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 11 d 214 n x x − − =  , determine the value of n. [3] NJC JC1 Promo 8865/2019/Q6 (Solutions) (a) ( )( ) ( ) 0 1 0 2 1 2 0 1 2 2 2 e e d ed e 22 1 e 1 2 2 2 e 2 xx x x xxx xx x −− − − − − − = =− +− − −  +=− + =   (b) ( ) ( ) ( ) ( )( ) ( ) 0 2 0 1 0 2 1 d 14 1 4 d 14 14 11 444 1 n n n x x xx x n − − =  − − − −− =  = −   −    ( ) ( ) ( ) ( ) 4 4 1 1 1 1 4 4 2 1 1 1 1 4 4 2 11 1 4 4 4 1 2 11 4 n n n n n − =− =− − =− = − = − − Section 2: Reverse process of Differentiation is Integration

7. NJC JC1 Promo 8865/2019/Q6(c) (c) Differentiate 2ee x x+ with respect to x. Hence, find 22e1 e6 e 3e d xxxxxx + + ++ . [4] NJC JC1 Promo 8865/2019/Q6 (Solutions) (c) ( ) ( ) 2 2 2 2e e e 1 e ed e 2e 1 2 e ed xx x x x x x x xxx ++ + + + + = += 2 2 2e 1 e e 12 e e d ex x x x x xx x C+ + + + + = + 22 22 2 e1 e e e e 2 6 e 3e d 3 2 e e d 3e x x x x x x x x x x xx xx C + + + ++ + + =+ =+   8. CJC JC2 Prelim 8865/2019/Q3(b) (b) (i) By first expressing 23 4 x x + + in the form 4 BA x+ + where A and B are constants, find d 2 3 d4 x xx +  + . [2] (ii) Hence find the exact value of ( ) ( ) 5 21 11 d244 xxx−  + ++  . [4] CJC JC2 Prelim 8865/2019/Q3(b) (Solutions) (b)(i) 23 44 2 3 ( 4) xB Axx x A x B + =+++ + = + + Method 1: Comparing coefficients, 2 4 3 3 4(2) 5 2 3 5 244 A A B B x xx = + =  = − =− + = −++ Method 2: 2 4 2 3 2 8 5 xx x ++ + − 2 3 5 244 x xx + = −++

( ) 2 d 2 3 d 5 2d 4 d 4 5 4 x x x x x x +    =−   ++    = + (b)(ii) ( ) ( ) ( ) 5 2 1 5 1 11 d244 1 2 3 1 ln 45 4 2 xxx x xx − −  + ++  + = + + +   1 13 1 1 1 1ln 9 ln 35 9 2 5 3 2 2 1 9 ln9 2 3 21 ln 392    = + − −       =+ =+ 9. NYJC JC1 Promo 8865/2019/Q4 (a) Differentiate 1 2 1e ln 2 3 x x   − −+ with respect to x. [3] (b) Differentiate ( ) 4 ln 2x with respect to x. Hence, find ( ) 32e 6 ln 2 d xxx xx +  . [4] (c) Given that 2 1 54 d 54 k x x x − = , find the value of k, where k is a real number. [5] NYJC JC1 Promo 8865/2019/Q4 (Solutions) (a) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 2 1 2 1 2 12 2 1 3 2 22 1 2 3 2 2 11e ln 2 e ln 233 1d e ln 2 3 d 112 1 e 123 e1 322 xx x x x xx x x xx xx − − −   − −− − − − − −   − − + = − +  −+ =− − − − − =+ − (b) ( ) ( ) ( ) ( ) 4 3 3 d ln 2 14 ln 2 2d2 4 ln 2 x xxx x x = =

( ) ( ) ( ) ( ) 332 2 3 2 42 e 6 ln 2 6 ln 2 d e d 4 ln 26e d d 4 13e ln 222 x x x x x x x xxxx xxx x xC + =+ =+ = + +   (c) ( ) 2 1 31 22 1 51 22 1 51 22 51 22 51 22 54 d 54 5 4 d 54 2 8 54 2 8 6 54 2 8 48 4 24 k k k x x x x x x xx kk kk kk − − = −=  −=   − − − =  −= −=   Let 1 2yk= 5 4 24yy−= Using G.C. 1 2 2yk== 4k = 10. DHS JC2 Prelim 8865/2019/Q2 (i) Differentiate 2 eln 1e x   + . [2] (ii) Hence find the exact value of the constant a such that 2 0 1 2 1 d.1e a x x− =+  [4] DHS JC2 Prelim 8865/2019/Q2 (Solutions) (i) ( ) 2 2 2 eln 1e ln e ln 1 e 11 ln 1 e2 x x x y =  + = − + = − + 22 22 d 1 2e e d 2 1 e 1 e xx xx y x  −=− = ++

(ii) 20 2 20 2 0 2 2 2 1 d1e e d1e eln 1e eeln ln 21e eeln ln 2 1e 1eln 2 a x xa x a x a a a x x −+ −=− + =−  + =− − + =− + +=   ( ) 2 2 2 2 1 2 1 e 1ln 22 1eln 1 2 1e e2 e 2e 1 ln 2e 1 a a a a a + =  + = + = =− =− 11. EJC JC1 Promo 8865/2019/Q6part (c) If 46ln(e 3)xy=+ , find d d y x . Hence evaluate 41 4 0 3e de3 x x x+ in terms of e . [5] EJC JC1 Promo 8865/2019/Q6 (Solutions) (c) 46ln(e 3)xy=+ 46ln(e 3)xy = + ( ) 4 4 44 d 6 24e 4ed e 3 e 3 x x xx y x == ++ 41 4 0 3e de3 x x x+ 41 4 0 1 24e d8 e 3 x x x= + 146 0 4 6 6 4 6 1 ln(e 3)8 1 ln(e 3) ln(4)8 1 e 3ln( )84 x=+  = + −  +=  or 146 0 4 4 1 6ln(e 3)8 3 ln(e 3) ln(4)4 3 e 3ln( )44 x=+  = + −  += 

12. TMJC JC2 Prelim 8865/2019/Q2 (i) Differentiate ( ) 2 5n 4l x x++ with respect to .x [2] (ii) Hence find 2 .5 2 d34 xx x x x+  + − +  [3] TMJC JC2 Prelim 8865/2019/Q2 (Solutions) (i) ( ) 2 2 d 45 45 24lnd x x xxxx +=++ ++ (ii) ( ) ( ) ( ) 3 2 2 3 2 2 2 2 d45 1 dd2 4 5 1 4 2 3 22 3 3ln 3 2 ln 2 52 1 452 xxx x x xx x x x x x xxC x x C x + − + −+  ++ =− ++ = + + =  −+++    Section 3: Application of Integration – Area 13. HCI JC2 Prelim 8865/2019/Q1 On the same axes, sketch the curves with the equations 2 2y x x= + + and 6 ln(2 1).yx= + − State clearly the equations of any asymptotes and the coordinates of the points where the two curves intersect. Find the area of the region bounded by the two curves, giving your answer correct to 3 decimal places. [5] HCI JC2 Prelim 8865/2019/Q1 (Solutions)