SAJC Check your Understanding_integration_tutor

Check your Understanding (Integration) Section 1: Integration Techniques 1. ACJC JC1 Promo 8865/2019/Q3 (i) Find 23e dx x−  . [2] (ii) Find 1 d 31 x x+ . [3] (iii) Without using a calculator, find 22 1 2 2 dxx x  + . [4] ACJC JC1 Promo 8865/2019/Q3 (Solutions) (i) 2 3 2 3 1 d 3 xxe x e c−− =− + (ii) 12 d 3 1331 x x c x = + + + (iii) 222 11 22 1 2 24 2 d 8 4 d 44ln 8 2 4ln 2 8(2) 2(2) 4ln1 8 2 4ln 2 14 x x x x xx xxx  + = + +   = + + = + + − − − = +  2. CJC JC1 Promo 8865/2019/Q6 Find (a) 4 2 d( 1) xx− , [3] (b) 125 ed15 x xx − + + . [4] CJC JC1 Promo 8865/2019/Q6 (Solutions) (a) ( ) 4 2 d 1 x x− ( ) 4 2 1 dxx − =− ( ) 3 12c 3 x − −=+  − ( ) 3 2 c 31x =− + −

(b) 125 ed15 x xx − + + ( ) 125ln 1 5 e c52 xx −+= + + − ( ) 121ln 1 5 e c 2 xx −= + − + 3. SAJC JC2 Prelim 8865/2019/Q1 Find the exact value of a such that 4 02 3 11 d d31 a xxx x=− . [3] SAJC JC2 Prelim 8865/2019/Q1 (Solutions) ( ) ( ) 4 02 3 12 12 4 02 3 ln 3 1 23 ln 3 1 043 31 1 3 11 d d31 a a x x a ae ea xxx x − =  − −= −= += =− 4. JPJC JC2 Prelim 8865/2019/Q2 (a) Find 2 2 1d x x  − . [3] (b) Find the exact value of n given that 2 0 11 d 14(7 2 ) n x x = − , where 7 2x . [3] JPJC JC2 Prelim 8865/2019/Q2 (Solutions) (a) 2 2 4 41 d 1 dxx xxx    − = − +       11 224 4 1 d 4ln 8x x x x x cx − = − + = − + + (b) 2 2 00 1 d (7 2 ) d (7 2 ) nn x x x x −=− − 1 00 (7 2 ) 1 1 1 1 1 ( 1)( 2) 2 7 2 2 7 2 7 n n x xn − −    = = = −    − − − −    

1 1 1 1 2 7 2 7 14 12 7 2 7 14 4 7 47 7 4 n n n n n  −= − =− −= = = 5. RI JC1 Promo 8865/2019/Q5 (a) Find    1 x − 2 x + e −3x   1 x + 2 x + e −3x d x. [4] (b) Show that 2 x 2 x + 3 = 2 x + 3 − k 2 x + 3 , where k is a positive integer to be determined. [1] Hence evaluate  −1 1 x 2 x + 3 d x, giving your answer in the form A + B 5 , where A and B are constants to be determined. [4] RI JC1 Promo 8865/2019/Q5 (Solutions) (a)    1 x − 2 x + e −3x   1 x + 2 x + e −3x d x =  1 x − (2 x + e −3x ) d x = lnx − x 2 + 1 3 e −3x + C (b) 2 x 2 x + 3 = (2 x + 3) − 3 2 x + 3 = 2 x + 3 − 3 2 x + 3 (shown)   −1 1 x 2 x + 3 d x = 1 2  −1 1 2 x + 3 − 3 2 x + 3 d x = 1 2    (2 x + 3) 3 2 3 2  2 − 3 (2 x + 3) 1 2 1 2  2 −1 1 = 1 2     (2 x + 3) 3 2 3 − 3 (2 x + 3) 1 2 −1 1 = 1 2       1 3 (5) 3 2 − 3 (5) 1 2 − 1 3 + 3

= 1 2   5 3 5 − 3 5 + 8 3 = 4 3 − 2 3 5 where A = 4 3 , B = − 2 3 6. NJC JC1 Promo 8865/2019/Q6(a)(b) (a) Show that ( )( ) 20 1 ee e d 2 xx x xx−− − +− = . [2] (b) If ( )0 2 1