TMJC H2 Chapter 11 Definite Integrals Extra Practice Solutions

Chapter 11 Definite Integrals TMJC 2024 Page 1 of 10 H2 Mathematics (9758) Chapter 11 Definite Integrals Extra Practice Solutions Qn 1 2009/VJC Prelim/1/10 ( ) 2 2 2 22 2 11e d e e d22 11 ee24 e 2 1 (shown)4 x x x xx x x x x x xc xc =− = − + = − +  (a) 48 44 04 11Required area e e d e e d22 xxx x x x x x   = − + −       = 7239.2 units2 (1 dp) (b) Required Volume ( )( ) ( ) 442 0 1 π 2e 4 π e d3 xxx=−  ( ) ( ) 482 0 88 83 16πe eπ 2 134 16πe e 1π 7 π3 4 4 43 ππe units12 4 x x= − −  = − −  =−  43 1 and 12 4AB==

Chapter 11 Definite Integrals TMJC 2024 Page 2 of 10 Qn 2 2011/TJC Prelim/1/10 ( )( ) ( )( ) 3 2 1 6 1Area 2 3 2 d 3 2 322 xx= + −    3 26ln x= ( )3 ln 3 ln 2=− units2 ( ) 22 2 2 4 3 0 2 1x y y x y+ − + =  + − = ( ) 1 2 2 2 2 1 2 1 20 Volume 2 π π d where 2 1 and 2 1y y x y x y x= − = + − = − − ( ) ( ) 221 22 0 2π 2 1 2 1 dx x x= + − − − −  39.5 units3 (3 s.f.) y x 𝑦 = 6 𝑥 𝑦 = 2𝑥 𝑦 = 3𝑥 ൫ξ2, 3ξ2൯ ൫ξ3, 2ξ3൯ (0,0) 𝑦 = 0 𝑥 = 0 y x 2 1 3 𝑥2 + 𝑦2 − 4𝑦 + 3 = 0

Chapter 11 Definite Integrals TMJC 2024 Page 3 of 10 Qn3 2013/VJC Prelim/1/10 (i) 1 1 1 2 112 2 0 20 0 1 2 2 0 cos d cos d 1 1 π 142 π1 1242 π1 1 4 2 2 xx x x x x x x −− = − −  −  = − −  = − −  = + −    (ii) 21 22 0 1 2 12 0 22 3 π1Volume π d π 2 2 ππ cos d 42 π π ππ 4 2 2 4 2 ππ unit 2 yx xx−  =−    =− = + − − =−   (iii) Required Equation: 1cos . 2yx −=− 21 12 0 Volume cos d 2 0.116 (3 s.f.) xx  −=−    Qn 4 2013/TJC Prelim/1/9 (i) ( )2 2 2 11e cos 4 d e cos 4 e 4sin 4 d22 x x x x x x x x= − − 221 e cos 4 + 2 e sin 4 d2 xx x x x=  ( )2 2 21 1 1e cos 4 + 2 e sin 4 e 4cos 4 d2 2 2 x x x x x x x=−   2 2 21 e cos 4 + e sin 4 4 e cos 4 d2 x x x x x x x=−  ( )22 15 e cos 4 d e cos 4 2sin 42 xx x x x x=+ + C 2e cos 4 dx xx = ( )21 e cos 4 2sin 4 + 10 x x x C+

Chapter 11 Definite Integrals TMJC 2024 Page 4 of 10 (ii) At the point of intersection, ex sin 2x = ex  ex (sin 2x − 1) = 0 either ex = 0 or sin 2x = 1 (reject, as ex > 0) or π2 2x= π 4x= (iii) ( ) ( ) ππ 44 22 00 Volume of solid generated π e d π e sin 2 dxx x x x=− ππ 44 2 2 2 00 π e d π e sin 2 dxx x x x=− ( ) ππ 44 π 4 22 00 22 0 1 cos 4π e d π e d 2 1 π e e cos 4 d2 xx xx xxx xx −=−  =+   ( ) ππ 4422 00 1 1 1π e + e cos 4 2sin 42