JUST THE MATHS SLIDES NUMBER 12.1 INTEGRATION 1 (Elementary - PDF document

“JUST THE MATHS” SLIDES NUMBER 12.1 INTEGRATION 1 (Elementary indefinite integrals) by A.J.Hobson 12.1.1 The definition of an integral 12.1.2 Elementary techniques of integration

UNIT 12.1 - INTEGRATION 1 ELEMENTARY INDEFINITE INTEGRALS 12.1.1 THE DEFINITION OF AN INTEGRAL In Differential Calculus, we are given functions of x and asked to obtain their derivatives. In Integral Calculus, we are given functions of x and asked what they are the derivatives of. The process of answering this question is called “integration” . Integration is the reverse of differentiation. DEFINITION Given a function f ( x ), another function , z , such that d z d x = f ( x ) is called an integral of f ( x ) with respect to x . Notes: (i) having found z , such that d z d x = f ( x ) , z + C is also an integral for any constant value, C . 1

(ii) We call z + C the “indefinite integral of f ( x ) with respect to x ” and we write � f ( x )d x = z + C. (iii) C is an arbitrary constant called the “constant of integration” . (iv) The symbol d x is a label, indicating the variable with respect to which we are integrating. (v) In any integration problem, the function being inte- grated is called the “integrand” . Result: Two functions z 1 and z 2 are both integrals of the same function f ( x ) if and only if they differ by a constant. Proof: (a) Suppose, firstly, that z 1 − z 2 = C, where C is a constant. Then, d d x [ z 1 − z 2 ] = 0 . 2

That is, d z 1 d x − d z 2 d x = 0 , or d z 1 d x = d z 2 d x . (b) Secondly, suppose that z 1 and z 2 are integrals of the same function. Then, d z 1 d x = d z 2 d x . That is, d z 1 d x − d z 2 d x = 0 , or d d x [ z 1 − z 2 ] = 0 . Hence, z 1 − z 2 = C where C may be any constant. 3

Any result encountered in differentiation could be re- stated in reverse as a result on integration. ILLUSTRATIONS 1. � 3 x 2 d x = x 3 + C. 2. � x 2 d x = x 3 3 + C. 3. � x n d x = x n +1 n + 1 + C Provided n � = − 1 . 4. � 1 � x − 1 d x = ln x + C. x d x i . e . 5. � e x d x = e x + C. 6. � cos x d x = sin x + C. 7. � sin x d x = − cos x + C. 4

Note: Basic integrals of the above kinds may be quoted from a table of standard integrals in a suitable formula booklet. More advanced integrals are obtainable using the rules which follow. 12.1.2 ELEMENTARY TECHNIQUES OF INTEGRATION (a) Linearity Suppose f ( x ) and g ( x ) are two functions of x while A and B are constants. Then � [ Af ( x ) + Bg ( x )]d x = A � f ( x )d x + B � g ( x )d x. The proof follows because differentiation is already linear. The result is easily extended to linear combinations of three or more functions. ILLUSTRATIONS 1. � ( x 2 + 3 x − 7)d x = x 3 3 + 3 x 2 2 − 7 x + C. 2. � (3 cos x + 4sec 2 x )d x = 3 sin x + 4 tan x + C. 5

(b) Functions of a Linear Function � f ( ax + b ) d x. (i) Inspection Method EXAMPLES 1. Determine the indefinite integral � (2 x + 3) 12 d x. Solution To arrive at (2 x + 3) 12 by differentiation, we must be- gin with a function related to (2 x + 3) 13 . In fact, d (2 x + 3) 13 = 13(2 x + 3) 12 . 2 = 26(2 x + 3) 12 . � � d x This is 26 times the function we are trying to integrate. Hence, � (2 x + 3) 12 = (2 x + 3) 13 + C. 26 2. Determine the indefinite integral � cos(3 − 5 x )d x. 6

Solution To arrive at cos(3 − 5 x ) by differentiation, we must begin with a function related to sin(3 − 5 x ). In fact d d x [sin(3 − 5 x )] = cos(3 − 5 x ) . − 5 = − 5 cos(3 − 5 x ) . This is − 5 times the function we are trying to integrate. Hence, � cos(3 − 5 x ) = − sin(3 − 5 x ) + C. 5 3. Determine the indefinite integral � e 4 x +1 d x. Solution To arrive at e 4 x +1 by differentiation, we must begin with a function related to e 4 x +1 . In fact, d e 4 x +1 = e 4 x +1 . 4 � � d x This is 4 times the function we are trying to integrate. Hence, � e 4 x +1 d x = e 4 x +1 + C. 4 7

4. Determine the indefinite integral 1 � 7 x + 3d x. Solution 1 To arrive at 7 x +3 by differentiation, we must begin with a function related to ln(7 x + 3). In fact, d 1 7 d x [ln(7 x + 3)] = 7 x + 3 . 7 = 7 x + 3 This is 7 times the function we are trying to integrate. Hence, 7 x + 3d x = ln(7 x + 3) 1 � + C. 7 Note: We treat the linear function ax + b like a single x , then divide the result by a . 8

(ii) Substitution Method In the integral � f ( ax + b )d x, we may substitute u = ax + b as follows: Suppose � f ( ax + b )d x. z = Then, d z d x = f ( ax + b ) . That is, d z d x = f ( u ) . But d z d u = d z d x. d x d u = f ( u ) . d x d u. Hence, � f ( u )d x z = d u d u. Note: The secret is to replace d x with d x d u . d u . 9

EXAMPLES 1. Determine the indefinite integral � (2 x + 3) 12 d x. z = Solution Putting u = 2 x + 3 gives d u d x = 2 and, hence, d x d u = 1 2 . Thus, � u 12 . 1 2d u = u 13 13 × 1 z = 2 + C. That is, z = (2 x + 3) 13 + C 26 as before. 2. Determine the indefinite integral � cos(3 − 5 x )d x. z = Solution Putting u = 3 − 5 x gives d u d x = − 5 and hence d x d u = − 1 5 . Thus, � cos u. − 1 5d u = − 1 z = 5 sin u + C. That is, z = − 1 5 sin(3 − 5 x ) + C, as before. 10

3. Determine the indefinite integral � e 4 x +1 d x. z = Solution Putting u = 4 x + 1 gives d u d x = 4 and, hence, d x d u = 1 4 . Thus, 4d u = e u � e u . 1 z = 4 + C. That is, z = e 4 x +1 + C, 4 as before 4. Determine the indefinite integral 1 � z = 7 x + 3d x. Solution Putting u = 7 x + 3 gives d u d x = 7 and, hence, d x d u = 1 7 Thus, � 1 u. 1 7d u = 1 z = 7 ln u + C. That is, z = 1 7 ln(7 x + 3) + C. as before. 11