1) The document discusses basic rules and concepts of integration, including that integration is the inverse process of differentiation and that the indefinite integral of a function f(x) is notated as ∫f(x) dx = F(x) + c, where F(x) is the primitive function and c is the constant of integration. 2) Methods of integration discussed include the substitution method, where a function is substituted for the variable, and integration by parts, which uses the product rule in reverse to solve integrals involving products. 3) Finding the constant of integration c requires knowing the value of the primitive function F(x) at a specific point, which eliminates the family of functions and isolates a

1. Lecturer: Farzad Javidanrad
Integral Calculus
(for MSc & PhD Business, Management & Finance Students)
(Autumn 2014-2015)
Basic Rules in
Integration

2. • For any operation in mathematics, there is always an inverse operation. For example, summation
and subtraction, multiplication and division. Even for a function 𝑓 there might be an inverse
function 𝑓−1, or for a non-singular square matrix 𝐴, 𝐴−1 can be defined as the inverse.
• For the process of differentiation the reverse process is defined as anti-differentiation or simply
integration.
• If 𝐹′
𝑥 = 𝑓(𝑥) or 𝑑 𝐹 𝑥 = 𝑓 𝑥 𝑑𝑥, then anti-derivative of 𝑓 𝑥 is defined as the indefinite
integral (or primitive function) of 𝑓 𝑥 and is mathematically expressed as:
𝑓 𝑥 𝑑𝑥 = 𝐹 𝑥 + 𝑐
• 𝑓 𝑥 is called the integrand and 𝑐 is the constant of integration and its
presence (in indefinite integration) introduces a family of functions which
have the same derivative in all points in their domain:
𝐹 𝑥 + 𝑐 ′
= 𝑓(𝑥)
The Concept of Integration
Adoptedand altered from http://cbse12math.syncacademy.com/2012/04/mathematics-ch-7-1integration-
as.html
integrand
Element of integration
Indefinite integral,
anti-derivative,
primitive function

3. • Find the indefinite integral for 𝑓 𝑥 = 𝑥4.
According to the definition if 𝐹 𝑥 is such a function we should have:
𝐹(𝑥) ′ = 𝑥4
Obviously, the function
𝑥5
5
satisfies the above equation but other functions such as
𝑥5
5
+ 1,
𝑥5
5
− 3 and
etc. can be considered as an answer so we can write the answer generally:
𝑥4 𝑑𝑥 =
𝑥5
5
+ 𝑐
• Using the definition of the indefinite integral we can find the integral of simple functions directly:
 0 𝑑𝑥 = 𝑐
 1 𝑑𝑥 = 𝑥 + 𝑐
 𝑥 𝑛
𝑑𝑥 =
𝑥 𝑛+1
𝑛+1
+ 𝑐 (𝑛 ≠ −1)
The Concept of Integration

4.  𝑥−1
𝑑𝑥 =
1
𝑥
𝑑𝑥 = 𝑙𝑛 𝑥 + 𝑐 (𝑥 ≠ 0)
 𝑒 𝑥
𝑑𝑥 = 𝑒 𝑥
+ 𝑐
 𝑎 𝑥
𝑑𝑥 =
𝑎 𝑥
𝐿𝑛𝑎
+ 𝑐 (𝑎 ≠ 1, 𝑎 > 0)
 𝑐𝑜𝑠𝑥 𝑑𝑥 = 𝑠𝑖𝑛𝑥 + 𝑐
 𝑠𝑖𝑛𝑥 𝑑𝑥 = −𝑐𝑜𝑐𝑥 + 𝑐
• Rules of Integration:
 The derivative of the indefinite integral is the integrand:
𝑓 𝑥 𝑑𝑥
′
= 𝑓(𝑥)
 The differential of the indefinite integral is equal to the element of the integration:
𝑑 𝑓 𝑥 𝑑𝑥 = 𝑓 𝑥 𝑑𝑥
Rules of Integration

5.  The indefinite integral of a differential of a function is equal to that function plus a constant:
𝑑 𝐹 𝑥 = 𝐹 𝑥 + 𝑐
 If 𝑎 ≠ 0 and is a constant, then:
𝑎. 𝑓 𝑥 𝑑𝑥 = 𝑎. 𝑓 𝑥 𝑑𝑥
 The indefinite integral of the summation (subtraction) of two integrable functions are the
summation (subtraction) of the indefinite integral for each one of them:
𝑓(𝑥) ± 𝑔(𝑥) 𝑑𝑥 = 𝑓 𝑥 𝑑𝑥 ± 𝑔 𝑥 𝑑𝑥
o This result can be extended to the finite number of integrable functions:
𝑓1 𝑥 ± 𝑓2 𝑥 ± ⋯ ± 𝑓𝑛(𝑥) 𝑑𝑥 = 𝑓1 𝑥 𝑑𝑥 ± 𝑓1 𝑥 𝑑𝑥 ± ⋯ ± 𝑓𝑛 𝑥 𝑑𝑥
Rules of Integration
A constant coefficient goes in and
comes out of the integral sign

6. o If they are all added, we can write:
𝑖=1
𝑛
𝑓𝑖 𝑥 𝑑𝑥 =
𝑖=1
𝑛
𝑓𝑖 𝑥 𝑑𝑥
 If 𝑓 𝑡 𝑑𝑡 = 𝐹 𝑡 + 𝑐, then:
𝑓 𝑎𝑥 + 𝑏 𝑑𝑥 =
1
𝑎
. 𝐹 𝑎𝑥 + 𝑏 + 𝑐
And if 𝑏 = 0, then
𝑓 𝑎𝑥 𝑑𝑥 =
1
𝑎
. 𝐹 𝑎𝑥 + 𝑐
• Using the last rule, we can easily calculate some integrals without applying a specific method:
 𝑒 𝛼𝑥 𝑑𝑥 =
1
𝛼
𝑒 𝛼𝑥 + 𝑐
Rules of Integration

7. 
𝑑𝑥
𝑥−𝑎
= 𝑙𝑛 𝑥 − 𝑎 + 𝑐
 sin 𝑚𝑥 =
−cos(𝑚𝑥)
𝑚
+ 𝑐
And some example for other rules:
 𝑥2
− 3𝑥 − 7 𝑑𝑥 = 𝑥2
𝑑𝑥 − 3 𝑥 𝑑𝑥 − 7 𝑑𝑥 =
𝑥3
3
−
3𝑥2
2
− 7𝑥 + 𝑐

𝑥− 𝑥 (𝑥+5)
4
𝑥
𝑑𝑥 =
𝑥2+5𝑥−𝑥 𝑥−5 𝑥
4
𝑥
𝑑𝑥 = 𝑥2−
1
4 𝑑𝑥 + 5 𝑥1−
1
4 𝑑𝑥 − 𝑥1+
1
2
−
1
4 𝑑𝑥 − 5 𝑥
1
2
−
1
4 𝑑𝑥
=
𝑥
11
4
11
4
+ 5 ×
𝑥
7
4
7
4
−
𝑥
9
4
9
4
− 5 ×
𝑥
5
4
5
4
+ 𝑐
= 4𝑥
4
𝑥3 𝑥2
11
+
5
7
− 4𝑥4
𝑥
𝑥
9
+ 1
= 4𝑥4
𝑥(
𝑥2 𝑥
11
+
5 𝑥
7
−
𝑥
9
− 1)
Rules of Integration

8.  Substitution Method: If the integrand is in the form of 𝑓(𝑔 𝑥 ). 𝑔′(𝑥), with substituting 𝑢 = 𝑔(𝑥)
we will have:
𝑓 𝑔 𝑥 . 𝑔′
𝑥 𝑑𝑥 = 𝑓 𝑢 . 𝑢′
𝑑𝑥 = 𝑓 𝑢 𝑑𝑢
And if 𝑓 𝑢 𝑑𝑢 = 𝐹 𝑢 + 𝑐, then:
𝑓 𝑔 𝑥 . 𝑔′ 𝑥 𝑑𝑥 = 𝐹 𝑔 𝑥 + 𝑐
o Find the indefinite integral
𝑥
1+𝑥2 𝑑𝑥.
Let 1 + 𝑥2 = 𝑢, then 2𝑥 𝑑𝑥 = 𝑑𝑢, and we will have:
𝑥
1 + 𝑥2 𝑑𝑥 =
1
2
𝑑𝑢
𝑢
=
1
2
𝑑𝑢
𝑢
=
1
2
𝑙𝑛 𝑢 + 𝑐 =
1
2
𝑙𝑛 1 + 𝑥2 + 𝑐
Methods of Integration
This method
corresponds to the
chain rule in
differentiation.

9. o Find 𝑥 𝑒 𝑥2
𝑑𝑥.
Let 𝑥2 = 𝑢, then 2𝑥 𝑑𝑥 = 𝑑𝑢, and:
𝑥 𝑒 𝑥2
𝑑𝑥 = 𝑒 𝑢
× 1
2 𝑑𝑢 =
1
2
𝑒 𝑢
+ 𝑐 =
1
2
𝑒 𝑥2
+ 𝑐
o Find
𝑑𝑥
𝑥 .𝑙𝑛𝑥
(𝑥 > 0).
Let 𝑙𝑛𝑥 = 𝑢, then
𝑑𝑥
𝑥
= 𝑑𝑢, and:
𝑑𝑥
𝑥 . 𝑙𝑛𝑥
=
𝑥. 𝑑𝑢
𝑥. 𝑢
=
𝑑𝑢
𝑢
= 𝑙𝑛 𝑢 + 𝑐 = 𝑙𝑛 𝑙𝑛 𝑥 + 𝑐
• Note that, having success with this method requires finding a relevant substitution, which comes
after lots of practices.
Methods of Integration

10.  Integration by Parts: This method corresponds to the product rule for differentiation. According to
the product rule, if 𝑢 and 𝑣 are continuous and differentiable functions in term of 𝑥, we have:
𝑑 𝑢𝑣 = 𝑣. 𝑑𝑢 + 𝑢. 𝑑𝑣
or
𝑢. 𝑑𝑣 = 𝑑 𝑢𝑣 − 𝑣. 𝑑𝑢
And we know that:
𝑑 𝐹 𝑥 = 𝐹 𝑥 + 𝑐
Therefore, using the integral notation for we have:
𝑢. 𝑑𝑣 = 𝑢𝑣 − 𝑣. 𝑑𝑢
Methods of Integration
A
A

11. o Find 𝑥. 𝑐𝑜𝑠𝑥 𝑑𝑥.
By choosing 𝑥 = 𝑢 and 𝑐𝑜𝑠𝑥 𝑑𝑥 = 𝑑𝑣, we have:
𝑥 = 𝑢
𝑐𝑜𝑠𝑥 𝑑𝑥 = 𝑑𝑣
⟹
𝑑𝑥 = 𝑑𝑢
𝑠𝑖𝑛𝑥 = 𝑣
Applying the formula, we have:
𝑥. 𝑐𝑜𝑠𝑥 𝑑𝑥 = 𝑥. 𝑠𝑖𝑛𝑥 − 𝑠𝑖𝑛𝑥 𝑑𝑥
= 𝑥. 𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥 + 𝑐
o Find 𝑥 𝑒 𝑥 𝑑𝑥.
By choosing 𝑥 = 𝑢 and 𝑒 𝑥
𝑑𝑥 = 𝑑𝑣, we have:
𝑥 = 𝑢
𝑒 𝑥
𝑑𝑥 = 𝑑𝑣
⟹
𝑑𝑥 = 𝑑𝑢
𝑒 𝑥
= 𝑣
Methods of Integration
No need to add the constant of
integration here when calculating
𝑐𝑜𝑠𝑥 𝑑𝑥. It need to be added just
once at the end

12. Applying the formula, we have:
𝑥 𝑒 𝑥
𝑑𝑥 = 𝑥 𝑒 𝑥
− 𝑒 𝑥
𝑑𝑥
= 𝑥 𝑒 𝑥
− 𝑒 𝑥
+ 𝑐 = 𝑒 𝑥
𝑥 − 1 + 𝑐
o Find 𝑥2 𝑙𝑛𝑥 𝑑𝑥.
By choosing 𝑙𝑛𝑥 = 𝑢 and 𝑥2
𝑑𝑥 = 𝑑𝑣, we have:
𝑙𝑛𝑥 = 𝑢
𝑥2
𝑑𝑥 = 𝑑𝑣
⟹
𝑑𝑥
𝑥 =𝑑𝑢
𝑥3
3 =𝑣
Applying the formula, we have:
𝑥2 𝑙𝑛𝑥 𝑑𝑥 =
𝑥3 𝑙𝑛𝑥
3
−
𝑥3 𝑑𝑥
3𝑥
=
𝑥3 𝑙𝑛𝑥
3
−
1
3
𝑥2 𝑑𝑥
=
𝑥3 𝑙𝑛𝑥
3
−
𝑥3
9
+ 𝑐
Methods of Integration

13. • Sometimes it is needed to use this method more than once to reach to the general solution (primitive
function).
o Find 𝑥2 𝑒 𝑥 𝑑𝑥.
By choosing 𝑥2 = 𝑢 and 𝑒 𝑥 𝑑𝑥 = 𝑑𝑣, we have:
𝑥2
= 𝑢
𝑒 𝑥
𝑑𝑥 = 𝑑𝑣
⟹
2𝑥. 𝑑𝑥 = 𝑑𝑢
𝑒 𝑥
= 𝑣
Applying the formula, we have:
𝑥2
𝑒 𝑥
𝑑𝑥 = 𝑥2 𝑒 𝑥 − 2 𝑥 𝑒 𝑥 𝑑𝑥
Here we need to use the method one more time for 𝑥 𝑒 𝑥
𝑑𝑥. We know from the last page the answer for this
part is 𝑒 𝑥 𝑥 − 1 + 𝑐, so the final answer is:
𝑥2 𝑒 𝑥 − 2 𝑒 𝑥 𝑥 − 1 + 𝑐
Methods of Integration

14. • There are many other methods such as integration by partial fractions, integration for trigonometric
functions, integration using series, but they are out of scope of this module.
How to find the constant of Integration?
• If the primitive function is passing through a point, then we have a single function out of the family
of functions. That point, which should belong to the domain of the function is called initial value or
initial condition.
o Find 𝑥2 𝑥3 − 5 𝑑𝑥, when 𝑦 0 = 2.
Through the substitution method the indefinite integral is:
𝑥2
𝑥3
− 5 𝑑𝑥 = 𝑥3
− 5
𝑑(𝑥3)
3
=
1
3
𝑢 − 5 𝑑𝑢
=
1
3
𝑢2
2
− 5𝑢 + 𝑐 =
𝑥6
6
−
5
3
𝑥3
+ 𝑐
As when 𝑥 = 0 then 𝑦 = 2, so 2 = 0 + 𝑐 ⟹ 𝑐 = −2
Therefore, the function will be: 𝑦 =
𝑥6
6
−
5
3
𝑥3 − 2
Initial Conditions in Integral Calculus

15. • There is a very important relation between the concept of indefinite integral of the function 𝑓(𝑥)
and the area under the curve of this function over the given interval.
• Imagine we are going to find the area under the curve 𝑦 = 𝑓(𝑥) over the interval [𝑎, 𝑏] (see Figure
2). One way to calculate this area is to divide the area into 𝑛 equal sub-intervals such as
𝑎, 𝑥1 , 𝑥1, 𝑥2 , … , [𝑥 𝑛−1, 𝑏], (each with the length equal to ∆𝑥) and construct rectangles (see the
figure 1).
• Obviously, the area under the curve can be estimated as 𝑅 = 𝑖=1
𝑛
𝑓 𝑥𝑖
∗
. ∆𝑥 (which is called a
Riemann Sum) and our approximation of this sum gets better and better if the number of sub-
intervals goes to infinity, which is equivalent
to say ∆𝑥 → 0, in this case the value of the
area approaches to a limit:
𝑆 = lim
𝑛→∞
𝑖=1
𝑛
𝑓 𝑥𝑖
∗
. ∆𝑥
The Definite Integral
Adoptedfrom Calculus Early Transcendental James Stewart p367

16. • We call this sum as a definite integral of y = 𝑓(𝑥) over the interval [𝑎, 𝑏] and it can be shown as
𝑥0=𝑎
𝑥 𝑛=𝑏
𝑓 𝑥 𝑑𝑥 or simply 𝑎
𝑏
𝑓 𝑥 𝑑𝑥. Therefore:
𝑎
𝑏
𝑓 𝑥 𝑑𝑥 = lim
𝑛→∞
𝑖=1
𝑛
𝑓 𝑥𝑖
∗
. ∆𝑥
• In this definition 𝑎 is the lower limit of the definite integral and 𝑏 is called the upper limit.
• The definite integral is a number so it is not sensitive to be represented by different variables. i.e.:
𝑎
𝑏
𝑓 𝑥 𝑑𝑥 =
𝑎
𝑏
𝑓 𝑧 𝑑𝑧 =
𝑎
𝑏
𝑓 𝜃 𝑑𝜃
The Definite Integral
• If 𝑓(𝑥) takes both positive and negative values, the area
under the curve and confined by the x-axis and lines
𝑥 = 𝑎 and 𝑥 = 𝑏 is the sum of areas above the x-axis
minus the sum of areas under the x-axis, i.e.:
𝑎
𝑏
𝑓 𝑥 𝑑𝑥 = Positive Areas − Negative Areas
Adoptedfrom Calculus Early Transcendental James Stewart p367

17. All of the properties of an indefinite integral can be extended into the definite integral (see slides
5&6), but there are some specific properties for definite integrals such as:
 If 𝑎 = 𝑏, then the area under the curve is zero:
𝑎
𝑎
𝑓 𝑥 𝑑𝑥 = 0
 If 𝑎 and 𝑏 exchange their position the new definite integral is the negative of the previous integral:
𝑏
𝑎
𝑓 𝑥 𝑑𝑥 = −
𝑎
𝑏
𝑓 𝑥 𝑑𝑥
 If 𝑦 = 𝑐 over the interval [𝑎, 𝑏] , then:
𝑎
𝑏
𝑓 𝑥 𝑑𝑥 =
𝑎
𝑏
𝑐 𝑑𝑥 = 𝑐(𝑏 − 𝑎)
Properties of Definite Integrals
AdoptedfromCalculusEarlyTranscendentalJamesStewartp373

18.  If 𝑓(𝑥) ≥ 0 and it is continuous over the interval [𝑎, 𝑏] and the interval can be divided into sub-
intervals such as 𝑎, 𝑥1 , 𝑥1, 𝑥2 , …, [𝑥 𝑛−1, 𝑏], then:
𝑎
𝑏
𝑓 𝑥 𝑑𝑥 =
𝑎
𝑥1
𝑓 𝑥 𝑑𝑥 +
𝑥1
𝑥2
𝑓 𝑥 𝑑𝑥 + ⋯ +
𝑥 𝑛−1
𝑏
𝑓 𝑥 𝑑𝑥 ≥ 0
 If 𝑓(𝑥) ≥ 𝑔(𝑥) and both are continuous over the interval [𝑎, 𝑏], then:
𝑎
𝑏
𝑓 𝑥 𝑑𝑥 ≥
𝑎
𝑏
𝑔 𝑥 𝑑𝑥
Properties of Definite Integrals
Adoptedfrom http://www.math24.net/properties-of-definite-integral.html
Adoptedfrom https://www.math.hmc.edu/calculus/tutorials/riemann_sums/

19. • The fundamental theorem of calculus asserts that there is a
specific relation between the area under a curve and the
indefinite integral of that curve.
• Another word, the value of the area under the continuous
curve 𝑦 = 𝑓 𝑥 in the interval [𝑎, 𝑏] can be calculated
directly through the difference between two boundary
values of any primitive function of 𝑓(𝑥); i.e. 𝐹 𝑏 − 𝐹(𝑎).
• Mathematically, we can express this fundamental
theorem as:
𝑎
𝑏
𝑓 𝑥 𝑑𝑥 = 𝐹 𝑏 − 𝐹(𝑎)
Where 𝐹(𝑥) is any anti-derivative (primitive) function
of 𝑓(𝑥).
The Fundamental Theorem of Calculus
𝑓(𝑥) = 2𝑥
𝐹 𝑏 − 𝐹 𝑎 = 9 − 1 = 8
Area=
𝒄
𝟐
𝒂 + 𝒃 = 𝟏 × 𝟐 + 𝟔 = 𝟖
Adoptedand altered from http://www.bbc.co.uk/schools/gcsebitesize/maths/algebra/transformationhirev1.shtml
𝐹 𝑥 = 𝑥2

20. o Evaluate the integral 1
3
2 𝑥
𝑑𝑥.
This is a continuous function in its whole domain, so there is no discontinuity in the interval [1,3].
One of the anti-derivative function for 𝑓 𝑥 = 2 𝑥 is 𝐹 𝑥 =
2 𝑥
𝑙𝑛2
, so, we have:
1
3
2 𝑥
𝑑𝑥 = 𝐹 3 − 𝐹 1 =
6
𝑙𝑛2
• We often use the following notation for its simplicity:
𝑎
𝑏
𝑓 𝑥 𝑑𝑥 = 𝐹(𝑥) 𝑎
𝑏 = 𝐹 𝑏 − 𝐹(𝑎)
o Evaluate the area between 𝑓 𝑥 = 𝑠𝑖𝑛𝑥
and 𝑔 𝑥 = 𝑐𝑜𝑠𝑥 in the interval [0,
𝜋
2
].
We know these functions cross each other
when 𝑥 =
𝜋
4
(see the graph)
Some Examples
Adoptedfromhttp://maretbccalculus2007-
2008.pbworks.com/w/page/20301409/Find%20The%20Area
%20of%20a%20Region

21. • We always need to find out which function is on the top and which is at the bottom(in any sub-
interval). In this example, in the interval [0,
𝜋
4
], the curve of 𝑔 𝑥 = 𝑐𝑜𝑠𝑥 is on the top and 𝑓 𝑥 =
𝑠𝑖𝑛𝑥 is at the bottom but in the second interval [
𝜋
4
,
𝜋
2
], it is vice-versa. So:
𝐴𝑟𝑒𝑎 =
0
𝜋
4
𝑐𝑜𝑠𝑥 − 𝑠𝑖𝑛𝑥 𝑑𝑥 +
𝜋
4
𝜋
2
𝑠𝑖𝑛𝑥 − 𝑐𝑜𝑠𝑥 𝑑𝑥
= 𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥 0
𝜋
4
+ −𝑐𝑜𝑠𝑥 − 𝑠𝑖𝑛𝑥 𝜋
4
𝜋
2
= 𝑠𝑖𝑛
𝜋
4
+ 𝑐𝑜𝑠
𝜋
4
− (𝑠𝑖𝑛0 + 𝑐𝑜𝑠0) + −𝑐𝑜𝑠
𝜋
2
− 𝑠𝑖𝑛
𝜋
2
− −𝑐𝑜𝑠
𝜋
4
− 𝑠𝑖𝑛
𝜋
4
= 2 2 − 1
Some Examples

22. • In definite integral 𝑎
𝑏
𝑓 𝑡 𝑑𝑡 if the upper limit replaced with 𝑥 (where 𝑥 varies in the interval [𝑎, 𝑏]) ,
in this case the area under the curve depends only on 𝑥 :
𝑎
𝑥
𝑓 𝑡 𝑑𝑡 = 𝐹 𝑥 − 𝐹 𝑎 = 𝑔 𝑥
• If 𝑥 varies, so the derivative of 𝑔 𝑥 with respect to 𝑥 is:
𝑎
𝑥
𝑓 𝑡 𝑑𝑡
′
= 𝑓 𝑥
Why?
• If 𝑎 or 𝑏 goes to infinity, then we are dealing with improper integrals. As infinity is not a number it
cannot be substituted for 𝑥. They must be defined as the limit of a proper definite integral.
−∞
𝑏
𝑓 𝑥 𝑑𝑥 = lim
𝑎→−∞ 𝑎
𝑏
𝑓 𝑥 𝑑𝑥 and
𝑎
+∞
𝑓 𝑥 𝑑𝑥 = lim
𝑏→+∞ 𝑎
𝑏
𝑓 𝑥 𝑑𝑥
Improper Integrals
Adoptedfrom Calculus Early Transcendental James Stewart p380

23. o Find 1
+∞ 5
𝑥2 𝑑𝑥.
1
+∞
5
𝑥2
𝑑𝑥 = lim
𝑏→+∞ 1
𝑏
5
𝑥2
𝑑𝑥 = lim
𝑏→+∞
−5
𝑥 1
𝑏
= lim
𝑏→+∞
−5
𝑏
+ 5 = 5
Improper Integrals