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# Integral calculus

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Introduction to Integral Calculus (Basics)

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### Integral calculus

1. 1. Lecturer: Farzad Javidanrad Integral Calculus (for MSc & PhD Business, Management & Finance Students) (Autumn 2014-2015) Basic Rules in Integration
2. 2. 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 πβ1, or for a non-singular square matrix π΄, π΄β1 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 πΉβ² π₯ = π(π₯) or π πΉ π₯ = π π₯ ππ₯, 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₯ ππ₯ = πΉ 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: πΉ π₯ + π β² = π(π₯) 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. 3. β’ Find the indefinite integral for π π₯ = π₯4. According to the definition if πΉ π₯ is such a function we should have: πΉ(π₯) β² = π₯4 Obviously, the function π₯5 5 satisfies the above equation but other functions such as π₯5 5 + 1, π₯5 5 β 3 and etc. can be considered as an answer so we can write the answer generally: π₯4 ππ₯ = π₯5 5 + π β’ Using the definition of the indefinite integral we can find the integral of simple functions directly: ο 0 ππ₯ = π ο 1 ππ₯ = π₯ + π ο π₯ π ππ₯ = π₯ π+1 π+1 + π (π β  β1) The Concept of Integration
4. 4. ο π₯β1 ππ₯ = 1 π₯ ππ₯ = ππ π₯ + π (π₯ β  0) ο π π₯ ππ₯ = π π₯ + π ο π π₯ ππ₯ = π π₯ πΏππ + π (π β  1, π > 0) ο πππ π₯ ππ₯ = π πππ₯ + π ο π πππ₯ ππ₯ = βππππ₯ + π β’ Rules of Integration: ο§ The derivative of the indefinite integral is the integrand: π π₯ ππ₯ β² = π(π₯) ο§ The differential of the indefinite integral is equal to the element of the integration: π π π₯ ππ₯ = π π₯ ππ₯ Rules of Integration
5. 5. ο§ The indefinite integral of a differential of a function is equal to that function plus a constant: π πΉ π₯ = πΉ π₯ + π ο§ If π β  0 and is a constant, then: π. π π₯ ππ₯ = π. π π₯ ππ₯ ο§ 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₯) ππ₯ = π π₯ ππ₯ Β± π π₯ ππ₯ o This result can be extended to the finite number of integrable functions: π1 π₯ Β± π2 π₯ Β± β― Β± ππ(π₯) ππ₯ = π1 π₯ ππ₯ Β± π1 π₯ ππ₯ Β± β― Β± ππ π₯ ππ₯ Rules of Integration A constant coefficient goes in and comes out of the integral sign
6. 6. o If they are all added, we can write: π=1 π ππ π₯ ππ₯ = π=1 π ππ π₯ ππ₯ ο§ If π π‘ ππ‘ = πΉ π‘ + π, then: π ππ₯ + π ππ₯ = 1 π . πΉ ππ₯ + π + π And if π = 0, then π ππ₯ ππ₯ = 1 π . πΉ ππ₯ + π β’ Using the last rule, we can easily calculate some integrals without applying a specific method: ο π πΌπ₯ ππ₯ = 1 πΌ π πΌπ₯ + π Rules of Integration
7. 7. ο ππ₯ π₯βπ = ππ π₯ β π + π ο sin ππ₯ = βcos(ππ₯) π + π And some example for other rules: ο π₯2 β 3π₯ β 7 ππ₯ = π₯2 ππ₯ β 3 π₯ ππ₯ β 7 ππ₯ = π₯3 3 β 3π₯2 2 β 7π₯ + π ο π₯β π₯ (π₯+5) 4 π₯ ππ₯ = π₯2+5π₯βπ₯ π₯β5 π₯ 4 π₯ ππ₯ = π₯2β 1 4 ππ₯ + 5 π₯1β 1 4 ππ₯ β π₯1+ 1 2 β 1 4 ππ₯ β 5 π₯ 1 2 β 1 4 ππ₯ = π₯ 11 4 11 4 + 5 Γ π₯ 7 4 7 4 β π₯ 9 4 9 4 β 5 Γ π₯ 5 4 5 4 + π = 4π₯ 4 π₯3 π₯2 11 + 5 7 β 4π₯4 π₯ π₯ 9 + 1 = 4π₯4 π₯( π₯2 π₯ 11 + 5 π₯ 7 β π₯ 9 β 1) Rules of Integration
8. 8. ο§ Substitution Method: If the integrand is in the form of π(π π₯ ). πβ²(π₯), with substituting π’ = π(π₯) we will have: π π π₯ . πβ² π₯ ππ₯ = π π’ . π’β² ππ₯ = π π’ ππ’ And if π π’ ππ’ = πΉ π’ + π, then: π π π₯ . πβ² π₯ ππ₯ = πΉ π π₯ + π o Find the indefinite integral π₯ 1+π₯2 ππ₯. Let 1 + π₯2 = π’, then 2π₯ ππ₯ = ππ’, and we will have: π₯ 1 + π₯2 ππ₯ = 1 2 ππ’ π’ = 1 2 ππ’ π’ = 1 2 ππ π’ + π = 1 2 ππ 1 + π₯2 + π Methods of Integration This method corresponds to the chain rule in differentiation.
9. 9. o Find π₯ π π₯2 ππ₯. Let π₯2 = π’, then 2π₯ ππ₯ = ππ’, and: π₯ π π₯2 ππ₯ = π π’ Γ 1 2 ππ’ = 1 2 π π’ + π = 1 2 π π₯2 + π o Find ππ₯ π₯ .πππ₯ (π₯ > 0). Let πππ₯ = π’, then ππ₯ π₯ = ππ’, and: ππ₯ π₯ . πππ₯ = π₯. ππ’ π₯. π’ = ππ’ π’ = ππ π’ + π = ππ ππ π₯ + π β’ Note that, having success with this method requires finding a relevant substitution, which comes after lots of practices. Methods of Integration
10. 10. ο§ 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£. ππ’ + π’. ππ£ or π’. ππ£ = π π’π£ β π£. ππ’ And we know that: π πΉ π₯ = πΉ π₯ + π Therefore, using the integral notation for we have: π’. ππ£ = π’π£ β π£. ππ’ Methods of Integration A A
11. 11. o Find π₯. πππ π₯ ππ₯. By choosing π₯ = π’ and πππ π₯ ππ₯ = ππ£, we have: π₯ = π’ πππ π₯ ππ₯ = ππ£ βΉ ππ₯ = ππ’ π πππ₯ = π£ Applying the formula, we have: π₯. πππ π₯ ππ₯ = π₯. π πππ₯ β π πππ₯ ππ₯ = π₯. π πππ₯ + πππ π₯ + π o Find π₯ π π₯ ππ₯. By choosing π₯ = π’ and π π₯ ππ₯ = ππ£, we have: π₯ = π’ π π₯ ππ₯ = ππ£ βΉ ππ₯ = ππ’ π π₯ = π£ Methods of Integration No need to add the constant of integration here when calculating πππ π₯ ππ₯. It need to be added just once at the end
12. 12. Applying the formula, we have: π₯ π π₯ ππ₯ = π₯ π π₯ β π π₯ ππ₯ = π₯ π π₯ β π π₯ + π = π π₯ π₯ β 1 + π o Find π₯2 πππ₯ ππ₯. By choosing πππ₯ = π’ and π₯2 ππ₯ = ππ£, we have: πππ₯ = π’ π₯2 ππ₯ = ππ£ βΉ ππ₯ π₯ =ππ’ π₯3 3 =π£ Applying the formula, we have: π₯2 πππ₯ ππ₯ = π₯3 πππ₯ 3 β π₯3 ππ₯ 3π₯ = π₯3 πππ₯ 3 β 1 3 π₯2 ππ₯ = π₯3 πππ₯ 3 β π₯3 9 + π Methods of Integration
13. 13. β’ Sometimes it is needed to use this method more than once to reach to the general solution (primitive function). o Find π₯2 π π₯ ππ₯. By choosing π₯2 = π’ and π π₯ ππ₯ = ππ£, we have: π₯2 = π’ π π₯ ππ₯ = ππ£ βΉ 2π₯. ππ₯ = ππ’ π π₯ = π£ Applying the formula, we have: π₯2 π π₯ ππ₯ = π₯2 π π₯ β 2 π₯ π π₯ ππ₯ Here we need to use the method one more time for π₯ π π₯ ππ₯. We know from the last page the answer for this part is π π₯ π₯ β 1 + π, so the final answer is: π₯2 π π₯ β 2 π π₯ π₯ β 1 + π Methods of Integration
14. 14. 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. ο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 ππ₯, when π¦ 0 = 2. Through the substitution method the indefinite integral is: π₯2 π₯3 β 5 ππ₯ = π₯3 β 5 π(π₯3) 3 = 1 3 π’ β 5 ππ’ = 1 3 π’2 2 β 5π’ + π = π₯6 6 β 5 3 π₯3 + π As when π₯ = 0 then π¦ = 2, so 2 = 0 + π βΉ π = β2 Therefore, the function will be: π¦ = π₯6 6 β 5 3 π₯3 β 2 Initial Conditions in Integral Calculus
15. 15. 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₯ πβ1, π], (each with the length equal to βπ₯) and construct rectangles (see the figure 1). β’ Obviously, the area under the curve can be estimated as π = π=1 π π π₯π β . βπ₯ (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 βπ₯ β 0, in this case the value of the area approaches to a limit: π = lim πββ π=1 π π π₯π β . βπ₯ The Definite Integral Adoptedfrom Calculus Early Transcendental James Stewart p367
16. 16. 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₯ ππ₯ or simply π π π π₯ ππ₯. Therefore: π π π π₯ ππ₯ = lim πββ π=1 π π π₯π β . βπ₯ 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₯ ππ₯ = π π π π§ ππ§ = π π π π ππ 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₯ ππ₯ = Positive Areas β Negative Areas Adoptedfrom Calculus Early Transcendental James Stewart p367
17. 17. ο±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: ο§ If π = π, then the area under the curve is zero: π π π π₯ ππ₯ = 0 ο§ If π and π exchange their position the new definite integral is the negative of the previous integral: π π π π₯ ππ₯ = β π π π π₯ ππ₯ ο§ If π¦ = π over the interval [π, π] , then: π π π π₯ ππ₯ = π π π ππ₯ = π(π β π) Properties of Definite Integrals AdoptedfromCalculusEarlyTranscendentalJamesStewartp373
18. 18. ο§ 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₯ πβ1, π], then: π π π π₯ ππ₯ = π π₯1 π π₯ ππ₯ + π₯1 π₯2 π π₯ ππ₯ + β― + π₯ πβ1 π π π₯ ππ₯ β₯ 0 ο§ If π(π₯) β₯ π(π₯) and both are continuous over the interval [π, π], then: π π π π₯ ππ₯ β₯ π π π π₯ ππ₯ 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. 19. β’ 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. β’ Another word, the value of the area under the continuous curve π¦ = π π₯ in the interval [π, π] can be calculated directly through the difference between two boundary values of any primitive function of π(π₯); i.e. πΉ π β πΉ(π). β’ Mathematically, we can express this fundamental theorem as: π π π π₯ ππ₯ = πΉ π β πΉ(π) Where πΉ(π₯) is any anti-derivative (primitive) function of π(π₯). The Fundamental Theorem of Calculus π(π₯) = 2π₯ πΉ π β πΉ π = 9 β 1 = 8 Area= π π π + π = π Γ π + π = π Adoptedand altered from http://www.bbc.co.uk/schools/gcsebitesize/maths/algebra/transformationhirev1.shtml πΉ π₯ = π₯2
20. 20. o Evaluate the integral 1 3 2 π₯ ππ₯. 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₯ = 2 π₯ is πΉ π₯ = 2 π₯ ππ2 , so, we have: 1 3 2 π₯ ππ₯ = πΉ 3 β πΉ 1 = 6 ππ2 β’ We often use the following notation for its simplicity: π π π π₯ ππ₯ = πΉ(π₯) π π = πΉ π β πΉ(π) o Evaluate the area between π π₯ = π πππ₯ and π π₯ = πππ π₯ in the interval [0, π 2 ]. We know these functions cross each other when π₯ = π 4 (see the graph) Some Examples Adoptedfromhttp://maretbccalculus2007- 2008.pbworks.com/w/page/20301409/Find%20The%20Area %20of%20a%20Region
21. 21. 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₯ = πππ π₯ is on the top and π π₯ = π πππ₯ is at the bottom but in the second interval [ π 4 , π 2 ], it is vice-versa. So: π΄πππ = 0 π 4 πππ π₯ β π πππ₯ ππ₯ + π 4 π 2 π πππ₯ β πππ π₯ ππ₯ = π πππ₯ + πππ π₯ 0 π 4 + βπππ π₯ β π πππ₯ π 4 π 2 = π ππ π 4 + πππ  π 4 β (π ππ0 + πππ 0) + βπππ  π 2 β π ππ π 2 β βπππ  π 4 β π ππ π 4 = 2 2 β 1 Some Examples
22. 22. β’ In definite integral π π π π‘ ππ‘ 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‘ ππ‘ = πΉ π₯ β πΉ π = π π₯ β’ If π₯ varies, so the derivative of π π₯ with respect to π₯ is: π π₯ π π‘ ππ‘ β² = 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. ββ π π π₯ ππ₯ = lim πβββ π π π π₯ ππ₯ and π +β π π₯ ππ₯ = lim πβ+β π π π π₯ ππ₯ Improper Integrals Adoptedfrom Calculus Early Transcendental James Stewart p380
23. 23. o Find 1 +β 5 π₯2 ππ₯. 1 +β 5 π₯2 ππ₯ = lim πβ+β 1 π 5 π₯2 ππ₯ = lim πβ+β β5 π₯ 1 π = lim πβ+β β5 π + 5 = 5 Improper Integrals