TMJC H2 Chapter 10 Integration Techniques Learning Package 2024

TTaammppiinneess MMeerriiddiiaann JJuunniioorr CCoolllleeggee 22002244 HH22 MMaatthheemmaattiiccss ((99775588)) CChhaapptteerr 1100 IInntteeggrraattiioonn TTeecchhnniiqquueess LLeeaarrnniinngg PPaacckkaaggee Resources  Core Concept Notes  Discussion Questions  Extra Practice Questions SLS Resources  Recordings on Core Concepts  Quick Concept Checks

RReefflleeccttiioonn oorr SSuummmmaarryy PPaaggee

Chapter 10 Integration Techniques TMJC 2024 Page 1 of 29 H2 Mathematics (9758) Chapter 10 Integration Techniques Core Concept Notes Success Criteria: Surface Learning Deep Learning Transfer Learning  Integrate derivatives to obtain the anti-derivatives / integrals of a function (i.e. indefinite integration is the reverse process of differentiation)  Evaluate definite integrals using graphing calculator  Integrate the following standard functions: ➢ Constant, a ➢ nx ,  \1n− ➢ 1x− ➢ ex ➢ xa  Recognise and integrate integrands of the form  f ( ) f ( ) n xx , where n  Recognise and integrate integrands of the form f ( )f '( )e xx  Integrate basic trigonometric functions  Evaluate definite integrals using anti-derivatives: f ( ) d F( ) F( ) b a x x b a=− , where d F( ) f ( )d xxx =  Integrate powers of basic trigonometric functions  Use MF27 integral formulas to integrate functions of the following standard forms: (a) 22 1 ax+ (b) 22 1 ax− (c) ( ) 22 1 ,ax xa −  (d) ( ) 22 1 ,xa xa −   Recognise and integrate integrands of the form f ( ) dg( ) x xx or f ( ) d g( ) x x x  Use a given substitution to simplify and integrate an expression  Use integration by parts to integrate an expression

Chapter 10 Integration Techniques TMJC 2024 Page 2 of 29 §1 Indefinite Integrals If two functions F(x) and f(x) are related as follows: ( )d F( ) f ( )d xxx = , then f(x) is the derivative of F(x) and F(x) is called an anti-derivative or integral of f(x). Illustration: Since 2d ( 1) 2d xxx += , 2 1x + is an anti-derivative of 2x. Since 2d ( 7) 2d xxx −= , 2 7x − is an anti-derivative of 2x. Since 2d ( ) 2d xxx = , 2x is also an anti-derivative of 2x. Observe that   ( )d d dF( ) F( ) f ( ) 0 f ( )d d d x C x C x xx x x + = + = + = where C is any arbitrary real constant. Recall that the process of finding f ( ) dxx for a given function f is called integration. We know that f ( ) dxx is actually equivalent to F(x) + C, the collection of all anti-derivatives of f(x), and we write: Note: Integration is the “reverse” of differentiation. i.e.: f ( )d F( )x x x C=+ because ( )d F( ) f ( )d x C xx += Example: (1) cos d sinx x x C=+ because ( )d sin cosd x C xx += (2) ( ) ( ) 9 10 15 d 5 10x x x C− = − + because ( ) ( ) ( ) 10 9 9d 1 10 5 5 5d 10 10 x C x xx  − + = − = − (3) 51e dx x+  = 511 e5 x C+ + because 5 1 5 1 5 1d 1 5 e e ed 5 5 x x x Cx + + + + = = Indefinite integral of f(x): Integral sign x : Variable of integration C : Constant of integration f : Integrand dx : Integrate with respect to x

Chapter 10 Integration Techniques TMJC 2024 Page 3 of 29 §2 Basic Properties 1. [f ( ) g( )] d f ( ) d g( ) dx x x x x x x+ = +   2. f ( ) g( ) d f ( ) d g( ) dx x x x x x x− = −   3. f ( )d f ( ) dk x x k x x= ,where k is any non-zero real constant. Quick Check: Are the following True/ False? Example 1 Find ( ) 22d ed xxx . Hence, find ( ) 2e 1 dxx x x + . Solution: ( ) ( ) 2 2 2 2 2 2d e 2 e 2 e 2 e 1d x x x xx x x x xx = + = + ( ) ( ) 2 2 2 2 2 2 2 e 1 d e 1e 1 d e , where 22 xx xx x x x x C Cx x x x B B + = +  + = + =   1. Is ? E.g: ? 2. Is E.g: ? 3. Is Eg: T / F T / F T / F Learning Point: Integration is the reverse process of Differentiation.

Chapter 10 Integration Techniques TMJC 2024 Page 4 of 29 §3 Computation of Definite Integrals The definite integral from a to b of f(x), where f(x) is continuous on the interval [a, b], is given by :  f ( ) d F( ) F( ) F( ), b b aa x x x b a= = − where d F( ) f ( )d xxx = a and b are called the limits of integration, where a is the lower limit and b is the upper limit. f ( ) d b a xx is called the definite integral from a to b of f(x) w.r.t x. Note: the constant of integration C is eliminated in the subtraction. Proof:  f ( ) d F( ) b b aa x x x C=+    F( ) F( )b C a C= + − + F( ) F( )ba=− Some Important Results • f ( ) d f ( ) d ba ab x x x x=− e.g. : 20 02 f ( ) d f ( ) dx x x x − − =− • f ( ) d f ( ) d f ( ) d b c c a b a x x x x x x+=   e.g. : 2 3 3 1 2 1 f ( ) d f ( ) d f ( ) dx x x x x x+=  

Chapter 10 Integration Techniques TMJC 2024 Page 5 of 29 §4 Computation of Definite Integrals using Graphing Calculator Example 2 Evaluate 3 2 1 2 dxx − . Steps/Keystrokes/Explanations Screen Display 1. Press ƒp and select 4: fnInt(. 2. Key in the lower and upper limits, integrand and variable of integration and press Í 3. To convert answer to exact form, press ƒo and select 4:►F◄►D to switch between fraction and decimal. Press Í Alternative Press » and select 1:►Frac to convert to fraction. Press Í. Solution: Using GC, 3 2 1 562 d 3xx − =

Chapter 10 Integration Techniques TMJC 2024 Page 6 of 29 §5 Integrals of Standard Functions [Not in MF27] ‘O’ Level 1 da x ax C=+ 2 For ,n a) 1n− , 1 d 1 n n xx x C n + =+ + b) 1 11, d dn x x x x −=− =  ln Cx=+ *There is a need to put modulus here to ensure the existence of the integral. 3 e d exx xC=+ 4 d ln x x aa x C a=+ ***VERY Important Result***: In general, if ( ) ( )f d Fx x x C=+ , then we have ( ) ( )1f d Fpx q x px q Cp+ = + + . (Result can be proven using integration by substitution.) Example 3 (a) 5 3 6 2 253d 15 2ln22 Cx xx xx x x +− = + + +  (b) 1 d43 1 n4 43l xx Cx= + + +  Golden Rule: Replace ( ) by 4 3xx + , so we need to divide by coefficient of x which is 4. Recall: 1 d lnxC xx =+ Golden Rule: When x is replaced by a linear form ( )px q+ , we divide the answer by coefficient of x.

Chapter 10 Integration Techniques TMJC 2024 Page 7 of 29 (c) ( ) ( ) 5 6 6 2 1 d 211 26 1 (2 1)12 xx x C xC − −=  + = − +  (d) 34 34 ed 1 e3 x x x C − −=+  §6 Integration involving the function f (x) and its derivative f '(x) 1 ( ) ( ) ( ) 1 ff ' f d 1 For , 1, n n xx x x C n nn + =+ +  −  Proof: Since   ( )  1d f ( ) 1 f ( ) f ( )d nn x n x xx + =+ and integration is a reverse process of differentiation, ( ) ( ) 1 1 1 [f ( )] f ( ) d [f ( )] [f ( )][f ( )] f ( ) d 1 nn n n n x x x x xx x x C n + + +=  =+ +   2   1 For , 1, f ( )f ( ) f ( ) d d ln f ( ) f ( ) nn xx x x x x C x −  =−  = = + Proof: Since d f ( )ln f ( )d f ( ) xxxx = and integration is a reverse process of differentiation f ( ) d ln f ( )f ( ) x x x Cx  =+ 3 f ( ) f ( )f ' ( )e d exxx x C =+ Proof: Since f ( ) f ( )d e e f ( )d xx xx = and integration is a reverse process of differentiation, f ( ) f ( )e f ( ) d exx x x C =+ Golden Rule: Replace ( ) by 3 4xx − , so we need to divide by coefficient of x which is 3. Recall: e d exx xC=+ Golden Rule: Replace ( ) by 2 1xx − , so we need to divide by coefficient of x which is 2. Recall: 1 d 1 n n xx x C n + =+ +

Chapter 10 Integration Techniques TMJC 2024 Page 8 of 29 Example 4 Concept: 1[f ( )]For , 1, f ' ( ) [f ( )] d 1 n n xn n x x x C n +  − = + + (a) 23(1 ) dx x x+