
𝑑𝑥 =∫𝑥
2
𝑑𝑥 − 5∫𝑥
3
𝑑𝑥
=
1
3
𝑥
3
−
5
4
𝑥
4
+ 𝑐.
(b) ∫
(
3𝑥 + 𝑥
2
− 6
)
𝑑𝑥 = 3
∫
𝑥𝑑𝑥 +
∫
𝑥
2
𝑑𝑥 − 6
∫
𝑑𝑥
=
3
2
𝑥
2
+
1
3
𝑥
3
− 6𝑥 + 𝑐
Miscellaneous Examples
Evaluate each of the following indefinite integrals:
1.
∫
√𝑥
2
3
𝑑𝑥.
Solution : ∫
√𝑥
2
3
𝑑𝑥 =
∫
(
𝑥
2
)
1
3
𝑑𝑥 = ∫
𝑥
2
3 𝑑𝑥 =
3
5
𝑥
5
3 + 𝑐
2. ∫
(
3 − 2𝑥
)
2
𝑑𝑥.
Solution : ∫
(3 − 2𝑥)
2
𝑑𝑥 =
∫
(9 − 12𝑥 + 4𝑥
2
)𝑑𝑥
= 9𝑥 − 6𝑥
2
+
4
3
𝑥
3
+ 𝑐.
3.
∫
𝑥
2
−
𝑥−
𝑑𝑥.
Solution :
∫
𝑥
2
−
𝑥−
𝑑𝑥 =
∫
(𝑥−2)(𝑥+2)
𝑥−
𝑑𝑥 =
∫
(
𝑥 + 2
)
𝑑𝑥 =
1
2
𝑥
2
+ 2𝑥 + 𝑐.
dx x
x
ln
1
kf x )( dx k xf )( dx ,
xf )([ g ( x )] dx xf )( dx g x )( dx
Definition 6.2 If , and f is a continuous function on the interval
[𝑎,𝑏] then the definite integral of from to is written as
and is given by
.
The numbers and are called the lower and upper limits of integration,
respectively.
Example: Evaluate the following definite integrals:
1..
Solution:.
2..
Solution: ∫
(3𝑥
2
+ 2)
2
𝑑𝑥 =
1
0
∫
(9𝑥
4
+ 12𝑥
2
+ 4)𝑑𝑥
1
0
= (
9
5
𝑥
5
+ 4𝑥
3
+ 4𝑥)
0
1
= (
9
5
(1)
5
+ 4(1)
3
+ 4
(
1
)
) − (
9
5
(0)
5
+ 4(0)
3
+ 4
(
0
)
)
=
9
5
+ 4 + 4 =
49
5
.
6.2 Properties of Definite Integrals
If and are continuous functions on the interval of integration , then
1.
Example: ∫
3𝑥
2
3
1
𝑑𝑥 = 3
∫
𝑥
2
𝑑𝑥
3
1
= 3(
1
3
)(𝑥
3
)
1
3
=(𝑥
3
)
1
3
=( 3 )
3
−( 1 )
3
= 27 − 1 = 26.
2.
Example: = ∫
2𝑥𝑑𝑥 + 3
∫
𝑑𝑥 =(𝑥
2
)
2
5
5
2
5
2
+ 3(𝑥)
2
5
=
(
5
)
2
−
(
2
)
2
+ 3
[
5 − 2
]
= 25 − 4 + 3(3)
= 30.
xf dx F x c
)( )(
xf )( x a x b
b
a
xf )( dx
xf )( dx F x )( c Fb )( Fa )(
b
a
b
a
a b
5
2
2( x )3 dx
2( )3 3 5( 5(3 )) 2( 2(3 )) 30
2 2
5
2
2
5
2
x dx x x
1
0
2
2
3 x 2 xdx
xf )( g x )( ba ],[
b
a
b
a
cf x )( dx c xf )( dx
b
a
b
a
b
a
xf )([ g ( x )] dx xf )( dx g x )( dx
5
2
2( x )3 dx
- , where.
Example: ∫
3𝑥
2
3
1
𝑑𝑥 =
(
𝑥
3
)
1
3
=
(
3
)
3
−
(
1
)
3
= 27 − 1 = 26.
∫
3𝑥
2
𝑑𝑥 +
∫
3𝑥
2
𝑑𝑥 =
3
2
2
1
(𝑥
3
)
1
2
+(𝑥
3
)
2
3
=
[(
2
)
3
−
(
1
)
3
]
+ [
(
3
)
3
−
(
2
)
3
]
=[8 − 1]+ [ 27 − 8]
= 7 +19 = 26 =
∫
3𝑥
2
3
1
𝑑𝑥.
4..
Example: ∫
3𝑥
2
3
1
𝑑𝑥 = 26.
∫ 3𝑥
2
1
3
𝑑𝑥 =
(
𝑥
3
)
3
1
=
(
1
)
3
−
(
3
)
3
= 1 −27 = − 26
= −
∫
3𝑥
2
3
1
𝑑𝑥.
5..
Example: ∫
1
𝑥
𝑑𝑥
4
4
=(ln 𝑥)
4
4
= ln4 − ln4 = 0
b
c
b
a
c
a
xf )( dx xf )( dx xf )( dx a c b
b
a
a
b
xf )( dx xf )( dx
)( 0
a
a
xf dx
MAT 1100 Integral Calculus I - 2020
Course: Introduction to mathematics (1100)
University: University of Zambia
