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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 𝑓 there might be an inverse function π‘“βˆ’1, or for a non-singular square matrix 𝐴, π΄βˆ’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 𝐹′ π‘₯ = 𝑓(π‘₯) or 𝑑 𝐹 π‘₯ = 𝑓 π‘₯ 𝑑π‘₯, then anti-derivative of 𝑓 π‘₯ is defined as the indefinite integral (or primitive function) of 𝑓 π‘₯ and is mathematically expressed as: 𝑓 π‘₯ 𝑑π‘₯ = 𝐹 π‘₯ + 𝑐 β€’ 𝑓 π‘₯ 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 π‘₯ + 𝑐 β€² = 𝑓(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 𝑓 π‘₯ = π‘₯4. According to the definition if 𝐹 π‘₯ is such a function we should have: 𝐹(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 βˆ’ 3 and etc. can be considered as an answer so we can write the answer generally: π‘₯4 𝑑π‘₯ = π‘₯5 5 + 𝑐 β€’ Using the definition of the indefinite integral we can find the integral of simple functions directly: οƒ˜ 0 𝑑π‘₯ = 𝑐 οƒ˜ 1 𝑑π‘₯ = π‘₯ + 𝑐 οƒ˜ π‘₯ 𝑛 𝑑π‘₯ = π‘₯ 𝑛+1 𝑛+1 + 𝑐 (𝑛 β‰  βˆ’1) The Concept of Integration
  4. 4. οƒ˜ π‘₯βˆ’1 𝑑π‘₯ = 1 π‘₯ 𝑑π‘₯ = 𝑙𝑛 π‘₯ + 𝑐 (π‘₯ β‰  0) οƒ˜ 𝑒 π‘₯ 𝑑π‘₯ = 𝑒 π‘₯ + 𝑐 οƒ˜ π‘Ž π‘₯ 𝑑π‘₯ = π‘Ž π‘₯ πΏπ‘›π‘Ž + 𝑐 (π‘Ž β‰  1, π‘Ž > 0) οƒ˜ π‘π‘œπ‘ π‘₯ 𝑑π‘₯ = 𝑠𝑖𝑛π‘₯ + 𝑐 οƒ˜ 𝑠𝑖𝑛π‘₯ 𝑑π‘₯ = βˆ’π‘π‘œπ‘π‘₯ + 𝑐 β€’ Rules of Integration:  The derivative of the indefinite integral is the integrand: 𝑓 π‘₯ 𝑑π‘₯ β€² = 𝑓(π‘₯)  The differential of the indefinite integral is equal to the element of the integration: 𝑑 𝑓 π‘₯ 𝑑π‘₯ = 𝑓 π‘₯ 𝑑π‘₯ Rules of Integration
  5. 5.  The indefinite integral of a differential of a function is equal to that function plus a constant: 𝑑 𝐹 π‘₯ = 𝐹 π‘₯ + 𝑐  If π‘Ž β‰  0 and is a constant, then: π‘Ž. 𝑓 π‘₯ 𝑑π‘₯ = π‘Ž. 𝑓 π‘₯ 𝑑π‘₯  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(π‘₯) Β± 𝑔(π‘₯) 𝑑π‘₯ = 𝑓 π‘₯ 𝑑π‘₯ Β± 𝑔 π‘₯ 𝑑π‘₯ o This result can be extended to the finite number of integrable functions: 𝑓1 π‘₯ Β± 𝑓2 π‘₯ Β± β‹― Β± 𝑓𝑛(π‘₯) 𝑑π‘₯ = 𝑓1 π‘₯ 𝑑π‘₯ Β± 𝑓1 π‘₯ 𝑑π‘₯ Β± β‹― Β± 𝑓𝑛 π‘₯ 𝑑π‘₯ 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: 𝑖=1 𝑛 𝑓𝑖 π‘₯ 𝑑π‘₯ = 𝑖=1 𝑛 𝑓𝑖 π‘₯ 𝑑π‘₯  If 𝑓 𝑑 𝑑𝑑 = 𝐹 𝑑 + 𝑐, then: 𝑓 π‘Žπ‘₯ + 𝑏 𝑑π‘₯ = 1 π‘Ž . 𝐹 π‘Žπ‘₯ + 𝑏 + 𝑐 And if 𝑏 = 0, then 𝑓 π‘Žπ‘₯ 𝑑π‘₯ = 1 π‘Ž . 𝐹 π‘Žπ‘₯ + 𝑐 β€’ Using the last rule, we can easily calculate some integrals without applying a specific method: οƒ˜ 𝑒 𝛼π‘₯ 𝑑π‘₯ = 1 𝛼 𝑒 𝛼π‘₯ + 𝑐 Rules of Integration
  7. 7. οƒ˜ 𝑑π‘₯ π‘₯βˆ’π‘Ž = 𝑙𝑛 π‘₯ βˆ’ π‘Ž + 𝑐 οƒ˜ sin π‘šπ‘₯ = βˆ’cos(π‘šπ‘₯) π‘š + 𝑐 And some example for other rules: οƒ˜ π‘₯2 βˆ’ 3π‘₯ βˆ’ 7 𝑑π‘₯ = π‘₯2 𝑑π‘₯ βˆ’ 3 π‘₯ 𝑑π‘₯ βˆ’ 7 𝑑π‘₯ = π‘₯3 3 βˆ’ 3π‘₯2 2 βˆ’ 7π‘₯ + 𝑐 οƒ˜ π‘₯βˆ’ π‘₯ (π‘₯+5) 4 π‘₯ 𝑑π‘₯ = π‘₯2+5π‘₯βˆ’π‘₯ π‘₯βˆ’5 π‘₯ 4 π‘₯ 𝑑π‘₯ = π‘₯2βˆ’ 1 4 𝑑π‘₯ + 5 π‘₯1βˆ’ 1 4 𝑑π‘₯ βˆ’ π‘₯1+ 1 2 βˆ’ 1 4 𝑑π‘₯ βˆ’ 5 π‘₯ 1 2 βˆ’ 1 4 𝑑π‘₯ = π‘₯ 11 4 11 4 + 5 Γ— π‘₯ 7 4 7 4 βˆ’ π‘₯ 9 4 9 4 βˆ’ 5 Γ— π‘₯ 5 4 5 4 + 𝑐 = 4π‘₯ 4 π‘₯3 π‘₯2 11 + 5 7 βˆ’ 4π‘₯4 π‘₯ π‘₯ 9 + 1 = 4π‘₯4 π‘₯( π‘₯2 π‘₯ 11 + 5 π‘₯ 7 βˆ’ π‘₯ 9 βˆ’ 1) Rules of Integration
  8. 8.  Substitution Method: If the integrand is in the form of 𝑓(𝑔 π‘₯ ). 𝑔′(π‘₯), with substituting 𝑒 = 𝑔(π‘₯) we will have: 𝑓 𝑔 π‘₯ . 𝑔′ π‘₯ 𝑑π‘₯ = 𝑓 𝑒 . 𝑒′ 𝑑π‘₯ = 𝑓 𝑒 𝑑𝑒 And if 𝑓 𝑒 𝑑𝑒 = 𝐹 𝑒 + 𝑐, then: 𝑓 𝑔 π‘₯ . 𝑔′ π‘₯ 𝑑π‘₯ = 𝐹 𝑔 π‘₯ + 𝑐 o Find the indefinite integral π‘₯ 1+π‘₯2 𝑑π‘₯. Let 1 + π‘₯2 = 𝑒, then 2π‘₯ 𝑑π‘₯ = 𝑑𝑒, and we will have: π‘₯ 1 + π‘₯2 𝑑π‘₯ = 1 2 𝑑𝑒 𝑒 = 1 2 𝑑𝑒 𝑒 = 1 2 𝑙𝑛 𝑒 + 𝑐 = 1 2 𝑙𝑛 1 + π‘₯2 + 𝑐 Methods of Integration This method corresponds to the chain rule in differentiation.
  9. 9. o Find π‘₯ 𝑒 π‘₯2 𝑑π‘₯. Let π‘₯2 = 𝑒, then 2π‘₯ 𝑑π‘₯ = 𝑑𝑒, and: π‘₯ 𝑒 π‘₯2 𝑑π‘₯ = 𝑒 𝑒 Γ— 1 2 𝑑𝑒 = 1 2 𝑒 𝑒 + 𝑐 = 1 2 𝑒 π‘₯2 + 𝑐 o Find 𝑑π‘₯ π‘₯ .𝑙𝑛π‘₯ (π‘₯ > 0). Let 𝑙𝑛π‘₯ = 𝑒, then 𝑑π‘₯ π‘₯ = 𝑑𝑒, and: 𝑑π‘₯ π‘₯ . 𝑙𝑛π‘₯ = π‘₯. 𝑑𝑒 π‘₯. 𝑒 = 𝑑𝑒 𝑒 = 𝑙𝑛 𝑒 + 𝑐 = 𝑙𝑛 𝑙𝑛 π‘₯ + 𝑐 β€’ Note that, having success with this method requires finding a relevant substitution, which comes after lots of practices. Methods of Integration
  10. 10.  Integration by Parts: This method corresponds to the product rule for differentiation. According to the product rule, if 𝑒 and 𝑣 are continuous and differentiable functions in term of π‘₯, we have: 𝑑 𝑒𝑣 = 𝑣. 𝑑𝑒 + 𝑒. 𝑑𝑣 or 𝑒. 𝑑𝑣 = 𝑑 𝑒𝑣 βˆ’ 𝑣. 𝑑𝑒 And we know that: 𝑑 𝐹 π‘₯ = 𝐹 π‘₯ + 𝑐 Therefore, using the integral notation for we have: 𝑒. 𝑑𝑣 = 𝑒𝑣 βˆ’ 𝑣. 𝑑𝑒 Methods of Integration A A
  11. 11. o Find π‘₯. π‘π‘œπ‘ π‘₯ 𝑑π‘₯. By choosing π‘₯ = 𝑒 and π‘π‘œπ‘ π‘₯ 𝑑π‘₯ = 𝑑𝑣, we have: π‘₯ = 𝑒 π‘π‘œπ‘ π‘₯ 𝑑π‘₯ = 𝑑𝑣 ⟹ 𝑑π‘₯ = 𝑑𝑒 𝑠𝑖𝑛π‘₯ = 𝑣 Applying the formula, we have: π‘₯. π‘π‘œπ‘ π‘₯ 𝑑π‘₯ = π‘₯. 𝑠𝑖𝑛π‘₯ βˆ’ 𝑠𝑖𝑛π‘₯ 𝑑π‘₯ = π‘₯. 𝑠𝑖𝑛π‘₯ + π‘π‘œπ‘ π‘₯ + 𝑐 o Find π‘₯ 𝑒 π‘₯ 𝑑π‘₯. By choosing π‘₯ = 𝑒 and 𝑒 π‘₯ 𝑑π‘₯ = 𝑑𝑣, we have: π‘₯ = 𝑒 𝑒 π‘₯ 𝑑π‘₯ = 𝑑𝑣 ⟹ 𝑑π‘₯ = 𝑑𝑒 𝑒 π‘₯ = 𝑣 Methods of Integration No need to add the constant of integration here when calculating π‘π‘œπ‘ π‘₯ 𝑑π‘₯. It need to be added just once at the end
  12. 12. Applying the formula, we have: π‘₯ 𝑒 π‘₯ 𝑑π‘₯ = π‘₯ 𝑒 π‘₯ βˆ’ 𝑒 π‘₯ 𝑑π‘₯ = π‘₯ 𝑒 π‘₯ βˆ’ 𝑒 π‘₯ + 𝑐 = 𝑒 π‘₯ π‘₯ βˆ’ 1 + 𝑐 o Find π‘₯2 𝑙𝑛π‘₯ 𝑑π‘₯. By choosing 𝑙𝑛π‘₯ = 𝑒 and π‘₯2 𝑑π‘₯ = 𝑑𝑣, we have: 𝑙𝑛π‘₯ = 𝑒 π‘₯2 𝑑π‘₯ = 𝑑𝑣 ⟹ 𝑑π‘₯ π‘₯ =𝑑𝑒 π‘₯3 3 =𝑣 Applying the formula, we have: π‘₯2 𝑙𝑛π‘₯ 𝑑π‘₯ = π‘₯3 𝑙𝑛π‘₯ 3 βˆ’ π‘₯3 𝑑π‘₯ 3π‘₯ = π‘₯3 𝑙𝑛π‘₯ 3 βˆ’ 1 3 π‘₯2 𝑑π‘₯ = π‘₯3 𝑙𝑛π‘₯ 3 βˆ’ π‘₯3 9 + 𝑐 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 𝑒 π‘₯ 𝑑π‘₯. By choosing π‘₯2 = 𝑒 and 𝑒 π‘₯ 𝑑π‘₯ = 𝑑𝑣, we have: π‘₯2 = 𝑒 𝑒 π‘₯ 𝑑π‘₯ = 𝑑𝑣 ⟹ 2π‘₯. 𝑑π‘₯ = 𝑑𝑒 𝑒 π‘₯ = 𝑣 Applying the formula, we have: π‘₯2 𝑒 π‘₯ 𝑑π‘₯ = π‘₯2 𝑒 π‘₯ βˆ’ 2 π‘₯ 𝑒 π‘₯ 𝑑π‘₯ Here we need to use the method one more time for π‘₯ 𝑒 π‘₯ 𝑑π‘₯. We know from the last page the answer for this part is 𝑒 π‘₯ π‘₯ βˆ’ 1 + 𝑐, so the final answer is: π‘₯2 𝑒 π‘₯ βˆ’ 2 𝑒 π‘₯ π‘₯ βˆ’ 1 + 𝑐 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. οƒ˜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 βˆ’ 5 𝑑π‘₯, when 𝑦 0 = 2. Through the substitution method the indefinite integral is: π‘₯2 π‘₯3 βˆ’ 5 𝑑π‘₯ = π‘₯3 βˆ’ 5 𝑑(π‘₯3) 3 = 1 3 𝑒 βˆ’ 5 𝑑𝑒 = 1 3 𝑒2 2 βˆ’ 5𝑒 + 𝑐 = π‘₯6 6 βˆ’ 5 3 π‘₯3 + 𝑐 As when π‘₯ = 0 then 𝑦 = 2, so 2 = 0 + 𝑐 ⟹ 𝑐 = βˆ’2 Therefore, the function will be: 𝑦 = π‘₯6 6 βˆ’ 5 3 π‘₯3 βˆ’ 2 Initial Conditions in Integral Calculus
  15. 15. β€’ There is a very important relation between the concept of indefinite integral of the function 𝑓(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 𝑦 = 𝑓(π‘₯) over the interval [π‘Ž, 𝑏] (see Figure 2). One way to calculate this area is to divide the area into 𝑛 equal sub-intervals such as π‘Ž, π‘₯1 , π‘₯1, π‘₯2 , … , [π‘₯ π‘›βˆ’1, 𝑏], (each with the length equal to βˆ†π‘₯) and construct rectangles (see the figure 1). β€’ Obviously, the area under the curve can be estimated as 𝑅 = 𝑖=1 𝑛 𝑓 π‘₯𝑖 βˆ— . βˆ†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 βˆ†π‘₯ β†’ 0, in this case the value of the area approaches to a limit: 𝑆 = lim π‘›β†’βˆž 𝑖=1 𝑛 𝑓 π‘₯𝑖 βˆ— . βˆ†π‘₯ The Definite Integral Adoptedfrom Calculus Early Transcendental James Stewart p367
  16. 16. β€’ We call this sum as a definite integral of y = 𝑓(π‘₯) over the interval [π‘Ž, 𝑏] and it can be shown as π‘₯0=π‘Ž π‘₯ 𝑛=𝑏 𝑓 π‘₯ 𝑑π‘₯ or simply π‘Ž 𝑏 𝑓 π‘₯ 𝑑π‘₯. Therefore: π‘Ž 𝑏 𝑓 π‘₯ 𝑑π‘₯ = lim π‘›β†’βˆž 𝑖=1 𝑛 𝑓 π‘₯𝑖 βˆ— . βˆ†π‘₯ β€’ In this definition π‘Ž is the lower limit of the definite integral and 𝑏 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‘Ž 𝑏 𝑓 π‘₯ 𝑑π‘₯ = π‘Ž 𝑏 𝑓 𝑧 𝑑𝑧 = π‘Ž 𝑏 𝑓 πœƒ π‘‘πœƒ The Definite Integral β€’ If 𝑓(π‘₯) takes both positive and negative values, the area under the curve and confined by the x-axis and lines π‘₯ = π‘Ž and π‘₯ = 𝑏 is the sum of areas above the x-axis minus the sum of areas under the x-axis, i.e.: π‘Ž 𝑏 𝑓 π‘₯ 𝑑π‘₯ = Positive Areas βˆ’ Negative Areas Adoptedfrom Calculus Early Transcendental James Stewart p367
  17. 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 π‘Ž = 𝑏, then the area under the curve is zero: π‘Ž π‘Ž 𝑓 π‘₯ 𝑑π‘₯ = 0  If π‘Ž and 𝑏 exchange their position the new definite integral is the negative of the previous integral: 𝑏 π‘Ž 𝑓 π‘₯ 𝑑π‘₯ = βˆ’ π‘Ž 𝑏 𝑓 π‘₯ 𝑑π‘₯  If 𝑦 = 𝑐 over the interval [π‘Ž, 𝑏] , then: π‘Ž 𝑏 𝑓 π‘₯ 𝑑π‘₯ = π‘Ž 𝑏 𝑐 𝑑π‘₯ = 𝑐(𝑏 βˆ’ π‘Ž) Properties of Definite Integrals AdoptedfromCalculusEarlyTranscendentalJamesStewartp373
  18. 18.  If 𝑓(π‘₯) β‰₯ 0 and it is continuous over the interval [π‘Ž, 𝑏] and the interval can be divided into sub- intervals such as π‘Ž, π‘₯1 , π‘₯1, π‘₯2 , …, [π‘₯ π‘›βˆ’1, 𝑏], then: π‘Ž 𝑏 𝑓 π‘₯ 𝑑π‘₯ = π‘Ž π‘₯1 𝑓 π‘₯ 𝑑π‘₯ + π‘₯1 π‘₯2 𝑓 π‘₯ 𝑑π‘₯ + β‹― + π‘₯ π‘›βˆ’1 𝑏 𝑓 π‘₯ 𝑑π‘₯ β‰₯ 0  If 𝑓(π‘₯) β‰₯ 𝑔(π‘₯) and both are continuous over the interval [π‘Ž, 𝑏], then: π‘Ž 𝑏 𝑓 π‘₯ 𝑑π‘₯ β‰₯ π‘Ž 𝑏 𝑔 π‘₯ 𝑑π‘₯ 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 𝑦 = 𝑓 π‘₯ in the interval [π‘Ž, 𝑏] can be calculated directly through the difference between two boundary values of any primitive function of 𝑓(π‘₯); i.e. 𝐹 𝑏 βˆ’ 𝐹(π‘Ž). β€’ Mathematically, we can express this fundamental theorem as: π‘Ž 𝑏 𝑓 π‘₯ 𝑑π‘₯ = 𝐹 𝑏 βˆ’ 𝐹(π‘Ž) Where 𝐹(π‘₯) is any anti-derivative (primitive) function of 𝑓(π‘₯). The Fundamental Theorem of Calculus 𝑓(π‘₯) = 2π‘₯ 𝐹 𝑏 βˆ’ 𝐹 π‘Ž = 9 βˆ’ 1 = 8 Area= 𝒄 𝟐 𝒂 + 𝒃 = 𝟏 Γ— 𝟐 + πŸ” = πŸ– Adoptedand altered from http://www.bbc.co.uk/schools/gcsebitesize/maths/algebra/transformationhirev1.shtml 𝐹 π‘₯ = π‘₯2
  20. 20. o Evaluate the integral 1 3 2 π‘₯ 𝑑π‘₯. 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 π‘₯ = 2 π‘₯ is 𝐹 π‘₯ = 2 π‘₯ 𝑙𝑛2 , so, we have: 1 3 2 π‘₯ 𝑑π‘₯ = 𝐹 3 βˆ’ 𝐹 1 = 6 𝑙𝑛2 β€’ We often use the following notation for its simplicity: π‘Ž 𝑏 𝑓 π‘₯ 𝑑π‘₯ = 𝐹(π‘₯) π‘Ž 𝑏 = 𝐹 𝑏 βˆ’ 𝐹(π‘Ž) o Evaluate the area between 𝑓 π‘₯ = 𝑠𝑖𝑛π‘₯ and 𝑔 π‘₯ = π‘π‘œπ‘ π‘₯ in the interval [0, πœ‹ 2 ]. We know these functions cross each other when π‘₯ = πœ‹ 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œ‹ 4 ], the curve of 𝑔 π‘₯ = π‘π‘œπ‘ π‘₯ is on the top and 𝑓 π‘₯ = 𝑠𝑖𝑛π‘₯ is at the bottom but in the second interval [ πœ‹ 4 , πœ‹ 2 ], it is vice-versa. So: π΄π‘Ÿπ‘’π‘Ž = 0 πœ‹ 4 π‘π‘œπ‘ π‘₯ βˆ’ 𝑠𝑖𝑛π‘₯ 𝑑π‘₯ + πœ‹ 4 πœ‹ 2 𝑠𝑖𝑛π‘₯ βˆ’ π‘π‘œπ‘ π‘₯ 𝑑π‘₯ = 𝑠𝑖𝑛π‘₯ + π‘π‘œπ‘ π‘₯ 0 πœ‹ 4 + βˆ’π‘π‘œπ‘ π‘₯ βˆ’ 𝑠𝑖𝑛π‘₯ πœ‹ 4 πœ‹ 2 = 𝑠𝑖𝑛 πœ‹ 4 + π‘π‘œπ‘  πœ‹ 4 βˆ’ (𝑠𝑖𝑛0 + π‘π‘œπ‘ 0) + βˆ’π‘π‘œπ‘  πœ‹ 2 βˆ’ 𝑠𝑖𝑛 πœ‹ 2 βˆ’ βˆ’π‘π‘œπ‘  πœ‹ 4 βˆ’ 𝑠𝑖𝑛 πœ‹ 4 = 2 2 βˆ’ 1 Some Examples
  22. 22. β€’ In definite integral π‘Ž 𝑏 𝑓 𝑑 𝑑𝑑 if the upper limit replaced with π‘₯ (where π‘₯ varies in the interval [π‘Ž, 𝑏]) , in this case the area under the curve depends only on π‘₯ : π‘Ž π‘₯ 𝑓 𝑑 𝑑𝑑 = 𝐹 π‘₯ βˆ’ 𝐹 π‘Ž = 𝑔 π‘₯ β€’ If π‘₯ varies, so the derivative of 𝑔 π‘₯ with respect to π‘₯ is: π‘Ž π‘₯ 𝑓 𝑑 𝑑𝑑 β€² = 𝑓 π‘₯ Why? β€’ If π‘Ž or 𝑏 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. βˆ’βˆž 𝑏 𝑓 π‘₯ 𝑑π‘₯ = lim π‘Žβ†’βˆ’βˆž π‘Ž 𝑏 𝑓 π‘₯ 𝑑π‘₯ and π‘Ž +∞ 𝑓 π‘₯ 𝑑π‘₯ = lim 𝑏→+∞ π‘Ž 𝑏 𝑓 π‘₯ 𝑑π‘₯ Improper Integrals Adoptedfrom Calculus Early Transcendental James Stewart p380
  23. 23. o Find 1 +∞ 5 π‘₯2 𝑑π‘₯. 1 +∞ 5 π‘₯2 𝑑π‘₯ = lim 𝑏→+∞ 1 𝑏 5 π‘₯2 𝑑π‘₯ = lim 𝑏→+∞ βˆ’5 π‘₯ 1 𝑏 = lim 𝑏→+∞ βˆ’5 𝑏 + 5 = 5 Improper Integrals

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