Integral Calculus 10 This unit is designed to introduce the learners to the basic concepts associated with Integral Calculus. Integral calculus can be classified and discussed into two threads. One is Indefinite Integral and the other one is Definite Integral . The learners will learn about indefinite integral, methods of integration, definite integral and application of integral calculus in business and economics. School of Business Blank Page Unit-10 Page-228 Bangladesh Open University Lesson-1: Indefinite Integral After completing this lesson, you should be able to: Describe the concept of integration; Determine the indefinite integral of a given function. Introduction The process of differentiation is used for finding derivatives and differentials of functions. On the other hand, the process of integration is used (i) for finding the limit of the sum of an infinite number of Reversing the process infinitesimals that are in the differential form f ′()x dx (ii) for finding of differentiation and finding the original functions whose derivatives or differentials are given, i.e., for finding function from the anti-derivatives. Thus, reversing the process of differentiation and derivative is called finding the original function from the derivative is called integration or integration. anti-differentiation. The integral calculus is used to find the areas, probabilities and to solve equations involving derivatives. Integration is also used to determine a function whose rate of change is known. In integration whether the object be summation or anti-differentiation, the sign ∫, an elongated S, the first letter of the word ‘sum’ is most generally used to indicate the process of the summation or integration. Therefore, is read the integral of f (x) with respect to x. ∫ f ()x dx ∫ f() x dx is read the integral of f (x) with Again, integration is defined as the inverse process of differentiation. respect to x. d Thus if g()x = f ()x dx then ∫ f ()x dx = g()x + c where c is called the constant of integration. Of course c could have any value and thus integral of a function is not unique! But we could say one thing here that any two integrals of the same function differ by a constant. Since c could also have the value zero, g()x is one of the values of ∫ f ()x dx . As c is unknown and indefinite, hence it is also referred to as Indefinite Integral. Some Properties of Integration The following two rules are useful in reducing differentiable expressions to standard forms. (i) The integral of any algebraic sum of differential expression equals the algebraic sum of the integrals of these expressions taken separately. i.e. ∫[ f ()x ± g(x)]dx = ∫ f ()x dx ± ∫ g()x dx (ii) A constant multiplicative term may be written either before or after the integral sign. i.e. ∫ cf ()x dx = c∫ f ()x dx ; where c is a constant. Business Mathematics Page-229 School of Business Some Standard Results of integration A list of some standard results by using the derivative of some well- known functions is given below: d i)( dx = x + c ∴ ()x = 1 ∫ dx x n+1 d x n+1 n = + n (ii ) ∫ x dx c ∴ = x , n ≠ −1 n +1 dx n +1 1 d 1 ( iii ) dx = log x + c ∴ (log )x = ∫ x dx x x x d x x (iv) e dx = e + c ∴ (e ) = e ∫ dx x x a d x x (v ) a dx = + c ∴ ( a ) = a log a ∫ log a dx d ( vi ) sin x dx = −cos x + c ∴ ( −cos )x = sin x ∫ dx d ( vii ) cos x dx = sin x + c ∴ (sin x) = cos x ∫ dx 2 d 2 ( viii ) sec x dx = tan x + c ∴ (tan x) = sec x ∫ dx 2 d 2 ( ix ) cosec x dx = −cot x + c ∴ (−cot x) = cosec x ∫ dx d )x( sec xtan x dx = sec x + c ∴ (sec )x = sec xtan x ∫ dx ( xi ) ∫cosecx cot x dx = −cosecx + c d ∴ (−cosecx) = cosex cot x dx 1 −1 d −1 1 ( xii ) ∫ dx = sin x + c ∴ (sin )x = 1- x2 dx 1 − x2 1 −1 d −1 1 ( xii ) dx = tan x + c ∴ (tan )x = ∫ 1 + x2 dx 1 + x2 1 −1 d −1 1 ( xiii ) ∫ dx = sec x + c ∴ (sec )x = x x2 1- dx x 1 + x2 d ( xiv ) tan x dx = log sec x + c ∴ (log sin )x = cot x ∫ dx d ( xv ) cot x dx = log sin x + c ∴ (log sec )x = tan x ∫ dx Unit-10 Page-230 Bangladesh Open University Illustrative Examples: Example -1: 2 Evaluate ∫ x dx Solution: 3 2 x x dx = + c ∫ 3 Example-2: Evaluate ∫ 5x dx Solution: 5 2 5x dx = x + c ∫ 2 Example-3: Evaluate ∫ − 2 dx Solution: ∫ − 2 dx = − 2x + c Example-4: Evaluate ∫ 2( x + )3 dx Solution: ∫ 2( x + )3 dx = ∫2xdx + ∫ 3dx 2 = x + 3x + c Example-5: 2 Evaluate ∫ 4( x − 7x + )6 dx Solution: 2 2 ∫ 4( x − 7x + )6 dx = ∫ 4x dx − ∫7xdx + ∫ 6dx 4 3 7 2 = x − x + 6x + c 3 2 Example-6: a b Evaluate + + 6dx ∫ x 3 x 2 Solution: a b a b + + 6dx = dx + dx + 6dx ∫ x 3 x 2 ∫ x 3 ∫ x 2 ∫ Business Mathematics Page-231 School of Business a b = − − + 6x + c 2x 2 x Example-7: dx Evaluate ∫ x −1 + x + 3 Solution: dx ( x − 1 − x + 3 ) dx ∫ = ∫ x −1 + x + 3 ( x − 1 + x + 3 )( x − 1 − x + 3 ) ( x − 1 − x + 3 ) dx = ∫ x − 1 − x − 3 −1 1 = x −1 dx + x + 3 dx 4 ∫ 4 ∫ 3 3 1 1 = x( + )3 2 − x( − )1 2 + c 6 6 Example-8: x −2 3 Evaluate e5( − x + ) dx; x ≠ 0 ∫ x Solution: x −2 3 x −2 3 5( e − x + ) dx = 5e dx − x dx + dx ∫ x ∫ ∫ ∫ x x 1 = e5 + + 3log x + c x Unit-10 Page-232 Bangladesh Open University Questions for Review: These questions are designed to help you assess how far you have understood and can apply the learning you have accomplished by answering (in written form) the following questions: 1. Integrate the following functions w. r. t. x 3 6 (i) 5( x + )dx ∫ x 3 1 (ii) dx ∫ x 1 (iii) ( x − )dx ∫ x 3 1 (iv) (x + )dx ∫ x 3 dx (v) ∫ x +1 + x −1 x x 1 (vi) 2( e + 7 + )dx ∫ x Business Mathematics Page-233 School of Business Lesson-2: Methods of Integration After studying this lesson, you should be able to: Describe the methods of integration; Determine the integral of any function. Introduction In comparing integral and differential calculus, most of the mathematicians would agree that the integration of functions is a more complicated process than the differentiation of functions. Functions can be differentiated through application of a number of relatively straightforward rules. This is not true in determining the integrals of functions. Integration is much less straightforward and often requires considerable ingenuity. Certain functions can be integrated quite simply by applying rules of Certain functions can be integrated quite integration. A natural question is that what happens when rules of simply by applying integration cannot be applied directly. Such functions require more rules of integration. complicated techniques. This lesson discusses four techniques that can be employed when the other rules do not apply and when the structure of the integrand is of an appropriate form. In general, experience is the best guide for suggesting the quickest and simplest method for integrating any given function. Methods of Integration The following are the four principal methods of integration: (i) Integration by substitution; (ii) Integration by parts; (iii) Integration by successive reductions; (iv) Integration by partial fraction. Integration by Substitution Substituting a new Substituting a new suitable variable for the given independent variable suitable variable for and integrating with respect to the substituted variable can often facilitate the given independent Integration. Experience is the best guide as to what substitution is likely variable and integra- to transform the given expression into another that is more readily ting with respect to the substituted vari- integrable. In fact this is done only for convenience. able can often The following examples will make the process clear. facilitate Integration. Example-1: 5 Evaluate ∫ (ax + b) dx Solution: Let, ax + b = z adx = dz 1 dx = dz a 5 5 1 (ax + b) dx = z . dz ∫ ∫ a Unit-10 Page-234 Bangladesh Open University 1 5 = z dz a ∫ 1 z 6 = . + c a 6 1 (ax + b)6 = . + c a 6 Example-2: 2 5 Evaluate ∫ (xx + )4 dx Solution: 2 Let, x + 4 = z 2xdx = dz 1 xdx = dz 2 2 5 5 1 (xx + )4 dx = z . dz ∫ ∫ 2 1 5 = z dz 2 ∫ 1 z 6 = . + c 2 6 1 2 6 = (x + )4 + c 12 Example-3: 2 x3 Evaluate ∫ x e dx Solution: 3 Let, x = z 2 3x dx = dz 1 x 2dx = dz 3 3 1 2ex x dx = ez . dz ∫ ∫ 3 1 = e z + c 3 1 3 = ex + c 3 Example-4: 2 Evaluate ∫ x x +1 dx . Solution: 2 Let, x + 1 = z 2xdx = dz Business Mathematics Page-235 School of Business 1 xdx = dz 2 2 1 ∫ x x +1 dx = z. dz ∫ 2 1 1 = z 2 dz 2 ∫ 3 1 z 2 = . + c 2 3 2 3 1 2 = (x + )1 2 + c 3 Integration by Parts Integration by parts is Integration by parts is a special method that can be applied in finding the a special method that integrals of a product of two integrable functions. This method of can be applied in integration is derived from the rule of differentiation of a product of two finding the integrals functions.