RI revision programme calculus 2 (hard)

RAFFLES INSTITUTION H2 Mathematics 9758 2025 Year 6 Term 3 Revision 4 (Zeta) Topic(s): Calculus II (Integration, Applications, DE) (Solutions) Y6 H2 Math Term 3 Revision Lecture 5: Calculus II Page 1 of 15 Source of Question: TJC Prelim 9758/2022/01/Q1 Solution: (a) 11sin 2 d 1.sin 2 dxx xx     1 2 1 12 2 1 2 2 1 1 2 12 2sin 2 . d 14 1sin 2 8 1 4 d4 141sin 2 . 4 1sin 2 1 4 2 xx x x x xx xxx x xx c xx x c                  (b) 222 22 2 2 1 1 11 d d 11 24 1 d 13 24 1 2 2 tan 33 2 22 = tan 33 xxxn x n xn n n x xn n xn c n n xn nn                  c  1 Find (a) 1sin 2 d ,xx  [3] (b) 22 1 d , where is a non-zero constant.xnxn x n [2]  1 2 dsin 2 1 d d1 2 d 12 vux x u vxx x    

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________________ Y6 H2 Math Term 3 Revision Lecture 5: Calculus II Page 2 of 15 Source of Question: TMJC Prelim 9758/2022/01/Q2(modified) 2 (a) Find  21 1 d 1 sin x xx    . [2] (b) Find 2 3 d23 x xxx    . [3] Solution: (a)     1 1 2 21 2 1 1 2 1 11 ds i n d 1s i n 1 sin 1 2 2s i n x xx xx x x C xC             (b) 2 33 dd23 (1 ) (3 ) 11 31 dd21 23 13ln 1 ln 322 xx xxxx xx x xxx x xC                 Alternatively,   22 2 2 2 31 2 2 4dd d23 2 23 23 11ln 2 3 4 d2 14 14 1 2ln ( 3)( 1) ln24 1 2 11ln 3 ln 1 ln 1 ln 322 13ln 1 ln 322 xx xx xxx xx xx xx x x xxx C x x xx x C xx C                                The integrand is of the form f( ) ( f ( ) ) nx x , where 1f( ) s i nx x and 1 2n .

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________________ Y6 H2 Math Term 3 Revision Lecture 5: Calculus II Page 3 of 15 Source of Question: NYJC Prelim 9758/2022/02/Q5 (modified) 3 (a) (i) Differentiate 2 4x  with respect to x. [1] (ii) Hence find 3 2 d 4 x x x    . [3] (b) Find (i) 3sin dmx x , where m is a non-zero constant, [2] (ii) 4cos 3 d .x x [4] Solution: (a)(i)   2 22 d2 4d 24 4 x xxx xx    (a)(ii) 2ux 2 d d 4 vx x x   d 2d u xx  2 4vx 3 22 2 2 d4 2 4 d 4 x x xx x x x x         3 2 2 22 24 4 3 x x xC     (b)(i)   32 2 2 3 sin d sin sin d sin 1 cos d sin sin cos d cos cos 3 mx x mx mx x mx mx x mx mx mx x mx mx Cmm          (b)(ii)  24 2 1cos 3 d 1 cos 6 d4 1c o s 6 1 cos 6 d424 1 cos 6 1 1 cos12 d424 2 1s i n 6 1s i n 1 2 41 2 8 9 6 3 sin 6 sin12 81 2 9 6 xx x x x xx xx x xxx xC xxxC               

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________________ Y6 H2 Math Term 3 Revision Lecture 5: Calculus II Page 4 of 15 Source of Question: NJC Prelim 9758/2022/01/Q1 4 (i) Given that  e 1 ln d n nIx x x for , 0 ,nn  show that 2 1 e 22 nn nII  for all .n   [2] (ii) Find the exact volume of the solid genera ted when the region bounded by the curve ln ,yx x the x-axis and the line ex is rotated completely about the x-axis. [3] Solution: (i)     e 1 e e22 1 11 2 e 1 1 2 1 ln d 1ln ln d22 e ln d22 e (shown)22 n n nn n n Ix x x xx xn x x x n xx x n I            (ii) Vo l u m e   e 2 1 2 2 1 22 0 e2 0 1 23 π ln d π e2π 22 ee 1π 22 2 ππ 22 2 π e1 u n i t s4             xxx I I I xI

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________________ Y6 H2 Math Term 3 Revision Lecture 5: Calculus II Page 5 of 15 Alternatively: Vo l u m e       e 2 1 e e22 2 11 2 e 1 e e22 2 11 e22 2 1 23 π ln d 1π ln 2 ln d22 eπ ln d2 e1π ln d22 2 eeπ 22 4 π e 1 units4 xxx xx x xx x xx x xx xx x x                     2 2 dln d d1 2 ln d2 vux x x ux xvxx   2 dln d d1 d2 vux x x ux vxx  

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________________ Y6 H2 Math Term 3 Revision Lecture 5: Calculus II Page 6 of 15 Source of Question: ACJC Prelim 9758/2022/01/Q4 (modified) 5 (a) The continuous function f( )x , where f( ) 0x  , is strictly decreasing for 1x  . Sketch the curve f( )yx for 1kxk where k is an integer and 1.k By comparing the areas of appropriate rectangles and the area under the curve f( ) ,yx show that for any integer 1,k  1 f( 1 ) f( )d f( ) k k kx x k     . [2] (b) Hence show that (i) 11 1ln10 1 .... 1 ln 923 9     . [3] (ii)   ln ( 1)! ln 1 ln !nn n n n    for n > 1. [4] Solution: (a) Area under curve =  1 fd k k x x   Area of rectangle A= f1 k x 1 Area of rectangle B= f k x 1 As seen from diagram:      1 f1 f d f k k kx x k    f(k) f(k+1) B y x k O k+1 A 1 y = f(x)

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________________ Y6 H2 Math Term 3 Revision Lecture 5: Calculus II Page 7 of 15 (b)(i) Let 1f( ) .x x 10 2 3 4 10 11 2 3 9 11 1 1 1d d d d .... d f(1)+f(2)+f(3)+.....+f(9) 111 1 .....123 9 xx x x xxx x x x         92 3 4 9 11 2 3 8 9 1 9 1 11 1 1 1d d d d .... d f(2)+f(3)+f(4)+.....+f(9) 11 1 1 ..... 123 9 11 1 11 d 1 ..... 23 9 k xx x x xxx x x x k xx                Combining both inequalities we have   910 9 11 1 10 9 11 111d1 d 11 1ln 1 ..... 1 ln23 9 11 1ln10 1 ..... 1 ln 923 9 k xxxkx x x           (b)(ii) We consider the function f(x) = ln x. This function is strictly increasing, so the inequality in (a) changes to     1 ff d f 1 k k kx x k    . We thus have                11 1 1 11 1 2 11 1 1 ff d f 1 ln1 ln 2 ... ln( 1) f d ... f d ln 2 ... ln ln ( 1)! ln d ln ! ln ( 1)! ln ln ! ln ( 1)! ln 1 ln ! nn n k kkk k n n n n kx x k nx x x x n nx x n nx x x n nn n n n                              

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________________ Y6 H2 Math Term 3 Revision Lecture 5: Calculus II Page 8 of 15 Source of Question: VJC Prelim 9758/2022/01/Q9 6 The curve C has equation 221 203 xy y . (i) Sketch C. [2] (ii) Use the substitution sinxp  to show that 2 22 0 π d 4 p ppxx , where p is a positive constant. [4] (iii) The region R is bounded by C, the line 3x and the x-axis. Find the exact area of R. [3] (iv) R is rotated completely about the y-axis. Find the exact volume of the solid obtained. [3] (v) Describe a pair of trans