Integral calculus

Lecturer: Farzad Javidanrad

Integral Calculus

(for MSc & PhD Business, Management & Finance Students)

(Autumn 2014-2015)

Basic Rules in Integration

• 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

Adopted and altered from http://cbse12math.syncacademy.com/2012/04/mathematics-ch-7-1integration-as.html

integrand

Element of integration

Indefinite integral,anti-derivative,

primitive function

• 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

5satisfies 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:

𝑥4𝑑𝑥 =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

𝑥𝑛 𝑑𝑥 =𝑥𝑛+1

𝑛+1+ 𝑐 (𝑛 ≠ −1)

The Concept of Integration

𝑥−1 𝑑𝑥 = 1

𝑥𝑑𝑥 = 𝑙𝑛 𝑥 + 𝑐 (𝑥 ≠ 0)

𝑒𝑥 𝑑𝑥 = 𝑒𝑥 + 𝑐

𝑎𝑥 𝑑𝑥 =𝑎𝑥

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

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


A constant coefficient goes in and comes out of the integral sign

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:

𝑒𝛼𝑥𝑑𝑥 =1

𝛼𝑒𝛼𝑥 + 𝑐


𝑑𝑥

𝑥−𝑎= 𝑙𝑛 𝑥 − 𝑎 + 𝑐

sin 𝑚𝑥 =−cos(𝑚𝑥)

𝑚+ 𝑐

And some example for other rules:

𝑥2 − 3𝑥 − 7 𝑑𝑥 = 𝑥2𝑑𝑥 − 3 𝑥 𝑑𝑥 − 7 𝑑𝑥 =𝑥3

3−

3𝑥2

2− 7𝑥 + 𝑐

𝑥− 𝑥 (𝑥+5)

4 𝑥𝑑𝑥 =

𝑥2+5𝑥−𝑥 𝑥−5 𝑥4 𝑥

𝑑𝑥 = 𝑥2−

1

4𝑑𝑥 + 5 𝑥1−

1

4𝑑𝑥 − 𝑥1+

1

2−1

4𝑑𝑥 − 5 𝑥1

2−1

4 𝑑𝑥

=𝑥11

4

11

4

+ 5 ×𝑥7

4

7

4

−𝑥9

4

9

4

− 5 ×𝑥5

4

5

4

+ 𝑐

= 4𝑥4𝑥3

𝑥2

11+

5

7− 4𝑥4 𝑥

𝑥

9+ 1

= 4𝑥4 𝑥(𝑥2 𝑥

11+

5 𝑥

7−

𝑥

9− 1)


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+𝑥2𝑑𝑥.

Let 1 + 𝑥2 = 𝑢, then 2𝑥 𝑑𝑥 = 𝑑𝑢, and we will have:

𝑥

1 + 𝑥2𝑑𝑥 =

12 𝑑𝑢

𝑢=1

2 𝑑𝑢

𝑢

=1

2𝑙𝑛 𝑢 + 𝑐 =

1

2𝑙𝑛 1 + 𝑥2 + 𝑐

Methods of Integration

This method corresponds to the

chain rule in differentiation.

o Find 𝑥 𝑒𝑥2𝑑𝑥.

Let 𝑥2 = 𝑢, then 2𝑥 𝑑𝑥 = 𝑑𝑢, and:

𝑥 𝑒𝑥2𝑑𝑥 = 𝑒𝑢 × 1

2 𝑑𝑢 =1

2𝑒𝑢 + 𝑐 =

1

2𝑒𝑥

2+ 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.


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


A

A

o Find 𝑥. 𝑐𝑜𝑠𝑥 𝑑𝑥.

By choosing 𝑥 = 𝑢 and 𝑐𝑜𝑠𝑥 𝑑𝑥 = 𝑑𝑣, we have:

𝑥 = 𝑢

𝑐𝑜𝑠𝑥 𝑑𝑥 = 𝑑𝑣⟹

𝑑𝑥 = 𝑑𝑢𝑠𝑖𝑛𝑥 = 𝑣

Applying the formula, we have:

𝑥. 𝑐𝑜𝑠𝑥 𝑑𝑥 = 𝑥. 𝑠𝑖𝑛𝑥 − 𝑠𝑖𝑛𝑥 𝑑𝑥

= 𝑥. 𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥 + 𝑐

o Find 𝑥 𝑒𝑥𝑑𝑥.

By choosing 𝑥 = 𝑢 and 𝑒𝑥𝑑𝑥 = 𝑑𝑣, we have:

𝑥 = 𝑢

𝑒𝑥𝑑𝑥 = 𝑑𝑣⟹

𝑑𝑥 = 𝑑𝑢𝑒𝑥 = 𝑣


No need to add the constant of integration here when calculating

𝑐𝑜𝑠𝑥 𝑑𝑥. It need to be added just

once at the end


𝑥 𝑒𝑥𝑑𝑥 = 𝑥 𝑒𝑥 − 𝑒𝑥𝑑𝑥

= 𝑥 𝑒𝑥 − 𝑒𝑥 + 𝑐 = 𝑒𝑥 𝑥 − 1 + 𝑐

o Find 𝑥2𝑙𝑛𝑥 𝑑𝑥.

By choosing 𝑙𝑛𝑥 = 𝑢 and 𝑥2𝑑𝑥 = 𝑑𝑣, we have:

𝑙𝑛𝑥 = 𝑢

𝑥2𝑑𝑥 = 𝑑𝑣⟹

𝑑𝑥𝑥 =𝑑𝑢

𝑥3

3 =𝑣


𝑥2𝑙𝑛𝑥 𝑑𝑥 =𝑥3𝑙𝑛𝑥

3−

𝑥3𝑑𝑥

3𝑥=𝑥3𝑙𝑛𝑥

3−1

3 𝑥2 𝑑𝑥

=𝑥3𝑙𝑛𝑥

3−

𝑥3

9+ 𝑐


• Sometimes it is needed to use this method more than once to reach to the general solution (primitive function).

o Find 𝑥2 𝑒𝑥𝑑𝑥.

By choosing 𝑥2 = 𝑢 and 𝑒𝑥𝑑𝑥 = 𝑑𝑣, we have:

𝑥2 = 𝑢𝑒𝑥𝑑𝑥 = 𝑑𝑣

⟹ 2𝑥. 𝑑𝑥 = 𝑑𝑢𝑒𝑥 = 𝑣


𝑥2 𝑒𝑥𝑑𝑥 = 𝑥2𝑒𝑥 − 2 𝑥 𝑒𝑥 𝑑𝑥

Here we need to use the method one more time for 𝑥 𝑒𝑥 𝑑𝑥. We know from the last page the answer for this

part is 𝑒𝑥 𝑥 − 1 + 𝑐, so the final answer is:

𝑥2𝑒𝑥 − 2𝑒𝑥 𝑥 − 1 + 𝑐


• 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 − 5𝑑(𝑥3)

3=1

3 𝑢 − 5 𝑑𝑢

=1

3

𝑢2

2− 5𝑢 + 𝑐 =

𝑥6

6−5

3𝑥3 + 𝑐

As when 𝑥 = 0 then 𝑦 = 2, so 2 = 0 + 𝑐 ⟹ 𝑐 = −2

Therefore, the function will be: 𝑦 =𝑥6

6−

5

3𝑥3 − 2

Initial Conditions in Integral Calculus

• 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 , … , [𝑥𝑛−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=1𝑛 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 = lim𝑛→∞

𝑖=1

𝑛

𝑓 𝑥𝑖∗ . ∆𝑥

The Definite Integral

Adopted from Calculus Early Transcendental James Stewart p367

• 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

𝑥𝑛=𝑏 𝑓 𝑥 𝑑𝑥 or simply 𝑎

𝑏𝑓 𝑥 𝑑𝑥. Therefore:

𝑎

𝑏

𝑓 𝑥 𝑑𝑥 = lim𝑛→∞

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


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


𝑏

𝑐 𝑑𝑥 = 𝑐(𝑏 − 𝑎)

Properties of Definite Integrals

Ad

op

ted

fro

m C

alcu

lus

Earl

y Tr

an

scen

den

tal J

am

es S

tew

art

p3

73

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 , …, [𝑥𝑛−1, 𝑏], then:

𝑎

𝑏


𝑥1

𝑓 𝑥 𝑑𝑥 + 𝑥1

𝑥2

𝑓 𝑥 𝑑𝑥 + ⋯+ 𝑥𝑛−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

Adopted from http://www.math24.net/properties-of-definite-integral.html

Adopted from https://www.math.hmc.edu/calculus/tutorials/riemann_sums/

• The fundamental theorem of calculus asserts that there is a specific relation between the area under a curve and theindefinite integral of that curve.

• Another word, the value of the area under the continuouscurve 𝑦 = 𝑓 𝑥 in the interval [𝑎, 𝑏] can be calculateddirectly through the difference between two boundaryvalues 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

𝟐𝒂 + 𝒃 = 𝟏 × 𝟐 + 𝟔 = 𝟖

Adopted and altered from http://www.bbc.co.uk/schools/gcsebitesize/maths/algebra/transformationhirev1.shtml

𝐹 𝑥 = 𝑥2

o Evaluate the integral 1

32𝑥𝑑𝑥.

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 = 2𝑥 is 𝐹 𝑥 =2𝑥

𝑙𝑛2, so, we have:

1

3

2𝑥𝑑𝑥 = 𝐹 3 − 𝐹 1 =6

𝑙𝑛2

• We often use the following notation for its simplicity:

𝑎

𝑏

𝑓 𝑥 𝑑𝑥 = 𝐹(𝑥) 𝑎𝑏 = 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

Ad

op

ted

fro

m h

ttp

://m

aret

bcc

alcu

lus2

007

-20

08.p

bw

ork

s.co

m/w

/pag

e/2

0301

409

/Fin

d%20

The%

20A

rea

%20

of%

20a%

20R

egio

n

• 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

𝜋4𝑐𝑜𝑠𝑥 − 𝑠𝑖𝑛𝑥 𝑑𝑥 +

𝜋4

𝜋2𝑠𝑖𝑛𝑥 − 𝑐𝑜𝑠𝑥 𝑑𝑥

= 𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥 0

𝜋4 + −𝑐𝑜𝑠𝑥 − 𝑠𝑖𝑛𝑥 𝜋

4

𝜋2

= 𝑠𝑖𝑛𝜋

4+ 𝑐𝑜𝑠

𝜋

4− (𝑠𝑖𝑛0 + 𝑐𝑜𝑠0) + −𝑐𝑜𝑠

𝜋

2− 𝑠𝑖𝑛

𝜋

2− −𝑐𝑜𝑠

𝜋

4− 𝑠𝑖𝑛

𝜋

4= 2 2 − 1

Some Examples

• In definite integral 𝑎

𝑏𝑓 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 = lim𝑎→−∞

𝑎

𝑏

𝑓 𝑥 𝑑𝑥 and 𝑎

+∞

𝑓 𝑥 𝑑𝑥 = lim𝑏→+∞

𝑎

𝑏

𝑓 𝑥 𝑑𝑥

Improper Integrals


o Find 1

+∞ 5

𝑥2𝑑𝑥.

1

+∞ 5

𝑥2𝑑𝑥 = lim

𝑏→+∞ 1

𝑏 5

𝑥2𝑑𝑥 = lim

𝑏→+∞

−5

𝑥1

𝑏

= lim𝑏→+∞

−5

𝑏+ 5 = 5

Improper Integrals

Integral calculus

Education