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Lecturer: Farzad Javidanrad
Integral Calculus
(for MSc & PhD Business, Management & Finance Students)
(Autumn 2014-2015)
Basic Rules in Integration
β’ For any operation in mathematics, there is always an inverse operation. For example, summation and subtraction, multiplication and division. Even for a function π there might be an inverse function πβ1, or for a non-singular square matrix π΄, π΄β1 can be defined as the inverse.
β’ For the process of differentiation the reverse process is defined as anti-differentiation or simply integration.
β’ If πΉβ² π₯ = π(π₯) or π πΉ π₯ = π π₯ ππ₯, then anti-derivative of π π₯ is defined as the indefinite
integral (or primitive function) of π π₯ and is mathematically expressed as:
π π₯ ππ₯ = πΉ π₯ + π
β’ π π₯ is called the integrand and π 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:
πΉ π₯ + π β² = π(π₯)
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
β’ Find the indefinite integral for π π₯ = π₯4.
According to the definition if πΉ π₯ is such a function we should have:
πΉ(π₯) β² = π₯4
Obviously, the function π₯5
5satisfies the above equation but other functions such as
π₯5
5+ 1,
π₯5
5β 3 and
etc. can be considered as an answer so we can write the answer generally:
π₯4ππ₯ =π₯5
5+ π
β’ Using the definition of the indefinite integral we can find the integral of simple functions directly:
0 ππ₯ = π
1 ππ₯ = π₯ + π
π₯π ππ₯ =π₯π+1
π+1+ π (π β β1)
The Concept of Integration
π₯β1 ππ₯ = 1
π₯ππ₯ = ππ π₯ + π (π₯ β 0)
ππ₯ ππ₯ = ππ₯ + π
ππ₯ ππ₯ =ππ₯
πΏππ+ π (π β 1, π > 0)
πππ π₯ ππ₯ = π πππ₯ + π
π πππ₯ ππ₯ = βππππ₯ + π
β’ Rules of Integration:
The derivative of the indefinite integral is the integrand:
π π₯ ππ₯β²= π(π₯)
The differential of the indefinite integral is equal to the element of the integration:
π π π₯ ππ₯ = π π₯ ππ₯
Rules of Integration
The indefinite integral of a differential of a function is equal to that function plus a constant:
π πΉ π₯ = πΉ π₯ + π
If π β 0 and is a constant, then:
π. π π₯ ππ₯ = π. π π₯ ππ₯
The indefinite integral of the summation (subtraction) of two integrable functions are the summation (subtraction) of the indefinite integral for each one of them:
π(π₯) Β± π(π₯) ππ₯ = π π₯ ππ₯ Β± π π₯ ππ₯
o This result can be extended to the finite number of integrable functions:
π1 π₯ Β± π2 π₯ Β± β―Β± ππ(π₯) ππ₯ = π1 π₯ ππ₯ Β± π1 π₯ ππ₯ Β±β―Β± ππ π₯ ππ₯
Rules of Integration
A constant coefficient goes in and comes out of the integral sign
o If they are all added, we can write:
π=1
π
ππ π₯ ππ₯ =
π=1
π
ππ π₯ ππ₯
If π π‘ ππ‘ = πΉ π‘ + π, then:
π ππ₯ + π ππ₯ =1
π. πΉ ππ₯ + π + π
And if π = 0, then
π ππ₯ ππ₯ =1
π. πΉ ππ₯ + π
β’ Using the last rule, we can easily calculate some integrals without applying a specific method:
ππΌπ₯ππ₯ =1
πΌππΌπ₯ + π
Rules of Integration
ππ₯
π₯βπ= ππ π₯ β π + π
sin ππ₯ =βcos(ππ₯)
π+ π
And some example for other rules:
π₯2 β 3π₯ β 7 ππ₯ = π₯2ππ₯ β 3 π₯ ππ₯ β 7 ππ₯ =π₯3
3β
3π₯2
2β 7π₯ + π
π₯β π₯ (π₯+5)
4 π₯ππ₯ =
π₯2+5π₯βπ₯ π₯β5 π₯4 π₯
ππ₯ = π₯2β
1
4ππ₯ + 5 π₯1β
1
4ππ₯ β π₯1+
1
2β1
4ππ₯ β 5 π₯1
2β1
4 ππ₯
=π₯11
4
11
4
+ 5 Γπ₯7
4
7
4
βπ₯9
4
9
4
β 5 Γπ₯5
4
5
4
+ π
= 4π₯4π₯3
π₯2
11+
5
7β 4π₯4 π₯
π₯
9+ 1
= 4π₯4 π₯(π₯2 π₯
11+
5 π₯
7β
π₯
9β 1)
Rules of Integration
Substitution Method: If the integrand is in the form of π(π π₯ ). πβ²(π₯), with substituting π’ = π(π₯)we will have:
π π π₯ . πβ² π₯ ππ₯ = π π’ . π’β²ππ₯ = π π’ ππ’
And if π π’ ππ’ = πΉ π’ + π, then:
π π π₯ . πβ² π₯ ππ₯ = πΉ π π₯ + π
o Find the indefinite integral π₯
1+π₯2ππ₯.
Let 1 + π₯2 = π’, then 2π₯ ππ₯ = ππ’, and we will have:
π₯
1 + π₯2ππ₯ =
12 ππ’
π’=1
2 ππ’
π’
=1
2ππ π’ + π =
1
2ππ 1 + π₯2 + π
Methods of Integration
This method corresponds to the
chain rule in differentiation.
o Find π₯ ππ₯2ππ₯.
Let π₯2 = π’, then 2π₯ ππ₯ = ππ’, and:
π₯ ππ₯2ππ₯ = ππ’ Γ 1
2 ππ’ =1
2ππ’ + π =
1
2ππ₯
2+ π
o Find ππ₯
π₯ .πππ₯(π₯ > 0).
Let πππ₯ = π’, then ππ₯
π₯= ππ’, and:
ππ₯
π₯ . πππ₯=
π₯. ππ’
π₯. π’=
ππ’
π’= ππ π’ + π = ππ ππ π₯ + π
β’ Note that, having success with this method requires finding a relevant substitution, which comes after lots of practices.
Methods of Integration
Integration by Parts: This method corresponds to the product rule for differentiation. According to the product rule, if π’ and π£ are continuous and differentiable functions in term of π₯, we have:
π π’π£ = π£. ππ’ + π’. ππ£
or
π’. ππ£ = π π’π£ β π£. ππ’
And we know that:
π πΉ π₯ = πΉ π₯ + π
Therefore, using the integral notation for we have:
π’. ππ£ = π’π£ β π£. ππ’
Methods of Integration
A
A
o Find π₯. πππ π₯ ππ₯.
By choosing π₯ = π’ and πππ π₯ ππ₯ = ππ£, we have:
π₯ = π’
πππ π₯ ππ₯ = ππ£βΉ
ππ₯ = ππ’π πππ₯ = π£
Applying the formula, we have:
π₯. πππ π₯ ππ₯ = π₯. π πππ₯ β π πππ₯ ππ₯
= π₯. π πππ₯ + πππ π₯ + π
o Find π₯ ππ₯ππ₯.
By choosing π₯ = π’ and ππ₯ππ₯ = ππ£, we have:
π₯ = π’
ππ₯ππ₯ = ππ£βΉ
ππ₯ = ππ’ππ₯ = π£
Methods of Integration
No need to add the constant of integration here when calculating
πππ π₯ ππ₯. It need to be added just
once at the end
Applying the formula, we have:
π₯ ππ₯ππ₯ = π₯ ππ₯ β ππ₯ππ₯
= π₯ ππ₯ β ππ₯ + π = ππ₯ π₯ β 1 + π
o Find π₯2πππ₯ ππ₯.
By choosing πππ₯ = π’ and π₯2ππ₯ = ππ£, we have:
πππ₯ = π’
π₯2ππ₯ = ππ£βΉ
ππ₯π₯ =ππ’
π₯3
3 =π£
Applying the formula, we have:
π₯2πππ₯ ππ₯ =π₯3πππ₯
3β
π₯3ππ₯
3π₯=π₯3πππ₯
3β1
3 π₯2 ππ₯
=π₯3πππ₯
3β
π₯3
9+ π
Methods of Integration
β’ Sometimes it is needed to use this method more than once to reach to the general solution (primitive function).
o Find π₯2 ππ₯ππ₯.
By choosing π₯2 = π’ and ππ₯ππ₯ = ππ£, we have:
π₯2 = π’ππ₯ππ₯ = ππ£
βΉ 2π₯. ππ₯ = ππ’ππ₯ = π£
Applying the formula, we have:
π₯2 ππ₯ππ₯ = π₯2ππ₯ β 2 π₯ ππ₯ ππ₯
Here we need to use the method one more time for π₯ ππ₯ ππ₯. We know from the last page the answer for this
part is ππ₯ π₯ β 1 + π, so the final answer is:
π₯2ππ₯ β 2ππ₯ π₯ β 1 + π
Methods of Integration
β’ 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?β’ 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 π₯2 π₯3 β 5 ππ₯, when π¦ 0 = 2.
Through the substitution method the indefinite integral is:
π₯2 π₯3 β 5 ππ₯ = π₯3 β 5π(π₯3)
3=1
3 π’ β 5 ππ’
=1
3
π’2
2β 5π’ + π =
π₯6
6β5
3π₯3 + π
As when π₯ = 0 then π¦ = 2, so 2 = 0 + π βΉ π = β2
Therefore, the function will be: π¦ =π₯6
6β
5
3π₯3 β 2
Initial Conditions in Integral Calculus
β’ There is a very important relation between the concept of indefinite integral of the function π(π₯)and the area under the curve of this function over the given interval.
β’ Imagine we are going to find the area under the curve π¦ = π(π₯) over the interval [π, π] (see Figure 2). One way to calculate this area is to divide the area into π equal sub-intervals such as π, π₯1 , π₯1, π₯2 , β¦ , [π₯πβ1, π], (each with the length equal to βπ₯) and construct rectangles (see the
figure 1).
β’ Obviously, the area under the curve can be estimated as π = π=1π π π₯π
β . βπ₯ (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 βπ₯ β 0, in this case the value of the
area approaches to a limit:
π = limπββ
π=1
π
π π₯πβ . βπ₯
The Definite Integral
Adopted from Calculus Early Transcendental James Stewart p367
β’ We call this sum as a definite integral of y = π(π₯) over the interval [π, π] and it can be shown as
π₯0=π
π₯π=π π π₯ ππ₯ or simply π
ππ π₯ ππ₯. Therefore:
π
π
π π₯ ππ₯ = limπββ
π=1
π
π π₯πβ . βπ₯
β’ In this definition π is the lower limit of the definite integral and π is called the upper limit.
β’ The definite integral is a number so it is not sensitive to be represented by different variables. i.e.:
π
π
π π₯ ππ₯ = π
π
π π§ ππ§ = π
π
π π ππ
The Definite Integral
β’ If π(π₯) takes both positive and negative values, the area under the curve and confined by the x-axis and lines π₯ = π and π₯ = π is the sum of areas above the x-axis minus the sum of areas under the x-axis, i.e.:
π
π
π π₯ ππ₯ = Positive Areas β Negative Areas
Adopted from Calculus Early Transcendental James Stewart p367
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 π = π, then the area under the curve is zero:
π
π
π π₯ ππ₯ = 0
If π and π exchange their position the new definite integral is the negative of the previous integral:
π
π
π π₯ ππ₯ = β π
π
π π₯ ππ₯
If π¦ = π over the interval [π, π] , then:
π
π
π π₯ ππ₯ = π
π
π ππ₯ = π(π β π)
Properties of Definite Integrals
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If π(π₯) β₯ 0 and it is continuous over the interval [π, π] and the interval can be divided into sub-intervals such as π, π₯1 , π₯1, π₯2 , β¦, [π₯πβ1, π], then:
π
π
π π₯ ππ₯ = π
π₯1
π π₯ ππ₯ + π₯1
π₯2
π π₯ ππ₯ + β―+ π₯πβ1
π
π π₯ ππ₯ β₯ 0
If π(π₯) β₯ π(π₯) and both are continuous over the interval [π, π], then:
π
π
π π₯ ππ₯ β₯ π
π
π π₯ ππ₯
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/
β’ The fundamental theorem of calculus asserts that there is a specific relation between the area under a curve and theindefinite integral of that curve.
β’ Another word, the value of the area under the continuouscurve π¦ = π π₯ in the interval [π, π] can be calculateddirectly through the difference between two boundaryvalues of any primitive function of π(π₯); i.e. πΉ π β πΉ(π).
β’ Mathematically, we can express this fundamental
theorem as:
π
π
π π₯ ππ₯ = πΉ π β πΉ(π)
Where πΉ(π₯) is any anti-derivative (primitive) function
of π(π₯).
The Fundamental Theorem of Calculus
π(π₯) = 2π₯
πΉ π β πΉ π = 9 β 1 = 8
Area=π
ππ + π = π Γ π + π = π
Adopted and altered from http://www.bbc.co.uk/schools/gcsebitesize/maths/algebra/transformationhirev1.shtml
πΉ π₯ = π₯2
o Evaluate the integral 1
32π₯ππ₯.
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 π π₯ = 2π₯ is πΉ π₯ =2π₯
ππ2, so, we have:
1
3
2π₯ππ₯ = πΉ 3 β πΉ 1 =6
ππ2
β’ We often use the following notation for its simplicity:
π
π
π π₯ ππ₯ = πΉ(π₯) ππ = πΉ π β πΉ(π)
o Evaluate the area between π π₯ = π πππ₯
and π π₯ = πππ π₯ in the interval [0,π
2].
We know these functions cross each other
when π₯ =π
4(see the graph)
Some Examples
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β’ 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 π π₯ = πππ π₯ is on the top and π π₯ =
π πππ₯ is at the bottom but in the second interval [π
4,π
2], it is vice-versa. So:
π΄πππ = 0
π4πππ π₯ β π πππ₯ ππ₯ +
π4
π2π πππ₯ β πππ π₯ ππ₯
= π πππ₯ + πππ π₯ 0
π4 + βπππ π₯ β π πππ₯ π
4
π2
= π πππ
4+ πππ
π
4β (π ππ0 + πππ 0) + βπππ
π
2β π ππ
π
2β βπππ
π
4β π ππ
π
4= 2 2 β 1
Some Examples
β’ In definite integral π
ππ π‘ ππ‘ if the upper limit replaced with π₯ (where π₯ varies in the interval [π, π]) ,
in this case the area under the curve depends only on π₯ :
π
π₯
π π‘ ππ‘ = πΉ π₯ β πΉ π = π π₯
β’ If π₯ varies, so the derivative of π π₯ with respect to π₯ is:
π
π₯
π π‘ ππ‘
β²
= π π₯
Why?
β’ If π or π goes to infinity, then we are dealing with improper integrals. As infinity is not a number it cannot be substituted for π₯. They must be defined as the limit of a proper definite integral.
ββ
π
π π₯ ππ₯ = limπβββ
π
π
π π₯ ππ₯ and π
+β
π π₯ ππ₯ = limπβ+β
π
π
π π₯ ππ₯
Improper Integrals
Adopted from Calculus Early Transcendental James Stewart p380
o Find 1
+β 5
π₯2ππ₯.
1
+β 5
π₯2ππ₯ = lim
πβ+β 1
π 5
π₯2ππ₯ = lim
πβ+β
β5
π₯1
π
= limπβ+β
β5
π+ 5 = 5
Improper Integrals