SAJC 2022 Chapter 7 Integration Techniques final(Tutor)

SAJC 2022 JC2 H1 Mathematics Chapter 7: Integration Techniques Page 1 of 17 Chapter 7 (Pure Mathematics) : Integration Techniques Objectives: At the end of the chapter, you should be able to (a) understand that integration is the reverse of differentiation (b) find the integral of nx , for any rational n, and, xe , together with constant multiples, sums and differences (c) find the integral of ( ) n ax b+ , for any rational n, and ()ax be + Content 7.1 Introduction 7.2 Definition 7.2.1 Basic Properties of Indefinite Integral 7.3 Integrate Basic Functions 7.3.1 Integrate Algebraic Functions 7.3.2 Integrate Exponential Function 7.4 Miscellaneous Examples References 1. New Additional Mathematics, Ho Soo Thong (Msc, Dip Ed), Khor Nyak Hiong (Bsc, Dip Ed) 2. New Syllabus Additional Mathematics (7th Edition), Shinglee Publishers Ptd Ltd

SAJC 2022 JC2 H1 Mathematics Chapter 7: Integration Techniques Page 2 of 17 7.1 Introduction If you have the gradient function d 25d y xx =+ and some related information, can you find the equation of the curve y ? The answer is yes. The process of getting y from d d y x is called Integration, also sometimes known as anti-differentiation. Example 1 Given the gradient function d 2d y xx = , find the general equation of y. Solution: We know that if we differentiate 2x , we will get 2x . Hence 2yx= . But this is not the complete picture , because if you differentiate 2 1x + , you will also get the same answer, 2x . The same is true for 2 2yx=+ . In fact if you differentiate 2x plus any constant, you get the same answer. Therefore 2d 2 d y x y x Cx =  = + , where C is an arbitrary constant. This reversal of differentiation, known as integration, is represented as follows: 22 dx x x C=+ C is the "Constant of Integration". It is there because of functions whose derivative are 2x is not unique. For example, 2d ( 4) 2d xxx −= ; 2d ( 20) 2d xxx += , etc… 7.2 Definition Integration is the reverse process of Differentiation. If ( ) ( )d Ffd xxx = , then ( )f ( ) d Fx x x C=+ . where ( )dF f ( )d x xx = and C is an arbitrary constant. Integral symbol Function we want to integrate Variable to integrate

SAJC 2022 JC2 H1 Mathematics Chapter 7: Integration Techniques Page 3 of 17 7.2.1 Basic Properties of Indefinite Integral Let f and g be two functions. Then, 1. dk x kx C=+ , where C is an arbitrary constant 2. Multiply function by constant: ( ) ( )f d f dk x x k x x= 3. Sum and Difference of functions: ( ) ( )f g dx x x= ( ) ( )f d g dx x x x 7.3 Integrate Basic Functions 7.3.1 Integrate Algebraic Functions Recall when we differentiate nx , we multiply by the number in the power and reduce the power by 1. i.e. 1d( ) d n nx nxx −= In integration, we raise the power by 1 and divide by 1n+ . See (a) Note: For (a) to (d) below, C represents an arbitrary constant. Common functions Function Integral (a) Power where 1n− dnxx 1 ,11 nx Cnn + = + −+ ( ) d n ax b x+ ( ) ( ) 1 ,11 n ax b Cnan + += + −+ (b) Constant 0ddk x kx x= kx C=+ Special Case (when power 1n=− ) (c) Power where 1n=− 1 1ddx x x x − = ln xC=+ (d) 1 1( ) d dax b x x ax b −+= + ln ax b Ca +=+

SAJC 2022 JC2 H1 Mathematics Chapter 7: Integration Techniques Page 4 of 17 Example 2 (Using results (a) to (b)) Find the following indefinite integrals. Solution: (i) 4 3 d 4 xx x C=+ , where C is an arbitrary constant (vi) ( )2dxx+ 2 22 x xC= + + OR ( )2dxx+ ( ) ( ) ( ) 22 22 2 1 2 xx CC++= + = + , where C is an arbitrary constant (ii) 02 d 2 d 2x x x x C= = + , where C is an arbitrary constant (vii) ( ) ( ) ( ) ( ) 7 6 7 353 5 d 73 35 ,21 xx x C x C −− = + −=+  where C is an arbitrary constant (iii) 1 2 3 2 3 2 dd 3 2 2 ,3 x x x x x C xC = =+ =+  where C is an arbitrary constant (viii) ( ) ( ) ( ) ( ) ( ) 3 3 2 2 2 d 2 3 4 d 34 342 23 1 , 3 3 4 x x x x x C C x − − =+ + +=+ − =− + +  where C is an arbitrary constant (iv) 1 2 1 2 1 2 1 dd 1 2 2, x x x x x C xC − = =+ =+  where C is an arbitrary constant (ix) ( ) ( ) ( ) 1 2 1 2 2 d 2 3 1 d 31 312 1 32 4 3 1 ,3 x x x x x C xC − =+ + +=+ = + +  where C is an arbitrary constant (v) 1 dxx x  + 31 222 2,3 x x C= + + where C is an arbitrary constant (x) ( ) ( ) 2 32 2 d 2 d 2.32 x x x x x x xx C − = − = − + 

SAJC 2022 JC2 H1 Mathematics Chapter 7: Integration Techniques Page 5 of 17 3 2 ,3 x xC= − + where C is an arbitrary constant Example 3 (using results (c) and (d)) Find the following indefinite integrals. Solution: (i) 3 3 1dd22 3 ln ,2 xxxx xC = =+  where C is an arbitrary constant (iii) 31 d 3 d1 2 1 2 ln 1 23 2 3 ln 1 2 ,2 xxxx x C xC =−− −=+ − =− − +  where C is an arbitrary constant (ii) ln 2 11 d,2 1 2 xxCx +=++ where C is an arbitrary constant (iv) ( ) ( ) ( ) ( ) ( ) ( ) 2 21 1 11 d4313 1 3 d 4 3 d ln 4 313 1 3 4 ln 4 31 ,3 1 3 4 xxx x x x x xx C x Cx −− −  + +− = − + + +−= + +−− += + +−   where C is an arbitrary constant

SAJC 2022 JC2 H1 Mathematics Chapter 7: Integration Techniques Page 6 of 17 7.3.2 Integrate Exponential Function Function Integral dxex xeC=+ , where C is an arbitrary constant dax bex+  1 ax bea += C+ , where C is an arbitrary constant Example 4 Find the following indefinite integrals. Solution: (i) 2 d 2 d 2, xx x e x e x eC = =+  where C is an arbitrary constant (iv) 74 74 d , 7 x x ee x C − − =+ where C is an arbitrary constant (ii) 2 2 2 2 1 dd 2 1 ,2 x x x x x e xe e C Ce − − = =+− =− +  where C is an arbitrary constant (v) ( ) 22 33dd 3 d 3 1 3 , xx x x x xx x x x x ee xxe e e e e x eeC eC e − − − =− =− = − + − = + +   where C is an arbitrary constant (iii) ( ) 1 2 2 2 2 dd d 1 2 2, xx x x x e x e x ex e C eC = = =+ =+   where C is an arbitrary constant (vi) 3 2 3 2( ) d ( ) dx x x xe e x e e x− = − 2 4 6 2 4 6 2 4 6 ( 2 ) d 2 2 4 6 ,2 2 6 x x x x x x x x x e e e x e e e C e e e C = − + = − + + = − + +  where C is an arbitrary constant

SAJC 2022 JC2 H1 Mathematics Chapter 7: Integration Techniques Page 7 of 17 Example 5 Show that 2d ln 2 1d xx  + 2 2 21 x x = + . Hence find 2 3 d 21 x x x + . Solution: 2d ln 2 1d xx  + = 2d1 ln(2 1)d2 xx  + 2 14 2 21 x x =  + 2 2 21 x x = + (Shown) 2 3 d 21 x x x + 2 32 d2 21 x x x = + 23 ln (2 1)2 xC= + + where C is an arbitrary constant Exercise 1 1. Find the following indefinite integrals. (i) ( ) 53 2 4 dx x x+− (ii) 1 dx x x −  (iii) ( )( )2 3 dt t t−+ (iv) ( ) 3 5 7 dxx+ (v) 32 dxex (vi) 35 dxex −  Solution: (i) ( ) 62 5 6 2 323 2 4 d 4 62 4 , 2 xxx x x x C x x x C + − = + − + = + − +  where C is an arbitrary constant (ii) 11 22 13 22 3 2 1 d d 13 22 22, 3 x x x x x x xx C x x C −− =−  = − + = − +  where C is an arbitrary constant (iii) 2 32 ( 2)( 3)d ( 6) d 6,32 t t t t t t tt tC − − = + − = + − +  where C is an arbitrary constant (iv)

SAJC 2022 JC2 H1 Mathematics Chapter 7: Integration Techniques Page 8 of 17 3 3 2 5 2 5 2 (5 7) d (5 7) d (5 7) 5 (5)2 2 (5 7) ,25 x x x x x C xC + = + +=+ = + +  where C is an arbitrary constant (v) 3 3 22d 3 x x ee x C=+ , where C is an arbitrary constant (vi) 35 35 d, 3 x x ee x C − − =+ where C is an arbitrary constant 2. DHS Prelim 8865/2018/Q2 (i) Differentiate 23(e 1)x+ with respect to x. (ii) Hence find 42 e (e e ) d . x x x x+ Solution: (i) ( ) 31