INTEGRATION
INTRODUCTION
In mathematics, we are familiar with many pairs of inverse operations: addition and
subtraction, multiplication and division, raising to powers and extracting roots, taking
logarithms and finding antilogarithms, and so on. In this chapter, we discuss the inverse
operation of differentiation, which we call anti-differentiation or more commonly integration.
Integration is therefore the process of finding a function given its derivative or differential,
i.e., moving from 𝑓𝑓′(𝑥𝑥) to 𝑓𝑓(𝑥𝑥). For instance: a physicist who knows the velocity of a particle
might wish to know its position at a given time. An engineer who can measure the variable
rate at which water is leaking from a tank wants to know the amount leaked over a certain
period of time. A biologist who knows the rate at which a bacteria population is increasing
might want to deduce what the size of the population will be at some future time. In each
case, the problem is to find a function 𝐹𝐹 whose derivative is a known function 𝑓𝑓. If such a
function F exists, it is called an antiderivative or an integral of 𝑓𝑓.
LEARNING OUTCOMES
On completion of this chapter, you will be able to:

Do anti-differentiation of basic algebraic functions and use the basic power and log
rules of integration with confidence.

Integrate all algebraic functions including rational functions.

Integrate transcendental functions.

Find the definite integral.

Apply integration to calculate areas and do word problems.
COMPILED BY T. PAEPAE
6.1
STANDARD INTEGRATION
Why it is important to understand: Standard Integration
“Engineering is all about problem solving and many problems in engineering can be solved
using calculus. Physicists, chemists, engineers, and many other scientific and technical
specialists use calculus in their everyday work; it is a technique of fundamental importance.
Both integration and differentiation have numerous applications in engineering and science
and some typical examples include determining areas, mean and rms values, volumes of
solids of revolution, centroids, second moments of area, differential equations and Fourier
series. Besides the standard integrals covered in this chapter, there are a number of other
methods of integration covered in later courses (semester 2). For any further studies in
engineering, differential and integral calculus are unavoidable”. Bird, J., 2017. Higher
engineering mathematics. Routledge.
SPECIFIC OUTCOMES
On completion of this study unit, you will be able to:

Understand that integration is the reverse process of differentiation.

Perform anti-differentiation of basic algebraic functions and explain the meaning of
indefinite integrals.

Apply the basic power rule of integration with confidence and apply the basic log rule
of integration with confidence.
1
6.1.1 The Process of Integration
In differentiation, if 𝑓𝑓(𝑥𝑥) = 2𝑥𝑥 2 then 𝑓𝑓′(𝑥𝑥) = 4𝑥𝑥. Thus, the integral of 4𝑥𝑥 is 2𝑥𝑥 2 . By similar
reasoning, the integral of 2𝑡𝑡 is 𝑡𝑡 2 . In describing this reverse process, an elongated 𝑆𝑆 (called
an integral sign), shown as
∫ is used to replace the words ‘the integral of’. Hence, from
above, ∫ 4𝑥𝑥 = 2𝑥𝑥 2 and ∫ 2𝑡𝑡 = 𝑡𝑡 2 .
Remark, the differential coefficient
𝑑𝑑𝑑𝑑
indicates that a function of 𝑥𝑥 is being differentiated
𝑑𝑑𝑑𝑑
with respect to 𝑥𝑥, the 𝑑𝑑𝑑𝑑 indicating that it is ‘with respect to 𝑥𝑥’. In integration, the variable of
integration is shown by adding 𝑑𝑑(𝑡𝑡ℎ𝑒𝑒 𝑣𝑣𝑣𝑣𝑣𝑣 𝑖𝑖 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) after the function to be integrated. Thus:
 ∫ 4𝑥𝑥 𝑑𝑑𝑑𝑑 means ‘the integral of 4𝑥𝑥 with respect to 𝑥𝑥’, and
 ∫ 2𝑡𝑡 𝑑𝑑𝑑𝑑 means ‘the integral of 2𝑡𝑡 with respect to 𝑡𝑡’.
The function to be integrated (the coefficient of either 𝑑𝑑𝑑𝑑 or 𝑑𝑑𝑑𝑑 above) is known as the
integrand.
The Constant of Integration
As stated above, the differential coefficient of 2𝑥𝑥 2 is 4𝑥𝑥, hence∫ 4𝑥𝑥 𝑑𝑑𝑑𝑑 = 2𝑥𝑥 2 . However, the
differential coefficient of 2𝑥𝑥 2 + 5 is also 4𝑥𝑥. Hence, ∫ 4𝑥𝑥 𝑑𝑑𝑑𝑑 is also equal to 2𝑥𝑥 2 + 5. To allow
for the possible presence of a constant, whenever the process of integration is performed, a
constant ‘𝐶𝐶’ is added to the result. Thus:
∫ 4𝑥𝑥 = 2𝑥𝑥 2 + 𝐶𝐶 and ∫ 2𝑡𝑡 = 𝑡𝑡 2 + 𝐶𝐶
Where ‘𝐶𝐶’ is called the constant of integration.
Note: Integration has one advantage that the result can always be checked by
differentiation. If the function obtained by integration is differentiated, we should get
back the original function (or the integrand).
Example 6.1
1.
Proving the Integrals
Prove that ∫ 4𝑥𝑥 3 𝑑𝑑𝑑𝑑 = 𝑥𝑥 4 + 𝐶𝐶
Solution:
Differentiating the right side gives:
𝑑𝑑 4
(𝑥𝑥 + 𝐶𝐶) = 4𝑥𝑥 3
𝑑𝑑𝑑𝑑
Since 4𝑥𝑥 3 is the coefficient of 𝑑𝑑𝑑𝑑, we have proved that
� 3𝑥𝑥 2 𝑑𝑑𝑑𝑑 = 𝑥𝑥 3 + 𝐶𝐶
2
2
3
Prove that ∫ √𝑥𝑥 𝑑𝑑𝑑𝑑 = √𝑥𝑥 3 + 𝐶𝐶
2.
Solution:
Differentiating the right side gives:
2 3 1
𝑑𝑑 2 3
� 𝑥𝑥 2 + 𝐶𝐶� = � 𝑥𝑥 2 � = √𝑥𝑥
3 2
𝑑𝑑𝑑𝑑 3
This does equal the coefficient of 𝑑𝑑𝑑𝑑, hence we have proved that
6.1.2 Basic Rules of Integration
2
� √𝑥𝑥 𝑑𝑑𝑑𝑑 = �𝑥𝑥 3 + 𝐶𝐶
3
As in example 6.1, we can apply differentiation to prove the following:



∫ 1 𝑑𝑑𝑑𝑑 = 𝑥𝑥 + 𝐶𝐶
𝑥𝑥 2
∫ 𝑥𝑥 𝑑𝑑𝑑𝑑 = 2 + 𝐶𝐶
𝑥𝑥 3
∫ 𝑥𝑥 2 𝑑𝑑𝑑𝑑 = 3 + 𝐶𝐶
Inspection of the above equations shows that the answer has an exponent one greater than
that of the differential and the denominator equals this exponent. This can be generalised as:
Basic power rule for integration:
𝑥𝑥 𝑛𝑛+1
∫ 𝑥𝑥 𝑛𝑛 𝑑𝑑𝑑𝑑 = 𝑛𝑛+1 + 𝐶𝐶
This rule is true when 𝑛𝑛 is fractional, zero, or a positive or negative integer, except for
𝑛𝑛 = −1, because ∫ 𝑥𝑥 −1 𝑑𝑑𝑑𝑑 =
𝑥𝑥 0
+ 𝐶𝐶 and division by zero is undefined.
0
The basic power rule can also be checked by differentiation:
(𝑛𝑛 + 1) 𝑥𝑥 𝑛𝑛+1−1
𝑑𝑑 𝑥𝑥 𝑛𝑛+1
= 𝑥𝑥 𝑛𝑛
�
�=
𝑛𝑛 + 1
𝑑𝑑𝑑𝑑 𝑛𝑛 + 1
When 𝑛𝑛 = −1, we must once again use differentiation to investigate this. From
𝑑𝑑
1
1
differentiation, 𝑑𝑑𝑑𝑑 (ln 𝑥𝑥) = 𝑥𝑥. We can therefore conclude that ∫ 𝑥𝑥 𝑑𝑑𝑑𝑑 = ln|𝑥𝑥| + 𝐶𝐶. This is known
as the basic log or negative-one power rule.
Basic log rule for integration:
1
∫ 𝑥𝑥 𝑑𝑑𝑑𝑑 = ln|𝑥𝑥| + 𝐶𝐶
Note: If 𝑥𝑥 < 0, then ln 𝑥𝑥 is not defined. This is why we take the absolute value of 𝑥𝑥, hence
ln|𝑥𝑥|.
3
Example 6.2
1.
1.1
Applying the Basic Integration Rules
Integrate the following:
∫ 1 𝑑𝑑𝑑𝑑
Solution:
� 1 𝑑𝑑𝑑𝑑 = � 𝑥𝑥 0 𝑑𝑑𝑑𝑑 =
1.2
∫ 𝑥𝑥 4 𝑑𝑑𝑑𝑑
𝑥𝑥 0+1
= 𝑥𝑥 + 𝐶𝐶
0+1
Solution:
� 𝑥𝑥 4 𝑑𝑑𝑑𝑑 =
1.3
1
∫ 𝑥𝑥 2 𝑑𝑑𝑑𝑑
𝑥𝑥 4+1
1
+ 𝐶𝐶 = 𝑥𝑥 5 + 𝐶𝐶
4+1
5
Solution:
�
𝑥𝑥 −2+1
1
1
-2
𝑑𝑑𝑑𝑑
=
�
𝑥𝑥
𝑑𝑑𝑑𝑑
=
+ 𝐶𝐶 = − + 𝐶𝐶
2
−2 + 1
𝑥𝑥
𝑥𝑥
𝑛𝑛
𝑚𝑚
Note: For fractional powers, it is necessary to appreciate √𝑎𝑎𝑚𝑚 = 𝑎𝑎 𝑛𝑛
1.4
∫ √𝑥𝑥 𝑑𝑑𝑑𝑑
Solution:
1
1
𝑥𝑥 2+1
2
� √𝑥𝑥 𝑑𝑑𝑑𝑑 = � 𝑥𝑥 2 𝑑𝑑𝑑𝑑 =
+ 𝐶𝐶 = �𝑥𝑥 3 + 𝐶𝐶
1.5
1
+1
2
1
∫ 3 𝑥𝑥 𝑑𝑑𝑑𝑑
3
√
Solution:
1
1
−
1
−
� 3 𝑑𝑑𝑑𝑑 = � 1 𝑑𝑑𝑑𝑑 = � 𝑥𝑥
√𝑥𝑥
𝑥𝑥 3
1.6
∫4
1
√𝑥𝑥 3
1
3 𝑑𝑑𝑑𝑑 =
𝑑𝑑𝑑𝑑
1
𝑥𝑥 −3+1
3 3
+ 𝐶𝐶 = �𝑥𝑥 2 + 𝐶𝐶
1
2
−3 + 1
Solution:
1
�4
𝑑𝑑𝑑𝑑 = � 3 𝑑𝑑𝑑𝑑 = � 𝑥𝑥
√𝑥𝑥 3
𝑥𝑥 4
3
4 𝑑𝑑𝑑𝑑 =
3
𝑥𝑥 −4+1
+ 𝐶𝐶 = 4 4√𝑥𝑥 + 𝐶𝐶
3
−4 + 1
4
6.1.3 Integration of Certain Combinations of Functions
A constant coefficient inside the integral sign can be moved outside the integral sign:

� 𝑎𝑎 𝑓𝑓(𝑥𝑥) 𝑑𝑑𝑑𝑑 = 𝑎𝑎 � 𝑓𝑓(𝑥𝑥) 𝑑𝑑𝑑𝑑
Example 6.3
1.
Taking Out a Constant Coefficient Inside the Integral Sign
Integrate the following:
∫ 4𝑥𝑥 2 𝑑𝑑𝑑𝑑
1.1
Solution:
� 4𝑥𝑥 2 𝑑𝑑𝑑𝑑 = 4 � 𝑥𝑥 2 𝑑𝑑𝑑𝑑 = 4 �
4
𝑥𝑥 2+1
� + 𝐶𝐶 = 𝑥𝑥 3 + 𝐶𝐶
3
2+1
∫ 2𝑥𝑥�1 − 𝑦𝑦 2 𝑑𝑑𝑑𝑑
1.2
Solution:
� 2𝑥𝑥�1 − 𝑦𝑦 2 𝑑𝑑𝑑𝑑 = 2�1 − 𝑦𝑦 2 � 𝑥𝑥 𝑑𝑑𝑑𝑑 = 𝑥𝑥 2 �1 − 𝑦𝑦 2 + 𝐶𝐶
The integral of a sum/difference equals the sum/difference of the integrals, meaning

term-by-term integration:
�(𝑓𝑓(𝑥𝑥) ± 𝑔𝑔(𝑥𝑥)±. . . ) 𝑑𝑑𝑑𝑑 = � 𝑓𝑓(𝑥𝑥) 𝑑𝑑𝑑𝑑 ± � 𝑔𝑔(𝑥𝑥) 𝑑𝑑𝑑𝑑±. ..
Example 6.4
Integrating Polynomial Functions
Integrate ∫(3𝑥𝑥 4 − 5𝑥𝑥 2 + 𝑥𝑥 + 2) 𝑑𝑑𝑑𝑑
Solution:
∫ (3x − 5 x + x + 2) dx = ∫ 3x dx − ∫ 5 x dx + ∫ xdx + ∫ 2dx
= 3∫ x dx − 5∫ x dx + ∫ xdx + 2 ∫ dx
4
2
=
4
2
4
2
3 5 5 3 1 2
x − x + x + 2x + C
5
3
2
Remark: constant of integration, 𝐶𝐶, is written only once.
5
Here is a warning that may prevent some common mistakes when integrating products and
quotients: the integral of the product of two functions is not equal to the product of the
integrals. The same goes for quotients. That is,
Wrong:
For example,
Wrong:
∫(𝒇𝒇(𝒙𝒙). 𝒈𝒈(𝒙𝒙)) 𝒅𝒅𝒅𝒅 = ∫ 𝒇𝒇(𝒙𝒙) 𝒅𝒅𝒅𝒅 . ∫ 𝒈𝒈(𝒙𝒙) 𝒅𝒅𝒅𝒅
𝑥𝑥 2 𝑥𝑥 2
∫ 𝑥𝑥 (𝑥𝑥 + 1) 𝑑𝑑𝑑𝑑 = (∫ 𝑥𝑥 𝑑𝑑𝑑𝑑) (∫(𝑥𝑥 + 1) 𝑑𝑑𝑑𝑑) = 2 � 2 + 𝑥𝑥� + 𝐶𝐶
𝑥𝑥 4
𝑥𝑥 3
= 4 + 2 + 𝐶𝐶
𝑥𝑥 3
𝑥𝑥 2
Correct:
∫ 𝑥𝑥 (𝑥𝑥 + 1) 𝑑𝑑𝑑𝑑 = ∫(𝑥𝑥 2 + 𝑥𝑥) 𝑑𝑑𝑑𝑑 = 3 + 2 + 𝐶𝐶
Wrong:
∫ 𝒈𝒈(𝒙𝒙) 𝒅𝒅𝒅𝒅 = ∫ 𝒈𝒈(𝒙𝒙) 𝒅𝒅𝒅𝒅
For example,
Wrong:
𝒇𝒇(𝒙𝒙)
∫
∫ 𝒇𝒇(𝒙𝒙) 𝒅𝒅𝒅𝒅
1
=
Correct:
∫
2
� � 𝑥𝑥 +𝑥𝑥
𝑥𝑥+1
∫(𝑥𝑥+1) 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 =
= 2 2 3 + 𝐶𝐶
𝑥𝑥
𝑥𝑥
𝑑𝑑𝑑𝑑
√
√
� � 𝑥𝑥 2
3
3√𝑥𝑥
3
+
+ 𝐶𝐶
4
2√𝑥𝑥
𝑥𝑥+1
1
2
𝑑𝑑𝑑𝑑 = ∫ �√𝑥𝑥 + 𝑥𝑥� 𝑑𝑑𝑑𝑑 = √𝑥𝑥 3 + 2√𝑥𝑥 + 𝐶𝐶
3
√𝑥𝑥
√
Example 6.5 below is integrated by first using basic algebra to simplify the integrand
before integrating. This is important as it will help us see the point of departure when we are
to integrate functions whose integrands cannot be simplified before integrating them.
Example 6.5
1.
Determine
Rewriting Before Integrating
 2x 2 + x − 4 
∫  3x  dx
Solution:
 2x 2 + x − 4 
 2x 2
x
4


dx
=
∫  3x  ∫  3x + 3x − 3x  dx
2
1
4 1
= ∫ x dx + ∫ dx − ∫ dx
3
3
3 x
1
1
4
= x 2 + x − ln x + C
3
3
3
6
2.
Determine
8x 3 − 1
∫ 2 x − 1 dx
Solution:
(
)
(2 x − 1) 4 x 2 + 2 x + 1 dx
8x 3 − 1
=
dx
∫
∫ 2x − 1
(2 x − 1)
(
)
= ∫ 4 x 2 + 2 x + 1 dx
= 4 ∫ x 2 dx + 2 ∫ x dx + ∫ dx
=
3.
Find
∫
4 3
x + x2 + x + C
3
(2 + x )2 dx
x
Solution:
∫
(2 + x )2 dx =
x
∫
4 + 4x + x 2
x
= 4∫ x
−
1
2
1
2
dx
1
2
3
2
dx + 4 ∫ x dx + ∫ x dx
 1
2 3  2 5
= 4 2 x 2  + 4  x 2  + x 2 + C


3  5
=8 x +
8 3 2 5
x +
x +C
3
5
ACTIVITY 1
Integrate the following:
1.
2


∫ x  3x + x 3 − 5  dx
 3 2 5 2

 x − x − 2 x + C 
2.
∫y
2

7
 7 y + C 
2
y dy
7
3.
∫  t − t  dt
2


3 ln t + t + C 
4.
∫ (x + 1)(3x − 2) dx
 3 1 2

 x + 2 x − 2 x + C 
5.
∫ x (x − 4) dx
3 3 7

4
3
 7 x − 3 x + C 
3
2
2
3
4 5 4 3

 5 t − 3 t + t + C 
6.
∫ (2t − 1) dt
7.
∫ 2 2x
8.
2x 4 − x 2
∫ x 3 dx
[x − ln x + C ]
9.
∫ (x − a )(x + a ) dx
 x3

2
 − a x + C
3

10.
x 2 + 5x + 6
∫ x 4 ( x + 2) dx
1
 1

− 2 x 2 − x 3 + C 


11.
∫
2
2
 1

+ C
−
2x


dx
3
2
2 3

 3 x + 2 x + 14 x + C 
x+2 x +7
dx
x
8
6.2
INTEGRATION USING ALGEBRAIC SUBSTITUTIONS
Why it is important to understand: Integration Using Algebraic Substitutions
“Most complex engineering problems cannot be solved without calculus. Calculus has
widespread applications in science, economics and engineering and can solve many
problems for which algebra alone is insufficient. For example, calculus is needed to calculate
the force exerted on a particle a specific distance from an electrically charged wire, and is
needed for computations involving arc length, centre of mass, work and pressure.
Sometimes the integral is not a standard one; in these cases, it may be possible to replace
the variable of integration by a function of a new variable. A change in variable can reduce
an integral to a standard form, and this is demonstrated in this unit”. Bird, J., 2017. Higher
engineering mathematics. Routledge.
SPECIFIC OUTCOMES
On completion of this study unit, you will be able to:

Explain transformation of the variable of basic algebraic functions.

Explain the meaning of indefinite integrals.

Use the general power rule for integration.

Use the general log rule for integration.

Integrate all algebraic functions including rational functions.
9
INTRODUCTION
Functions which require integrating are not always in the ‘standard form’ or ‘integrand cannot
be easily simplified’ as was the case in all the exercises above. However, it is often possible
to change a function into a form which can be integrated by using algebraic substitution or
transformation of variables. This is a technique in which we change the variable of
integration and manipulate the integral until the expression fits one of the standard forms.
6.2.1 Transformation of the Variable
We will use “u” as a dummy variable, that is, u stands for any variable of interest. The
substitution usually made is to let u be equal to f (x) such that f (u ) du is a standard
integral. Say we for instance want to integrate the following functions:
(a)
2
∫ x(x + 1) dx
and
∫ x(x + 1) dx
1
2
2
(b)
Integrating (a) is fairly basic as all we need to do is to first multiply the bracket out and then
integrate each term separately as follows:
∫ x(x + 1) dx = ∫ (x + x ) dx
2
3
=
1 4 1 2
x + x +C
2
4
Integrating (b) is however not too straight forward as the power 1 2 prohibits multiplying out
the bracket. This therefore requires integration using algebraic substitution and this is how
we go about it:
Step 1:
Let u = x
Step 2:
Differentiate u with respect to x and then make dx the subject. That is:
2
+1
du
= 2x
dx
Step 3:
dx =
⇔
du
2x
We then write the integral in term of u as:
∫ x(x + 1) dx = ∫ x(u ) 2 x
2
1
2
1
2
1
= ∫u 2
du
1
du 1 2
=
u du
2 2∫
The original variable ‘ x ’ has been completely removed and the integral is now only in
terms of u and is a standard integral. We can now integrate:
10
1
3
1 2
1 2 2 
=
u
du
u

+C
2∫
2 3 
3
1
= u2 +C
3
Having started with x as the variable of integration, we should get the final
Step 4:
answer in terms of x . We do this by the initial substitution.
3
1 32
1
u + C = (x 2 + 1)2 + C
3
3
Example 6.1
∫x
Determine
2
Change of Variables
x 3 + 1 dx .
Solution:
We make the substitution u = x 3 + 1 because its differential is du = 3 x 2 dx , which, apart from
the constant factor 3 , occurs in the integral. Thus, using dx = du 3x 2 and using the
substitution rule, we have:
∫x
2
x 3 + 1 dx = ∫ x 2 u
du
3x 2
1
1
= ∫ u 2 du
3
3
1 2 
=  u2 +C
3 3 
3
2
x3 + 1 + C
=
9
(
)
Notice that at the final stage we returned to the original variable x . The idea behind the
Substitution Rule is to replace a relatively complicated integral by a simpler integral. This is
accomplished by changing from the original variable x to a new variable u that is a function
of x . Thus, in the Example above, we replaced the integral
∫x
2
x 3 + 1 dx by the simpler
1 12
integral
u du . The main challenge in using the substitution rule is to think of an
3∫
appropriate substitution. You should try to choose u to be some function in the integrand
whose differential also occurs (except for a constant factor). If that is not possible, try
choosing u to be some complicated part of the integrand (perhaps the inner function in a
composite function). The steps used for integration by substitution are summarized in the
following guidelines:
11
GUIDELINES FOR MAKING A CHANGE OF VARIABLES
1.
Choose a substitution u = f ( x ) . Usually, it is best to choose the inner part of a
composite function, such as a quantity raised to a power.
du
.
f ' ( x)
2.
Compute dx =
3.
Rewrite the integral in terms of the variable u ( x must cancel out completely).
4.
Find the resulting integral in terms of u .
5.
Replace u by f (x) to obtain an antiderivative in terms of x .
6.
Check your answer by differentiating (this step may be omitted).
Example 6.2
1.
Change of Variables
Find ∫ 2 x − 1 dx .
Solution:
First, let u be the inner function, u = 2 x − 1 . Then calculate the differential du to be
du = 2 dx . Now, using
2 x −1 = u and dx = du 2 , substitute to obtain:
 du 
∫ 2 x − 1 dx = ∫ u  2 
1
1
= ∫ u 2 du
2
3
1 2 2 
=  u +C
2 3 
3
1
= u2 +C
3
1
(2 x − 1)3 + C
=
3
2.
Find
∫ sin 3x cos 3x dx
2
Solution:
Because sin 2 3 x = (sin 3 x ) , you can let u = sin 3 x .
2
Then
du
= 3 cos 3 x
dx
⟹
12
dx =
1
du
3 cos 3 x
Substituting u and dx in the original integral yields:


∫ sin 3x cos 3x dx = ∫ (u ) cos 3x  3 cos 3x du 
2
2
1
1 2
u du
3∫
1 1 
=  u3  + C
3 3 
1
3
= (sin 3 x ) + C
9
1
= sin 3 3 x + C
9
=
3.
Determine ∫ x
2 x − 1 dx .
Solution:
As in example 1, let u = 2 x − 1 and obtain dx = du 2 . Because the integrand contains
a factor of x , you must also solve for x in terms of u as shown:
u = 2x −1
x=
⇒
u +1
2
Now, using substitution, you obtain:
 u +1
 du 
∫ x 2 x − 1 dx = ∫  2  u  2 
1
1
= ∫ (u + 1) u 2 du
4
=
1
3

1  2
2 

u
u
+
∫
 du

4 

=
5
3
1 2 2 2 2 
u
u
+

+C
4 5
3 
=
1 52 1 32
u + u +C
10
6
=
1
10
(2 x − 1)5 + 1 (2 x − 1)3 + C
6
13
6.2.2 The General Power and Log Rules for Integration
One of the most common u -substitutions involves quantities in the integrand that are raised
to a power. Because of the importance of this type of substitution, it is given a special name:
General Power Rule for Integration. With algebraic substitutions, the substitution usually
made is to let u be equal to f (x) such that f (u ) du is a standard integral. It is found that
integrals of the forms,
∫ f ' ( x)[ f ( x)] dx
n
n ≠ −1
∫ f ' ( x)[ f ( x)] dx
n
and
n = −1
can both be integrated by substituting u for f (x) as follows:

For
∫ f ' ( x)[ f ( x)] dx
n
n ≠ −1
If we let u = f ( x ) . Differentiating u with respect to x gives du = f ' ( x ) dx . Substituting
back results in:
∫ [ f ( x)] f ' ( x) dx = ∫ [u ] du
n
n
u n +1
+C
n +1
[ f (x )]n+1 + C
=
n +1
=
The general power rule becomes:
∫
The general power rule:

For
f ' (x )
∫ [ f (x )] dx
n
n +1
[
f ( x)]
f ' ( x)[ f ( x)] dx =
+C
n
n +1
n ≠ −1
n =1
If we let u = f ( x ) . Differentiating u with respect to x gives du = f ' ( x ) dx . Thus, the
given integral can be written as:
∫ [ f (x )] f ' (x ) dx = ∫ u du
1
1
n
= ln u + C
= ln f ( x ) + C
The general log rule becomes:
14
f ' ( x)
∫ [ f ( x)] dx = ln f ( x) + C
The general log rule:
Example 6.3
1.
n
n =1
Using the General Power and Log Rules of Integration
Find the integral of
∫ x − 2 dx
(use the general power rule)
Solution:
∫
x − 2 dx = ∫ ( x − 2 ) 2 dx
1
Let f ( x) = x − 2 , then f ' ( x) = 1
n
Therefore,
1

2
 



−
1
x
2
∫       dx
f (x)
f ' ( x)
According to the general power rule:
∫
2.
Find integral of
+1
(
2
x − 2)2
(x − 2)3 + C
(1)(x − 2) dx =
+C =
1
1
2
1
+1
2
∫ 5x 1 − 4 x dx
2
3
(use the general power rule)
Solution:
[
]
1
2
2
∫ 5x 1 − 4 x dx = 5∫ x 1 − 4 x 2 dx
Let f ( x) = 1 − 4 x 2 , then f ' ( x) = −8 x
To correctly apply the general power rule, we need − 8 x as the derivative. We
therefore need to multiply the integrand by − 8 since we already have x given. After
multiplying by the constant ( − 8 ), we immediately put the reciprocal of this constant
in front of the integral sign as follows:
15
1
2
−8 x
1 − 4 x 2  dx
5∫ x 1 − 4 x 2  dx = 5∫
−8
1
2
n
f ' ( x)
1

2
5  
2
=
− 4 x  dx
−8 x 1
−8 ∫


f (x)
According to the generalised power rule,
[
]
(
)
3
2
5
5 2
2 2
2 
x
x
dx
x
−
−
=
−
8
1
4
1
4
 +C
−8 ∫
− 8  3

1
=−
3.
Find the integral of
5
12
(1 − 4 x ) + C
2
x
∫ x − 2 dx
2
Solution:
∫ x − 2 dx = ∫ x [x − 2] dx
x
2
−
1
2
Let f ( x) = x 2 − 2 , then f ' ( x) = 2 x
2
∫
−
1
1
−
2
2x
1
2
2
 x 2 − 2  dx =


dx = ∫
2
x
x
2
−
∫


2
2
2
x −2
x
1

1
2
2
=
2 x − 2  + C
2

(
=
)
x2 − 2 + C
16
4.
Find the integral of
∫ (3x + 1) x dx
3
2
Solution:
∫ x (3x + 1) dx
2
3
Let f ( x) = 3 x 2 + 1 , then f ' ( x) = 6 x
dx
6 x ( 3 x + 1)
∫ x ( 3x + 1) =
6∫
2
1
3
2
3
dx
= ∫ f '( x) [ f ( x) ] dx
n
(
)
(
)
4
 2

1  3x + 1 
=
+C

6
4


=
5.
4
1
3x 2 + 1 + C
24
2x 2
Find the integral of ∫
dx
1 − x3
Solution:
Let f ( x) = 1 − x 3 , then f ' ( x) = −3 x 2
2x2
x2
2 −3 x 2
2 f '( x)
2
2
dx
=
dx
=
dx =
− ∫
dx =
− ln 1 − x 3 + C
3
∫ 1 − x3
∫ 1 − x3
∫
3 f ( x)
3
−3 1 − x
(
6.
Find the integral of
x +1
∫ x + 2 x dx
2
Solution:
Let f ( x) = x 2 + 2 x , then f ' ( x) = 2 x + 2 = 2 ( x + 1)
x +1
1 2( x + 1)
1 f '( x)
=
∫ x 2 + 2 x dx 2=
∫ x 2 + 2 x dx 2 ∫ f ( x) dx
1
=
ln x 2 + 2 x + C
2
17
)
Note:
Some integrals whose integrands involve quantities raised to powers cannot be
found by either substitution or the General Power Rule. Consider the two integrals:
∫ x (x + 1) dx and ∫ (x + 1) dx
2
2
2
2
The substitution u = x 2 + 1 works in the first integral but not in the second. In the second, the
substitution fails because the integrand lacks the factor x needed for f ' ( x) . Fortunately, for
(
) = x + 2 x + 1 and use the
this particular integral, you can expand the integrand as x 2 + 1
2
2
4
Basic Power Rule to integrate each term. This is how you solved number 6 in Activity 1.
ACTIVITY 2
Integrate the following:
8

1 2
16 (x + 3) + C 
1.
∫ x (x + 3) dx
2.
∫ (5 x − 3) dx
4

 5 ln 5 x − 3 + C 
3.
∫ 2 x + 3 dx
1
 3
4.
∫ (x + 1) dx
5.
∫ x − 1 dx
 2
− 3
6.
∫ (sin θ ) cosθ dθ
1 6

 6 sin θ + C 
7.
∫ (e + 1) e dx
4

1 x
(
)
e
+
+ C
1
 4

8.
∫x
 1
− 15
7
2
4
2
1− x
5
−3
3

2


+ C
+
+
x
ln
1
2

x +1

2x
x
(2 x + 3)3 + C 
x
3 + 5 x −2 dx
18
(x − 1)3 + C 

(3 + 5 x ) + C 
−2 3

[ln 1 + sin x + C ]
sin 2 x
9.
∫ 1 + sin x dx
10.
∫ x 2 x + 1 dx
1
10
11.
∫ x ax + b dx
 2
15a 2
2
2
(2 x + 1)5 − 1 (2 x + 1)3 + C 

6
(ax + b )3 (3ax − 2b ) + C 

6.2.3 Integrating Rational Functions
Recall that rational functions are functions that can be written in the form
f (x )
, where f ( x )
g (x )
and g ( x ) are polynomials. Some integration problems will arise in forms that look nearly the
same as the rules we have learned thus far, but many will not. In order to apply the rules,
you’ll often need to manipulate or reshape integrands to fit the rules we have learned. The
kinds of manipulation that tend to be useful are expansions, factoring, long division,
completing the square, and rationalisation. In this section, we will deal with rational functions
that involve the general power rule, general log rule, long division and rationalisation.
6.2.3.1
Rational Functions Involving the General Power Rule
You will note that the problems we will deal with here are not too different from the ones we
have already dealt with in section 6.2.2. Although these integrands are given in fractional
form, you will note that the exponent of the denominator in each case is not 1.
Example 6.4
1.
Find:
1.1
∫
Rational Functions Involving the General Power Rule
1
x2 1+ 1
dx
x
Solution:
∫
1
x2
(
)
1
1
dx = ∫ 2 1 + 1 2 dx
x
x
1+ 1
x
Let f ( x) = 1 +
19
1
1
, then f ' ( x) = − 2
x
x
= −∫ −
(
)
1
1
1 + 1 2 dx
2
x
x
= −2 1 + 1 + C
x
1.2
e3x
∫ (e + 2) dx
3x
2
Solution:
∫ (e + 2) dx = ∫ e (e + 2) dx
e3x
3x
3x
−2
3x
Let f ( x) = e 3 x + 2 , then f ' ( x) = 3e 3 x
2
=
(
)
−2
1
3x
3x
+
e
e
dx
3
2
3∫
1
= − 3x
+C
3(e + 2 )
Or else factorise to create the derivative.
Example 6.5
1.
1.1
Factorising the Integrand - Creating the Derivative
Find:
∫ x − 2 x dx
2
4
Solution:
∫
[ (
)] dx
(
) dx
x 2 − 2 x 4 dx = ∫ x 2 1 − 2 x 2
= ∫ x 1 − 2x 2
1
2
1
2
(
1
= − ∫ −4 x 1 − 2 x 2
4
=−
1
6
(a b )m = a m b m
Let f ( x) = 1 − 2 x 2 , then f ' ( x) = − 4 x
1
2
) dx
(1 − 2 x )3 + C
20
1.2
1
∫ x + 2 x + 1 dx
2
Solution:
1
1
∫ x + 2 x + 1 dx = ∫ (x + 1) dx
Factorise the denominator
∫ (x + 1) dx = ∫ (x + 1) dx
Let f ( x) = x + 1 , then f ' ( x) = 1
2
2
1
−2
2
= − ( x + 1) + C
−1
=−
1
+C
x +1
Or else multiply out (when either substitution or general power rule does not work).
Example 6.6
1.
1.1
Multiplying Out Before Integrating
Find:
∫ (x + 3) dx
2
2
Solution:
∫ (x + 3) dx = ∫ (x + 6 x + 9)dx
2
2
4
2
= ∫ x 4 dx + 6 ∫ x 2 + 9 ∫ 1 dx
=
Example 6.7
1.
1.1
x5
+ 2x3 + 9x + C
5
A Disguised Form of the General Power Rule
Find:
ln 2 x
∫ x dx
Solution:
ln 2 x
1
2
∫ x dx = ∫ x (ln x )
Let f ( x) = ln x , then f ' ( x) =
21
1
x
ln 2 x
1
2
n
∫ x dx = ∫ x (ln x ) = ∫ f ' ( x) [ f ( x)]
1
(ln x )3 + C
3
1
= ln 3 x + C
3
=
1.2
∫ cot x ln(sin x ) dx
Solution:
Let f ( x) = ln (sin x ) , then f ' ( x) =
∫ cot x [ln(sin x )] dx
cos x
= cot x
sin x
∫ cot x [ln(sin x )] dx = ∫ f ' ( x) [ f ( x)]
n
1
[ln(sin x )]2 + C
2
1
= ln 2 (sin x ) + C
2
=
1.3
∫ (2 x + 3) x − 4 dx
Solution:
Here, the integrand is in the form of a product, which can be easily reduced to the
standard form, as indicated below:
∫ (2 x + 3) x − 4 dx = ∫ (2 x − 8 + 11) x − 4 dx
= ∫ [2 ( x − 4 ) + 11] x − 4 dx
1
= 2 ∫ (x − 4 ) 2 dx + 11∫ (x − 4 ) 2 dx
3
= 2×
=
4
5
5
3
2
(x − 4) 2 + 11 × 2 (x − 4) 2
5
3
(x − 4)5 + 22 (x − 4)3 + C
3
22
6.2.3.2
Rational Functions Involving the General Log Rule
Example 6.8
Using the Log Rule for Integration
Determine:
1.
1
∫ (x + 2)ln( x + 2) dx
Solution:
∫ (x + 2)ln( x + 2) dx = ∫ (x + 2) [ln( x + 2)] dx
1
1
−1
Let f ( x) = ln ( x + 2 ) , then f ' ( x) =
1
x+2
∫ (x + 2)ln( x + 2) dx = ∫ (x + 2) [ln( x + 2)] dx
1
1
−1
= ln ln(x + 2 ) + C
2.
dx
∫ x (1 − x )
Solution:
∫ x (1 − x ) = ∫ x (1 − x ) dx
1
dx
−1
Let f ( x) = 1 −
x , then f ' ( x) = −
∫ x (1 − x ) dx = −2∫ − 2 x (1 − x ) dx
1
−1
−1
1
= −2 ln 1 − x + C
3.
sin 3θ
∫ 1 − cos 3θ dθ
Solution:
sin 3θ
∫ 1 − cos 3θ dθ
Let f (θ ) = 1 − cos 3θ , then f ' (θ ) = 3 sin 3θ
23
1
2 x
sin 3θ
1
3 sin 3θ
∫ 1 − cos 3θ dθ = 3 ∫ 1 − cos 3θ dθ
1
= ln 1 − cos 3θ + C
3
In solving the above problems, it has been possible to evaluate the integrals in the form of
quotients and products of functions, simply because the integrands can be converted to
standard forms, by applying certain algebraic operations. In fact, there are different methods
for handling integrals involving quotients and products and so on. For example, consider the
following integrals.
6.2.3.3
Rational Functions Involving Long Division
A good rule of thumb when integrating a rational function whose numerator has a degree
greater or equal to that of the denominator is to do long division. This results in a
polynomial plus a simpler rational function as a remainder.
Example 6.9
Using Long Division Before Integrating
Determine:
1.
x 3 + 3x 2 − x − 3
dx
∫
x+2
Solution:
An improper fraction; apply long division
x2 + x − 3
x + 2 x 3 + 3x 2 − x − 3
− x3 − 2x 2
x2 − x
− x 2 − 2x
− 3x − 3
3x + 6
3
x 3 + 3x 2 − x − 3
3 

dx = ∫  x 2 + x − 3 +
 dx
∫
x+2
x + 2

24
=
2.
1 3 1 2
x + x − 3 x + 3 ln x + 2 + C
3
2
x +1
∫ x − 1 dx
Solution:
If the degree of numerator and denominator is same, then creating the same factor as
the denominator (as shown below) is a quicker method than actual division.
x +1
x −1+ 2
∫ x − 1 dx = ∫ x − 1 dx
2 
 x −1
= ∫
+
 dx
 x −1 x −1
2 

= ∫ 1 +
 dx
x −1

= x + 2 ln x − 1 + C
6.2.3.4
Rational Functions Involving Factorisation or Multiplication by One
This type of integrals does not appear to fit any of the basic rules. However, factorising the
denominator transforms the integral into a form that can be integrated.
Example 6.10
Integrating Functions Involving Multiplication by One
Determine:
1.
1
∫ x + x dx
Solution:
1
∫ x + x dx = ∫
=∫
=∫
1

 1
x  x 2 + 1


1
2
x
−
dx
1
2
1
2
1
x 2 +1
x
1
2
−
1
2
x +1
−
dx =
−
1
Let f ( x) = x + 1 , then f ' ( x) = 0,5 x 2
dx
1
2
1 0,5 x
dx
0,5 ∫ 12
x +1
25
= 2 ln x + 1 + C
We could have integrated this problem by first multiplying the numerator and
denominator by x
−
1
2
to transform this integral into a form that can be integrated. This
procedure is referred to as multiplication by one and will be applied in example 2.
2.
2
∫ x + x dx
3
Solution:
 − 13 
x 
dx
2
dx
2
=
∫ x+3 x
∫  1   −1 
 x + x 3   x 3 




Multiplication by one
1
3
= 2  ∫
2
2 −3
x
3
dx
2
3
1
2 −
Let f ( x) = x + 1 , then f ' ( x) = x 3
3
2
x 3 +1
= 3 ln 3 x 2 + 1 + C
Some functions (constant and a variable in the denominator) will require change of variable
method and at times together with long division. These types of problems are generally
solved by letting the term with the square root be equal to u and then manipulating the
integral to be in terms of u alone.
Example 6.11
Functions Involving Change of a Variable & Long Division
Integrate the following:
1.
1
∫ 1 + 2 x dx
Solution:
Let u =
2x
[
]
⇒
du 1
1
−1
= (2 x ) 2 × 2 =
dx 2
2x
⇒
dx = 2 x du
⇒
26
dx = u du
Now, using substitution, we obtain:
 1 
1
u
∫ 1 + 2 x dx = ∫  1 + u  u du = ∫ u + 1 du
Improper fraction; long division
u +1−1
du
u +1
1 
 u +1
= ∫
−
 du
 u +1 u +1
=∫
1 

= ∫ 1 −
 du
 u + 1
= u − ln u + 1 + C
= 2 x − ln 2 x + 1 + C
2.
x
∫ x + 1 dx
Solution:
Let u =
x
⇒
du 1 − 12
1
= (x ) =
dx 2
2 x
⇒
dx = 2 x du
⇒
dx = 2u du
Now, using substitution, we obtain:
∫
u
u2
dx = ∫
2u du = 2 ∫
du
u +1
u +1
x +1
x
Applying long division:
u −1
u + 1 u 2 + 0u + 0
− u2 − u
−u +0
+ u +1
1
Therefore:
1 
u2

2∫
du = 2 ∫  u − 1 +
 du
u +1
u +1

27
Improper fraction

u 2
= 2  − u + ln u + 1  + C
2

2
= u − 2u + 2 ln u + 1 + C
= x − 2 x + 2 ln 2 x + 1 + C
ACTIVITY 3
Integrate the following:
[ x + 2x − 4 + C ]
x +1
1.
∫ x + 2 x − 4 dx
2.
∫ x dx
3.
∫ (x + 6 x − 1) dx
4.
∫
5.
∫ x − 2 x + 1 dx
6.
∫ 2 x − 2 x + 3 dx
7.
∫ (4 x − 12 x + 9) dx
1

4
 8 (2 x − 3) + C 
8.
∫ 2 x − 3x dx
1
 6
2
2
(


−1
+ C
 2

 2 x + 6x − 1
x+3
(
2
2
ln x + 3
dx
x
2
3

2
∫ x + 3x − 1 dx
10.
∫ x ln x dx
4
(2 x − 3) + C 
3
2

[ln x + 3x − 1 + C ]
2 x + 3x
3
9.

2
3
4
(ln x + 3)3 + C 
[ln 2x − 2x + 3 + C ]
2x −1
2
2
)
3 3

5
 5 ( x − 1) + C 

2
3
)
 ln 10

2
 2 log x + C 


log x
4
2
2
1

 3 ln ln x + C 
1
3
28
11.
x2 + x +1
∫ 1 + x 2 dx
12.
e2x
∫ e 2 x + 1 dx
13.
sec 2 x
∫ tan 2 2 x dx
14.
∫ 1 + 2 x dx
x
x 1

 2 − 4 ln 1 + 2 x + C 


15.
log 2 x
∫ x dx
 ln 10

3
 3 log x + C 
1+ x
3
4

+
x
+ C
1
 3

16.
∫
(
[ln e + 1 + C ]
2x
 1

− 2 cot 2 x + C 
(
dx
x
)
1


2
 x + 2 ln x + 1 + C 
)
17.
∫ t ln t dt
ln(ln t )
1 2

 2 ln (ln t ) + C 
18.
x2 + x +1
∫ 1 + x 2 dx
1


2
 x + 2 ln 1 + x + C 


19.
∫
1 + u −1
du
u2
20.
∫x
21.
3x 2 − 7 x
∫ 3x + 2 dx
22.
23.
n −1
(
2
 3
n
4
4

(a + bx ) + C 
n 3

2

n
 nb a + bx + C 
∫ a + bx dx
∫ 1− a x
−1 3
 x2

 − 3 x + 2 ln 3 x + 2 + C 
2

x n −1
b3 x3
(1 + u ) + C 
 2
 3nb

a + bx n dx
)

 b3
4 4
− 4 1 − a x + C 
 2a

dx
29
24.
x3 + 4x 2 − 1
∫ x + 2 dx
 x3

2
 + x − 4 x + 7 ln x + 2 + C 
3

25.
∫ (x − 1) x + 1 dx
2
 5
6.3
INTEGRATING TRANSCENDENTAL FUNCTIONS
(x + 1)5 − 4 (x + 1)3 + C 
3

Why it is important to understand: Integrating Transcendental Functions
So far, we have used only algebraic functions as examples when finding integrals, that is,
functions that can be built up by the usual algebraic operations of addition, subtraction,
multiplication, division, and raising to constant powers. Both in theory and practice there are
other functions, called transcendental, that are very useful. Most important among these are
the trigonometric functions, the inverse trigonometric functions, exponential functions, and
logarithms.
SPECIFIC OUTCOMES
On completion of this study unit, you will be able to:

Integrate exponential functions.

Integrate trigonometric functions.

Integrate composite or mixed functions.
30
6.3.1 Integration of Exponential Functions
These are functions where the independent variable or f (x) is an exponent.
Given y = a f ( x )
dy
= a f ( x ) . f ' ( x) . ln a
dx
dy = a f ( x ) . f ' ( x) . ln a dx
(
)
∫ dy = ∫ (a . f ' ( x) . ln a ) dx
y = ∫ (a ( ) . f ' ( x) . ln a ) dx = a
f (x)
f x
f ( x)
+C
(
)
∴ ∫ a f ( x ) . f ' ( x) . ln a dx = a f ( x ) + C
Given y = e f ( x )
(
)
y = ∫ (e ( ) . f ' ( x) ) dx = e
y = ∫ e f ( x ) . f ' ( x) . ln e dx = e f ( x ) + C
f x
f ( x)
+C
(
)
∴ ∫ e f ( x ) . f ' ( x) dx = e f ( x ) + C
Example 6.1
Integrating Exponential Functions
Find the integrals:
1.
∫ 9e
3x
dx
Solution:
∫ 9e
3x
dx = 9 ∫ e 3 x dx
9 ∫ e 3 x dx =
Let f ( x) = 3 x , then f ' ( x) = 3
9 3x
e × 3 dx
3∫
= 3e 3 x + C
ax
∫ xe dx
2
2.
Solution:
ax
ax
∫ xe dx = ∫ e . x dx
2
2
Let f ( x) = ax 2 , then f ' ( x) = 2ax
31
=
=
3.
2
1
e ax × 2ax dx
∫
2a
1 ax 2
e +C
2a
∫ (ae + be ) dx
−x
x
Solution:
∫ (ae + be ) dx = a ∫ e dx + b ∫ e dx
a ∫ e dx + b ∫ e dx = a ∫ e dx − b ∫ e (−1) dx
−x
x
−x
x
−x
x
−x
x
= ae x − be − x + C
4.
∫ 10
2x
dx
Solution:
∫ 10
∫ 10
Let f ( x) = 2 x , then f ' ( x) = 2
2x
dx
2x
dx =
=
5.
1
10 2 x × (2 ln 10 ) dx
2 ln 10 ∫
10 2 x
+C
2 ln 10
∫ (e + 1) dx
x
2
Solution:
Multiply out
∫ (e + 1) dx
∫ (e + 1) dx = ∫ (e + 2e + 1) dx
∫ (e + 2e + 1) dx = ∫ e dx + 2∫ e dx + ∫ dx
x
2
x
2
2x
2x
x
x
2x
x
1
∫ e dx + 2∫ e dx + ∫ dx = 2 ∫ e × 2 dx + 2∫ e dx + ∫ dx
2x
x
=
2x
x
1 2x
e + 2e x + x + C
2
6.3.2 Integration of Trigonometric Functions
Since integration is the reverse process of differentiation, trigonometric integration rules
(standard integrals) may be deduced and readily checked by differentiation. For instance:
32
In differentiation:
If y = sin x ,
dy
= cos x
dx
Therefore:
dy = cos x dx
∫ dy = ∫ cos x dx
y = ∫ cos x dx = sin x + C
Standard Integrals of Trigonometric Functions
∫ cos 𝑥𝑥 𝑑𝑑𝑑𝑑 = sin 𝑥𝑥 + 𝐶𝐶
∫ sin 𝑥𝑥 𝑑𝑑𝑑𝑑 = − cos 𝑥𝑥 + 𝐶𝐶
∫ csc 𝑥𝑥 cot 𝑥𝑥 𝑑𝑑𝑑𝑑 = − csc 𝑥𝑥 + 𝐶𝐶
∫ sec 𝑥𝑥 tan 𝑥𝑥 𝑑𝑑𝑑𝑑 = sec 𝑥𝑥 + 𝐶𝐶
∫ csc 2 𝑥𝑥 𝑑𝑑𝑑𝑑 = − cot 𝑥𝑥 + 𝐶𝐶
∫ sec 2 𝑥𝑥 𝑑𝑑𝑑𝑑 = tan 𝑥𝑥 + 𝐶𝐶
Since you are familiar with the logarithmic function, we can add four more basic formulae,
which involve the natural log, to the list:
Additional Formulae
� tan 𝑥𝑥 𝑑𝑑𝑑𝑑 = ln|sec 𝑥𝑥| + 𝐶𝐶
� cot 𝑥𝑥 𝑑𝑑𝑑𝑑 = ln|sin 𝑥𝑥| + 𝐶𝐶
� csc 𝑥𝑥 𝑑𝑑𝑑𝑑 = ln|csc 𝑥𝑥 − cot 𝑥𝑥| + 𝐶𝐶
� sec 𝑥𝑥 𝑑𝑑𝑑𝑑 = ln|sec 𝑥𝑥 + tan 𝑥𝑥| + 𝐶𝐶
Two of these additional formulae ( tan x and cot x ) can be derived as done below,
Find:
∫ tan x dx
Solution:
sin x
∫ tan x dx = ∫ cos x dx
Let u = cos x ; then du = − sin x dx
33
⇒∫
sin x
du
dx = − ∫
= − ln u + C
cos x
u
= − ln cos x + C = ln
1
+ C = ln sec x + C
cos x
While the other two ( sec x and cos ecx ) are derived by substitution together with partial
fractions as follows:

∫ sec x – by trickery
The standard trick used to integrate sec x is to multiply the integrand by 1 =
(
sec x + tan x
and
sec x + tan x
)
then substitute u = sec x + tan x , du = sec x tan x + sec 2 x dx .
∫ sec x dx = ∫ sec x

sec x + tan x
sec 2 x + sec x tan x
1
dx = ∫
dx = ∫ du = ln u + C
sec x + tan x
sec x + tan x
u
= ln sec x + tan x + C
∫ sec x – by partial fractions
(will be discussed in detail in semester 2)
Another method of integrating
∫ sec x dx , that is more tedious, but less dependent on a
memorised trick, is to convert
∫ sec x dx into the integral of a rational function using the
substitution u = sin x , du = cos x dx and then use partial fractions.
1
cos x
1
cos x
∫ sec x dx = ∫ cos x dx = ∫ cos x × cos x dx = ∫ cos x dx
2
Since sin 2 x + cos 2 x = 1 , it follows that:
cos x
cos x
∫ cos x dx = ∫ 1 − sin x dx
2
2
Let u = sin x , du = cos x dx . Substituting gives,
cos x
1
1
∫ 1 − sin x dx = ∫ 1 − u du = ∫ (1 + u )(1 − u ) du
2
2
1 1
1 
du
=∫ 
+
2 1 + u 1 − u 
1
= [ln (1 + u ) − ln (1 − u )] + C
2
1 u +1
1 sin x + 1
= ln
+ C = ln
+C
2 u −1
2 sin x − 1
To see that this answer is really the same as the one above, note that
34
1 sin x + 1
1 sin x + 1  sin x + 1 
ln
+ C = ln
×
 +C
2 sin x − 1
2 sin x − 1  sin x + 1 
1 (sin x + 1)
1 (sin x + 1)
1 (sin x + 1)
= ln
= ln
+ C = ln
+C
2
2
2
2
2
sin x − 1
cos 2 x
− cos x
2
= ln
Example 6.2
2
2
sin x + 1
+ C = ln sec x + tan x + C
cos x
Integrating Basic Trigonometric Functions
Find the integrals of the following functions:
1.
∫ 2 sin x dx
Solution:
∫ 2 sin x dx = 2∫ sin x dx
= −2 cos x + C
2.
∫ 4 sec x tan x dx
Solution:
∫ 4 sec x tan x dx = 4∫ sec x tan x dx
= 4 sec x + C
3.
∫ 3 cot x dx
Solution:
∫ 3 cot x dx = 3∫ cot x dx
= 3 ln sin x + C
4.
∫ 5 sec x dx
Solution:
∫ 5 sec x dx = 5∫ sec x dx = 5 ln sec x + tan x + C
The basic integration rules for trigonometric functions can be generalised to enable
integrating functions-of-a-function as follows:
In differentiation, we said if y = sin f ( x) , using the chain rule, the differential coefficient
dy
= f ' ( x) cos f ( x) , therefore:
dx
dy = f ' ( x) cos f ( x) dx
35
∫ dy = ∫ f ' ( x) cos f ( x) dx
y = ∫ f ' ( x) cos f ( x) dx = sin x + C
Integral Formulae for Trigonometric Functions-of-a-Function
� 𝑓𝑓′(𝑥𝑥) sin 𝑓𝑓 (𝑥𝑥)𝑑𝑑𝑑𝑑 = − cos 𝑓𝑓 (𝑥𝑥) + 𝐶𝐶
� 𝑓𝑓′(𝑥𝑥) cos 𝑓𝑓 (𝑥𝑥)𝑑𝑑𝑑𝑑 = sin 𝑓𝑓 (𝑥𝑥) + 𝐶𝐶
� 𝑓𝑓′(𝑥𝑥) tan 𝑓𝑓 (𝑥𝑥)𝑑𝑑𝑑𝑑 = ln|sec 𝑓𝑓 (𝑥𝑥)| + 𝐶𝐶
� 𝑓𝑓′(𝑥𝑥) cot 𝑓𝑓 (𝑥𝑥)𝑑𝑑𝑑𝑑 = ln|sin 𝑓𝑓 (𝑥𝑥)| + 𝐶𝐶
� 𝑓𝑓′(𝑥𝑥) csc 𝑓𝑓(𝑥𝑥) 𝑑𝑑𝑑𝑑 = ln|csc 𝑓𝑓(𝑥𝑥) − cot 𝑓𝑓 (𝑥𝑥)| + 𝐶𝐶
� 𝑓𝑓′(𝑥𝑥) csc 2 𝑓𝑓(𝑥𝑥) 𝑑𝑑𝑑𝑑 = − cot 𝑓𝑓 (𝑥𝑥) + 𝐶𝐶
� 𝑓𝑓′(𝑥𝑥) sec 𝑓𝑓(𝑥𝑥) 𝑑𝑑𝑑𝑑 = ln|sec 𝑓𝑓(𝑥𝑥) + tan 𝑓𝑓(𝑥𝑥)| + 𝐶𝐶
� 𝑓𝑓′(𝑥𝑥) sec 𝑓𝑓 (𝑥𝑥) tan 𝑓𝑓 (𝑥𝑥)𝑑𝑑𝑑𝑑 = sec 𝑓𝑓 (𝑥𝑥) + 𝐶𝐶
� 𝑓𝑓′(𝑥𝑥) sec 2 𝑓𝑓 (𝑥𝑥)𝑑𝑑𝑑𝑑 = tan 𝑓𝑓 (𝑥𝑥) + 𝐶𝐶
� 𝑓𝑓′(𝑥𝑥) csc 𝑓𝑓(𝑥𝑥) cot 𝑓𝑓 (𝑥𝑥)𝑑𝑑𝑑𝑑 = − csc 𝑓𝑓(𝑥𝑥) + 𝐶𝐶
� 𝑓𝑓′(𝑥𝑥) sec 𝑓𝑓 (𝑥𝑥) tan 𝑓𝑓 (𝑥𝑥)𝑑𝑑𝑑𝑑 = sec 𝑓𝑓 (𝑥𝑥) + 𝐶𝐶
36
Example 6.3
Integrating Trigonometric Functions
Find the integrals of the following functions:
1.
∫ cos 3x dx
Solution:
∫ cos 3x dx
Let f ( x) = 3 x , then f ' ( x) = 3
1
∫ cos 3x dx = 3 ∫ 3 cos 3x dx
1
= sin 3 x + C
3
2.
∫ sec(π − t ) tan(π − t ) dt
Solution:
Let f (t ) = π − t , then f ' (t ) = −1
∫ sec(π − t ) tan(π − t ) dt
∫ sec(π − t ) tan(π − t ) dt = −∫ − sec(π − t ) tan(π − t ) dt
= − sec (π − t )
3.
π
∫ cos ec 2 x dx
2
Solution:
π
∫ cos ec 2 x dx
2
π
Let f ( x) =
2 π
π
2
x , then f ' ( x) =
π
2
π
∫ cos ec 2 x dx = π ∫ 2 cos ec 2 x dx
2
=−
4.
2
π
cot
2
π
2
x+C
∫ (3 sin 5 x )(2 + cos 5 x ) dx
10
Solution:
∫ (3 sin 5 x )(2 + cos 5 x ) dx
10
Let f ( x) = 2 + cos 5 x , then f ' ( x) = −5 sin 5 x
3∫ sin 5 x (2 + cos 5 x ) dx = −
10
3
10
− 5 sin 5 x (2 + cos 5 x ) dx
5∫
3
11
= − (2 + cos 5 x ) + C
55
37
∫
5.
x cos 3 x − 5 cos 3 x
dx
x−5
Solution:
∫
(x − 5) cos 3x dx
x cos 3 x − 5 cos 3 x
dx = ∫
( x − 5)
x−5
Take out the common factor
= ∫ cos 3 x dx
=
1
1
3 cos 3 x dx = sin 3 x + C
∫
3
3
Integration Using Trigonometric Identities
The main problem in evaluating integrals lies in converting the integrand to some standard
form. When the integrand involves trigonometric functions, it is sometimes possible to
convert the integrand into a standard form, by applying algebraic operations and/or
trigonometric identities. In such cases, the integrand can be changed to a standard form
without changing the variable of integration. Once this is done, we can easily write the final
result, using the standard formulas.
Note the guidelines below for integrating the following squared trigonometric functions.
∫ sin f ( x) dx
use sin 2 x =
1
(1 − cos 2 x )
2
∫ cos f ( x) dx
use cos 2 x =
1
(1 + cos 2 x )
2
∫ tan f ( x) dx
use 1 + tan 2 x = sec 2 x
∫ cot f ( x) dx
use 1 + cot 2 x = cos ec 2 x
2
2
2
2
Example 6.4
Integrating Trigonometric Functions
Find the integrals of the following function:
1.
∫ cos 2 x dx
2
1
(1 + cos 2 x )
2
1 1
= + cos 2 x
2 2
Use cos 2 x =
Solution:
1
1

∫ cos 2 x dx = ∫  2 + 2 cos 4 x  dx
2
38
1
1
1 dx + ∫ cos 4 x dx
∫
2
2
1 1
1

= x +  sin 4 x  + C
2
2 4

1
1
= x + sin 4 x + C
2
8
=
2.
∫ 1 + sin 2 x dx
Solution:
Here, the integrand is not in the standard form. We consider,
1 + sin 2 x = sin 2 x + cos 2 x + 2 sin x cos x
= sin 2 x + 2 sin x cos x + cos 2 x
= (sin x + cos x )
2
∫ (sin x + cos x ) dx = ∫ (sin x + cos x ) dx
2
= − cos x + sin x + C
= sin x − cos x + C
3.
sin 3 x
∫ cos 3x dx
2
Solution:
Method 1 – Using the General Power Rule
∫ cos 3x dx = ∫ sin 3x (cos 3x )
sin 3 x
−2
2
=−
=−
Let f ( x) = cos 3 x , then f ' ( x) = −3 sin 3 x
dx
1
−2
− 3 sin 3 x (cos 3 x ) dx
3∫
[
]
1
1
1
−1
− (cos 3 x ) = ×
3
3 cos 3 x
1
= sec 3 x + C
3
Method 2 – Using Trigonometric Identities
sin 3 x

1
  sin 3 x 
∫ cos 3x dx = ∫  cos 3x   cos 3x  dx
2
39
= ∫ sec 3 x tan 3 x dx
1
3 sec 3 x tan 3 x dx
3∫
1
= sec 3 x + C
3
=
4.
1
∫ 1 + cos x dx
Solution:
1 − cos x
 1 − cos x 
dx
∫ 1 + cos x ×  1 − cos x  = ∫ 1 − cos x dx
2
1 − cos x
dx
sin 2 x
cos x 
 1
= ∫ 2 −
 dx
2
 sin x sin x 
=∫
= ∫ cos ec 2 x dx − ∫ cos ecx cot x dx
= − cot x + cos ecx + C
Composite or Mixed Functions
This section of work includes integrating functions including algebraic, exponential and
trigonometric functions.
Example 6.5
Integrating Composite Functions
Integrate the following functions:
1.
2 tan 2 x
∫ cos 2 2 x dx
Solution:
(
)
2 tan 2 x
tan 2 x
2
∫ cos 2 2 x dx = ∫ 2 sec 2 x dx
∫2
tan 2 x
(sec 2 x ) dx = 2 ln1 2 ∫ 2
2
tan 2 x
Let f ( x) = tan 2 x , then f ' ( x) = 2 sec 2 x
2
(2 sec 2 x ) ln 2 dx
2
 1  tan 2 x
=
+C
2
 2 ln 2 
40
2.
1
∫ 1 + e dx
x
Solution:
Method 1 – Multiplication by One
 e−x 
e−x
dx


=
×
∫ 1 + e x  e − x  ∫ e − x + 1 dx
Let f ( x) = e
−x
+ 1 , then f ' ( x) = −e − x
− e−x
e−x
∫ e − x + 1 dx = −∫ e − x + 1 dx
= − ln e − x + 1 + C = ln
ex
+C
ex +1
Method 2 – A Disguised Form of the Log Rule
1+ ex − ex
1
∫ 1 + e x dx = ∫ 1 + e x dx
1+ ex
ex 
 dx
−
= ∫ 
x
1 + e x 
1+ e
ex
dx = x − ln 1 + e x + C
= ∫ dx − ∫
x
1+ e
The two answers are the same/equal, except only for appearing in different forms.
3.
∫2
x −1
e ln cos (2 ) dx
x
Solution:
x −1 ln cos (2 )
dx = ∫ 2 x . 2 −1 cos 2 x dx
∫2 e
x
=
1 x
2 cos 2 x dx
∫
2
Let f ( x) = 2 , then f ' ( x) = 2 . ln 2
x
1 x
1 1

cos 2 x . 2 x . ln 2 dx 
2 cos 2 x dx = 
∫
∫
2  ln 2
2

1
sin 2 x + C
2 ln 2
1
=
sin 2 x + C
ln 4
=
41
x
ACTIVITY 4
Integrate the following:
1.
∫ sin 2 p cos x dx
[sin 2 p sin x + C ]
2.
∫e
 1 ax +b

+ C
a e


3.
∫ x2
ax + b
dx
x 2 −1

 1  2 x 2 −1 
 + C
 


 2  ln 2 
dx
4.
∫
(e + 1) dx
[e + 2 x − e + C ]
5.
∫ 3e
[− 3e
6.
e x
∫ x 2 dx
7.
∫ (x + 1)e
8.
 x 1 
∫  e + e x  dx
 1 2x

e −2 x
+ C
 e + 2x −
2
2

9.
∫ cot(4 x + 1) dx
1

 4 ln sin (4 x + 1) + C 

10.
∫ x tan3x dx
1

2
 6 ln sec 3 x + C 

11.
∫
(1 − secθ ) ⋅ tan θ dθ
cos θ
1 2


sec θ − 2 sec θ + C 
12.
∫ sin 2 x dx
1
1

 2 x − 8 sin 4 x + C 
13.
∫ 5 cot 2θ cos ec2θ dθ
 5

− 2 cos ec 2θ + C 


14.
∫ cot 2 x dx
1
− cot 2 x − x + C
2
15.
∫ sin x cos x dx
16.
∫
2
x
e
1− x
−x
x
x
dx
1
1− x
+C
]
− e 1 x + C 


x 2 + 2 x +1
 1 x 2 + 2 x +1

+ C
2 e


dx
2
2
2
2
1 2
 1


 1

2
 2 sin x + C  or − 2 cos x + C  or − 4 cos 2 x + C 
[− cos(ln x) + C ]
sin(ln x)
dx
x
42
2

3
 3 sin t + C 
17.
∫ sin t cos t dt
18.
∫ 4 x + 1 dx
19.
∫
cos x
dx
x
20.
∫
sin x
dx
x
21.
∫ x sec (3x − 1) dx
1

2
 6 tan 3 x − 1 + C 
22.
∫e

 1 tan 3 x
+ C
3 e


23.
∫ cosθ sin (sin θ ) dθ
24.
∫ sec x + tan x dx
25.
1
 24
x
2
tan 3 x
(4 x + 1)3 − 1 4 x + 1 + C 

8
[2 sin x + C ]
[− 2 cos x + C ]
(
2
sec 2 3 x dx
)
[− cos(sin θ ) + C ]
1
[ln 1 + sin x + C ]
∫ ln(cos x ) dx
tan x
[− ln(ln(cos x )) + C ]
26.
∫ cot x [ln (sin x )] dx
1 2

 2 ln (sin x ) + C 

27.
∫
sin (1 x ) cos (1 x )
dx
x2

 1 21
− 2 sin x + C 
28.
∫ x sec x dx
1
3

 1
− 12 tan x 4 + C 

e x
[2e + C ]
2
5
3
4
29.
∫ x dx
30.
∫ cos 2 x dx
1

 2 tan x + C 
tan 2 x
[2 tan x − 2 x + C ]
31.
x
1
2
∫
x
dx
[− ln 10 cot (log x ) + C ]
1
32.
∫ x sin (log x ) dx
33.
∫ 1 + sin x dx
2
[sec x − tan x + x + C ]
sin x
43
34.
∫ tan 3x sec 3x dx
2

3
 9 tan 3 x + C 

35.
∫ tan x sec dx
1

3
 3 tan x + C 


36.
cos ec 2 x
∫ cot 2 x dx
[tan x + C ]
37.
2 sin 2 x
∫ sec 2 x dx
 1

sin 2 x
+ C
 ln 4 . 2

2
2
2
( x ) dx
cos π

 1
π 
− π sin  x  + C 
 


38.
∫
39.
∫ sin nx dx
1
1

 2 x − 4n sin 2nx + C 
40.
1 + cos 2 x
∫ cos 2 x dx
[tan x + x + C ]
41.
∫ (1 + tan x )dx
[tan x + C ]
cos ec 2 1

 1
cot  x  + C 
  

x
2
2
2
( x ) dx
42.
∫
43.
∫ sec x (sec x + tan x ) dx
[tan x + sec x + C ]
44.
∫ (tan x + cot x ) dx
[tan x − cot x + C ]
45.
sin θ + sin θ tan 2 θ
dθ
∫
sec 2 θ
[− cos x + C ]
46.
∫ sin x dx
sin 2 x
[2 sin x + C ]
47.
∫ 1 + cos x dx
[− ln 1 + cos x + C ]
48.
∫ sin t sec (cos t ) dt
[− tan x + C ]
49.
∫ tan x ln(cos x) dx
 1 2

− 2 ln cos x + C 


50.
∫ sec x + tan x dx
tan x
[sec x − tan x + x + C ]
x
2
2
sin 2 x
2
2
2
44
6.4
DEFINITE INTEGRALS
Why it is important to understand: Definite Integrals
In the previous lesson we have discussed the anti-derivative, i.e., integration of a function.
The very word integration means to have some sort of summation or combining of results.
Now the question arises: Why do we study this branch of Mathematics? In fact, the
integration helps to find the areas under various laminas (and volumes of solids of
revolution) when we have definite limits of it. This branch finds applications in a variety of
other problems in Engineering, Statistics, Physics, Biology, Commerce and many more. In
this section, we will define and interpret definite integrals geometrically, evaluate definite
integrals using properties and apply definite integrals to find area of a bounded region.
SPECIFIC OUTCOMES
On completion of this study unit, you will be able to:

Explain the difference between indefinite and definite integration.

Explain the meaning of upper and lower limits.

Evaluate definite integrals.
45
6.4.1 Difference Between Indefinite and Definite Integrals
Integrals containing an arbitrary constant C in their results are called indefinite integrals
since their precise value cannot be determined without further information. Definite
integrals are those with fixed limits, resulting in a numerical number. If an expression is
written as [x ]a , ' ' b' ' is called the upper limit and ' ' a ' ' the lower limit. The operation of
b
applying the limits is defined as [x ]a = (b) − (a ) .
b
b
∫ f ( x) dx
A definite integral has the form:
a
where a and b are usually numbers or values. By definition, the value of definite integral is
b
∫ f ( x) dx = [F ( x) + C ]
b
a
a
= [F (b) + C ] − [F (a ) + C ]
= F (b) − F (a )
Properties of the Definite Integral

∫ f (x ) dx = 0

∫ f (x ) dx = −∫ f (x ) dx

∫ f (x ) dx + ∫ f (x ) dx = ∫ f (x ) dx
a
a
a
b
b
a
c
b
b
a
c
a
where a < c < b
6.4.2 Evaluation of Definite Integrals
Being familiar with all the previous integration and understanding the meaning of definite
integrals with its upper and lower limits, we can evaluate some examples.
Example 6.1
1.
1.1
Evaluating a Definite Integral
Evaluate each definite integral:
∫ (x − 3) dx
2
2
1
Solution:
46
1
2
 x3

x − 3 dx =  − 3 x 
3
1
∫ (
2
)
2
2
8
 1

=  − 6  −  − 3 = −
3
3
 3

3x 2 − 1
∫2 x 3 − x dx
3
1.2
Solution:
[
]
3
3x 2 − 1
3
dx
x
x
ln
=
−
∫2 x 3 − x
2
3
= ln 27 − 3 − ln 8 − 2 = 1,386
1.3
2
∫ e
2x
0
dx
Solution:
2
∫ e
2
2x
0
1

dx =  e 2 x 
2
0
=
π
1.4
(
)
1 4
e − 1 = 26,799
2
4
∫ sec x dx
2
0
Solution:
π
π
4
2
∫ sec x dx = tan x]04
0
= 1− 0
=1
47
ACTIVITY 5
1.
Evaluate the following integrals:
x 2 − 3x + 2
1
1.1
∫
x
0
12 
5
 
dx
∫ 2 x + 1 dx
 26 
 3 
2t 2 + t 2 t − 1
dt
∫1
t2
[32,444]
3
∫ x + 4 dx
[6,85]
2
[0,0714]
4
1.2
0
9
1.3
1.4
5x
2
−1
1.5
dx
∫ (3 − 5 x )
2
1
 98 
 3 

4
1.6
∫ x x + 9 dx
2
0
e
1.7
1
 2 
ln x
∫ x dx
1
∫ (e + e )dt
t
[9,11]
u−2
du
2
 
3
∫ 1 + x ln x  dx
1 
[5,36]
1
∫ (2t − 1) dt
1 
3
 
x
[sin x]
1
1.8
2
0
4
1.9
∫
u
1
e2
1.10

e
1.11
2
0
1.12
∫ cos t dt
0
π
∫ x sin (x )cos(x )dx
1
 32 

4
1
 2 
2
3
1.13
2
2
0
π
1.14
∫ tan x sec x dx
2
0
48
π
1.15
1 − sin 2 θ
∫0 cos 2 θ dθ
π
1.16
π 
4
 
4
π 
4
 
2
∫ sin x dx
2
0
π
1.17
2
sin 2 x
∫
1
1 − sin 2 x
4
0
π
1.18
[1,07]
dx
4
[0,881]
∫ 1 + tan x dx
2
0
5
1.19
x
∫ 2 x − 1 dx
Hint: let u =
16 
 3 
2x − 1
1
π
1.20
[0]
∫ sin x cos x dx
π
2
4
1
2.
Given that
0
∫ 3x x + 4 dx = 5 5 − 8 , what is ∫ 3u u + 4 du ?
2
2
0
3.
If
[8 − 5 5 ]
1
9
9
9
0
0
0
∫ f ( x) dx = 37 and ∫ g ( x) dx = 16 , find ∫ [2 f ( x) + 3g ( x)]dx
[122]
b
4.
Write as a single integral in the form
∫ f ( x) dx :
a
2
5
−1
−2
2
−2
5

 ∫ f ( x) dx 
 −1

∫ f ( x) dx + ∫ f ( x) dx − ∫ f ( x) dx
5.
If
5
5
4
1
4
1
∫ f ( x) dx = 12 and ∫ f ( x) dx = 3,6 , find ∫ f ( x) dx
49
6.5
APPLICATIONS OF INTEGRATION
Why it is important to understand: Applications of Integration
“Engineering is all about problem solving and many problems in engineering can be solved
using integral calculus. One important application is to find the area bounded by a curve;
often such an area can have a physical significance like the work done by a motor, or the
distance travelled by a vehicle. Other examples can involve position, velocity, force, charge
density, resistivity and current density. Electrical currents and voltages often vary with time
and engineers may wish to know the average or mean value of such a current or voltage
over some particular time interval. An associated quantity is the root mean square (rms)
value of a current which is used, for example, in the calculation of the power dissipated by a
resistor. Mean and rms values are required with alternating currents and voltages, pressure
of sound waves, and much more. Revolving a plane figure about an axis generates a
volume, called a solid of revolution, and integration may be used to calculate such a volume.
There are many applications in engineering, and particularly in manufacturing. Centroids of
basic shapes can be intuitive − such as the centre of a circle; centroids of more complex
shapes can be found using integral calculus − as long as the area, volume or line of an
object can be described by a mathematical equation. Centroids are of considerable
importance in manufacturing, and in mechanical, civil and structural design engineering”.
Bird, J., 2017. Higher engineering mathematics. Routledge.
SPECIFIC OUTCOMES
On completion of this study unit, you will be able to:

Calculate the magnitude of an area under a curve.

Calculate the magnitude of an area bounded by the curve and the given axes.

Apply integration to motion (displacement, velocity and acceleration).
50
6.5.1 Determining the Area under a Curve
Up to now, we are all able to calculate the area of common objects (regions with straight
sides) like a square, triangle, cylinder, sphere or polygons. How about finding the area of the
region 𝑆𝑆 that lies under the curve 𝑦𝑦 = 𝑓𝑓(𝑥𝑥) from 𝑎𝑎 to 𝑏𝑏.
𝑦𝑦
𝑦𝑦 = 𝑓𝑓(𝑥𝑥)
S
a
𝑥𝑥
b
Five rectangles, as shown below, can be used to determine this required area. The base of
each rectangle equals ∆𝑥𝑥 =
𝑦𝑦
𝑏𝑏−𝑎𝑎
5
𝑦𝑦1
𝑦𝑦2
𝑦𝑦3
𝑦𝑦4
𝑦𝑦 = 𝑓𝑓(𝑥𝑥)
𝑦𝑦5

a
b
∆x
𝑥𝑥
The heights are 𝑦𝑦1 ; 𝑦𝑦2 , ... and so on. Under these conditions, the approximate area is:
A = y1 ∆ x + y2 ∆ x + y3 ∆ x + y4 ∆ x + y5 ∆ x
= ( y1 + y2 + y3 + y4 + y5 ) ∆ x
Notice that this approximation appears more accurate as the number of strips (rectangles)
increases. In fact, the approximation becomes exact as 𝑛𝑛 → ∞. In summary:
∞
𝑏𝑏
𝐴𝐴 = (𝑦𝑦1 + 𝑦𝑦2 + 𝑦𝑦3 + 𝑦𝑦4 + 𝑦𝑦5 )∆𝑥𝑥 = � 𝑦𝑦𝑛𝑛 ∆𝑥𝑥 = � 𝑓𝑓(𝑥𝑥)𝑑𝑑𝑑𝑑 = 𝐹𝐹(𝑏𝑏) − 𝐹𝐹(𝑎𝑎)
𝑛𝑛=1
51
𝑎𝑎
Hence, we can think of this definite integral as the summation of an indefinite number of
rectangles. This is referred to as the Fundamental Theorem of integration. The following is
a general procedure for calculating the area bounded by the curve and the axis.
1.
Sketch a graph, if not drawn already.
2.
Find any intercepts of 𝑓𝑓(𝑥𝑥) in [𝑎𝑎, 𝑏𝑏]. These divide the total region into sub-regions.
3.
Sketch a representative rectangle and calculate the area as follows:
Area enclosed between the curve and the x-axis:

𝑥𝑥=𝑏𝑏
𝐴𝐴 = �
𝑥𝑥=𝑎𝑎
𝑏𝑏
𝑦𝑦 𝑑𝑑𝑑𝑑 = � 𝑓𝑓(𝑥𝑥) 𝑑𝑑𝑑𝑑
𝑎𝑎
Area enclosed between the curve and the y-axis:

𝑦𝑦=𝑏𝑏
𝐴𝐴 = �
𝑦𝑦=𝑎𝑎
𝑏𝑏
𝑥𝑥 𝑑𝑑𝑑𝑑 = � 𝑓𝑓(𝑦𝑦) 𝑑𝑑𝑑𝑑
𝑎𝑎
Note: Sometimes we may integrate over a region where part of the graph lies above and
part below the x-axis. In this case, the positive area above the axis and the negative
area below the axis must be calculated separately and then only the magnitudes
(absolute values) added to find the required area.
Example 6.1
Finding the Area under a Curve
Compute the area of the figure bounded by the curve y = sin x and the x-axis from
1.
x = 0 to x = π .
∆x
Solution:
π
∫ sin x dx = [ − cos x]
A=
π
0
0
=
− [ cos π − cos 0]
=
− cos π + cos 0
=− (−1) + 1 =1 + 1
= 2 square units
52
Find the area bounded by the lines 𝑥𝑥 = 0, 𝑦𝑦 = 0, 𝑦𝑦 = 1, and 𝑦𝑦 = 𝑥𝑥 2 , as shown in the
2.
figure below.
𝑦𝑦
𝑥𝑥
Solution:
1
Since 𝑥𝑥 = 𝑦𝑦 2
1
1
𝐴𝐴 = � 𝑦𝑦 2 𝑑𝑑𝑦𝑦
0
2 3 1
= � 𝑦𝑦 2 �
3
0
2
= (1 − 0)
3
=
3.
2
3
Find the area bounded by the curve 𝑦𝑦 = cos 𝑥𝑥 between 𝑥𝑥 = 0 and 𝑥𝑥 = 2𝜋𝜋.
Solution:
π 2
A1 = ∫ cos x dx
0
=
sin x ]0
1
[=
π 2
53
3π 2
∫ cos x dx
A2 =
π 2
=
sin x ]π 2
2
[=
3π 2
2π
A3 = ∫ cos x dx
3π 2
=
sin x ]3π 2 1
[=
2π
Required area of the shaded region = A1 + A2 + A3
=1 + 2 + 1
= 4 square units
4.
− x + 4 and y = 0 (i.e., x-axis).
Compute the area bounded by the curve y =
2
Solution:
We have to compute the area of the shaded region that is symmetrical with respect to the
y-axis. We shall compute the area of the region situated in the first quadrant, that is, half of
the total area in question.
2
(
)
1
A = ∫ − x 2 + 4 dx
2
0
2
 x3

=−
+ 4x

 3
0
 8  16
=  − + 8 =
 3  3
A=
32
square units
3
54
Note: In this example, we have chosen to compute half of the area in question (which is
above the x-axis) taking a vertical strip as the elementary area. This approach is definitely
convenient. However, we may as well choose the elementary horizontal strip to compute the
area of the region, which is on the right-hand side of the y-axis. With this approach, the area
𝑦𝑦=𝑏𝑏
in question will be given by the definite integral ∫𝑦𝑦=𝑎𝑎 𝑥𝑥𝑥𝑥𝑥𝑥, (where x is to be replaced in terms
of y and the limits 𝑎𝑎 and 𝑏𝑏, must be found out).
−x + 4
The given curve is y =
2
∴
x2 =
4− y
∴
x=
(4 − y )2
1
To find the limits of the integral, we have, for x = 0 , y = 4 , and for x = 2 , y = 0. (Now we
integrate from lower value of y to upper value of y . ) Thus, half of the area A is given by
4
1
=
A
2
1
∫ ( 4 − y ) 2 dy
0
4
3


2
2
4
−
y
(
)


= −


3

 0
2
 16
=
−  ( −8 )  =
3
 3
A=
32
square units
3
55
ACTIVITY 6
1.
2.
Calculate the shaded area in each of the following:
1.1
y = x e x between x = 0 and x = 1 .
1.2
y=
1.3
x = y3
1.4
y=
1.5
=
x y2 − 2
1.6
y=
2
ln x
between x = 1 and x = 4 .
x
2x
between x = −1 and x = 6 .
x +1
2
1
between x = 3 and x = 8 .
x ln x
Each of the regions A , B , and C bounded by the graph of f and the x -axis has area
2
∫ [ f ( x) + 2 x + 5]dx
3 square units. Find the value of:
[15]
−4
3.
It can be shown that the volume V of a sphere of radius r is given by the integral
∫ π (r − x ) dx
r
2
−r
Show that:
V =
4
π r3
3
56
2
6.5.2 Motion – Distance, Velocity, and Acceleration
For a particle in straight line motion the displacement, s , velocity, v , and acceleration, a ,
are related as follows:
v(t ) =
ds
.......................... (1)
dt
dv d 2 s
=
............... (2 )
dt dt 2
a (t ) =
We can integrate both sides of these equations to give:
From (1): s (t ) = ∫ v (t ) dt
From (2): v(t ) = ∫ a (t ) dt
Remember that, when integrating, we obtain a constant of integration, C and we can usually
find the value of C if we are given some conditions associated with the problem. Let’s look
at some examples to illustrate how we can find the value of C , given some initial conditions.
Example 6.2
1.
Solving Particle Motion Problems
Calculate the displacement that a body undergoes between 𝑡𝑡 = 2 seconds and
𝑡𝑡 = 5 seconds if it is moving according to the equation 𝑣𝑣 = 5𝑡𝑡 − 𝑡𝑡 2 .
Solution:
5
s (t ) = ∫ v(t )dt
2
5
(
)
= ∫ 5t − t 2 dt
2
5
1 
5
=  t2 − t3
3 2
2
= 13,5 m
2.
The acceleration, 𝑎𝑎, of a particle is given by 𝑎𝑎 = cos(3𝑡𝑡). At 𝑡𝑡 = 0, 𝑠𝑠 =
obtain an expression for the displacement, 𝑠𝑠.
Solution:
We first obtain the velocity by using v(t ) = ∫ a (t ) dt with a = cos (3t ) :
∴ v (t ) = ∫ cos (3t ) dt
v(t ) =
sin (3t )
+C
3
57
1
𝑑𝑑𝑑𝑑
and 𝑑𝑑𝑑𝑑 = 0,
3
To find C we use the initial condition, v(0) = 0 . By substitution we get:
∴0 =
sin (0 )
+C
3
∴C = 0
1
⇒ v (t ) = sin (3t )
3
Then we have to integrate v(t ) to get the displacement s (t ) .
s (t ) = ∫ v (t ) dt
sin (3t )
dt
3
cos(3t )
s (t ) = −
+D
9
=∫
To find D we use the initial condition, s (0) =
1
. By substitution we get:
3
1
cos(0 )
∴ =−
+D
3
9
∴D =
4
9
1
4
⇒ s (t ) = − cos(3t ) +
9
9
3.
A particle moves in a straight line from the origin (at t = 0 ) with given velocity
v(0) = 2 ms −1 and the acceleration a = 32 ms −2 . Find the distance at any time.
Solution:
v(t ) = ∫ a (t )dt = ∫ 32dt = 32t + C
But v(0) = 2
v(t ) = 32 (0) + C = 2
∴C = 2
v(t ) = 32t + 2
s (t ) = ∫ v(t )dt = ∫ (32t + 2 ) dt = 16t 2 + 2t + K
But s (0) = 0
0 = 16(0) 2 + 2(0 ) + K
s (t ) = 16t 2 + 2t
58
∴K = 0