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 2 10 xx xx    =+        ππ 0221 1 1π e 1 e e2 2 10     = − + − −         π 21 π 2e 310 =−   units3 y x (0,0) (0,1) ቀπ 2 , 0ቁ 𝑦 = e𝑥 sin 𝑥 𝑦 = e𝑥 π 4 using (i) answer

Chapter 11 Definite Integrals TMJC 2024 Page 5 of 10 Qn 5 2017/MJC Promo/4a 3 1 33 2 3 2 3 2 0 0 1 134 3 2 4 3 2 01 5 4 d 5 4 d 5 4 d 5 4 5 4 4 3 2 4 3 2 7 27 7 12 4 12 95 12 x x x x x x x x x x x x x x x x x x − + = − + − − +    = − + − − +       = − − − =    Qn 6 N2018/RVHS Promo/9 (i) Let 2 d 2(ln )(ln ) = 1 dd uxu x v v x x xx=  =  =  Let d1ln = 1 dd uu x v v x x xx=  =  =  (ii) Volume ( ) ( ) 4e2 42 1 π 4 e π dxy=−  ( ) 4e 24 1 16π e π ln dyy=−  ( ) 4e24 1 16π e π ln 2 ln 2x x x x x= − − +  ( ) 4 4 2 4 416π e π e 4 8e 2e 2= −  − + − 4416πe π 10e 2= − −  436π e 2π units=+ ( ) 2 ln dxx ( ) ( ) ( ) 22 2 2lnln d ln d ln 2 ln d xx x x x x x x x x x x =− =−   ( ) ( ) ( ) ( ) 22 2 2 ln d ln 2 ln d 1ln 2 ln d ln 2 ln 2 x x x x x x x x x x x x x x x x x x C =− = − −  = − + +   y x O (0,1) 𝑦 = e𝑥 x = 4

Chapter 11 Definite Integrals TMJC 2024 Page 6 of 10 Qn7 2018/YJC Promo/11 ( ) ( ) ( ) 3 3 21.031852 0 3e From GC, 1.03185, 3.09554 1Volume π 3.09554 1.03186 π e d3 = 6.17 x x xx xy xx = == =−  y x O 𝑦 = 𝑥e𝑥3 𝑦 = 3𝑥

Chapter 11 Definite Integrals TMJC 2024 Page 7 of 10 Qn 8 N2009/P1/11 (i) (ii) ( ) 2 2 2 2 2 22 2 2 2 2 2 2 2 2 2 1 2 1 2 f ( ) e f ( ) e e ( 2 ) e 2 e e 2 e 0 e 1 2 0 11e 0 or 2 2 (N.A e 0) 1 1 1When , e 2 2 2e 1 1 1When , e 2 2 2e x x x x x xx x x x xx x x x x x x xx xy xy − − − − − −− − − − − −− =  = + − = − −= −= = =  =  = = = =− =− =− Coordinates of the turning points are 1 1 1 1, and , 2 2e 2 2e     −−       (iii) ( ) 2 2 2 2 2 2 00 0 0 0 0 f ( )d e d 1ed 2 1 ed2 1 e2 1 ee2 1 (1 e )2 nn x n u n u nu n n x x x x uu u u − − − − − − = = = =−  = − + =−    Area of region between C and the positive x-axis 2 0 1lim f ( ) d (1 0) 0.5 units2 n n xx → = = − = y (0,0) x x x 2 2 d1 d 2 When 0 0 When xu x x u u u xu x n u n =  =  = =  = =  =

Chapter 11 Definite Integrals TMJC 2024 Page 8 of 10 (iv) 2 22 20 2 4 f ( ) d 2 f ( )d (by symmetry) 12 (1 e )2 1e x x x x − − − = =  − =−  (v) Vol of revolution ( ) ( ) 2 1 2 0 1 2 0 3 π f ( ) d π e d 0.11570π 0.36349 0.363 units x xx xx− = = = =    Qn 9 2016/RVHS/Promo/11 (a) 2 2 22 dd , 3tan 3sec d( 9) xx xx x  =  = + , 2 2 22 9 tan 3sec d (9 tan 9)   = + 2 2 4 9 tan 3sec d 81sec   = 2 2 1 tan d3 sec   =  21 sin d3 =  1 (1 cos 2 ) d6 =− 1 sin 2 62 c= − + 1 sin 2 6 12 1 2sin cos 6 12 c c   = − + = − + 1 22 1 1 3tan (2)6 3 12 99 xx c xx −= − + ++ 1 2 1 tan63 2( 9) xx c x −= − + + tan3=x  3tan x=  9 sin 2 + = x x  9 3cos 2 + = x 

Chapter 11 Definite Integrals TMJC 2024 Page 9 of 10 (b)(i) When a = 3, Volume generated 23 20 2 π d 9 x x x =  + 3 1 2 0 12π tan63 2( 9) xx x −=−  + 1 π32π 6 4 36 =  − 211ππ12 6=− units3 (ii) In the actual hourglass, the neck connecting the two glass bulbs constitutes to a volume which is not accounted for in the theoretical working. (iii)       + a xx x 0 2 2 d9 π2 2112( π π)12 6=  − 6 1π12 1d9 0 2 2 −=     + a xx x 6 1π12 1 )9(23tan6 1 2 1 −=    +−− a ox xx 06 1π12 1 )9(23tan6 1 2 1 =+−+−− a aa Using GC, 4.8600148 4.86a=

Chapter 11 Definite Integrals TMJC 2024 Page 10 of 10 Qn 10 (a) Using GC, intersection between ( ) 2 21yx− = + and 26yx+= occur when 53 or  4xx== . Also, ( ) 2 2 1 2 1y x y x− = +  =  + ( ) ( ) ( ) ( ) 5 22 33 24 511 4 Volume generated π 2 1 d π 6 2 d π 2 1 d 78.5725 78.6 3 s.f. x x x x x x −− = + + + − − − + ==    (b) Points of intersection of curves are (−5, 9) and (0, 4). ( ) ( ) 29 4 9 222 0 0 4 16Volume π 2 d π 2 d π d 13 466.52 8.3775 107.66 350.48 350 (3 s.f) yy y y y y  −= − − − − + −  = − − = =   