M2_Bk2_Sol_Ch09_E.pdf

# M2_Bk2_Sol_Ch09_E.pdf - 9 New Progress in Senior...

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© Hong Kong Educational Publishing Co. 56 New Progress in Senior Mathematics Module 2 Book 2 (Extended Part) Solution Guide 9 9 pp.87 p.87 1. No 2. Yes 3. Yes 4. Yes pp.60 – 90 9.1 First we divide the interval 1 0 x into p.60 n subintervals of width x Δ . Thus, . 1 0 1 n n x = = Δ If we choose i z as the right end point of each subinterval, then we have n i n i x i a z i = + = Δ + = 1 0 = →∞ = n i n n n i dx x 1 2 1 0 2 1 lim + + + = →∞ 3 2 3 2 3 2 2 1 lim n n n n n ) 2 1 ( 1 lim 2 2 2 3 n n n + + + = →∞ 6 ) 1 2 )( 1 ( 1 lim 3 + + = →∞ n n n n n 3 6 ) 1 2 )( 1 ( lim n n n n n + + = 6 1 2 1 1 lim + + = →∞ n n n 6 ) 0 2 )( 0 1 ( + + = 3 1 = 9.2 First we divide the interval 2 0 π x into p.61 n subintervals of width x Δ . Thus, . 2 0 2 n n x π = π = Δ n i n i x i a z i 2 2 0 π = π + = Δ + = π 2 0 cos tdt π π = = n i n n n i 1 2 2 cos lim π + + π + π π = →∞ n n n n n n 2 cos 2 2 cos 2 cos 2 lim π π + π = →∞ 2 1 4 sin 2 2 2 1 sin 2 lim n n n n n π π + π π = →∞ 2 1 4 sin 2 4 2 sin 2 lim n n n n π π π = →∞ 2 1 4 sin 2 4 cos 2 lim n n n n π π π π = n n n n n 4 4 cos 4 sin 4 lim 1 0 1 1 = = 9.3 (a) + 5 1 )] ( 5 ) ( 7 [ dx x g x f p.64 + = 5 1 5 1 ) ( 5 ) ( 7 dx x g dx x f + = 5 1 5 1 ) ( 5 ) ( 7 dx x g dx x f = 1 5 5 1 ) ( 5 ) ( 7 dx x g dx x f 36 ) 4 ( 5 ) 8 ( 7 = = Definite Integrals

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57 © Hong Kong Educational Publishing Co. Definite Integrals (b) 5 3 )] ( 3 ) ( 5 [ dx x g x f = 5 3 5 3 ) ( 3 ) ( 5 dx x g dx x f = 5 3 5 3 ) ( 3 ) ( 5 dx x g dx x f ] ) ( ) ( [ 3 ] ) ( ) ( [ 5 3 1 5 1 3 1 5 1 = dx x g dx x g dx x f dx x f + = 1 3 1 5 3 1 5 1 ] ) ( ) ( [ 3 ] ) ( ) ( [ 5 dx x g dx x g dx x f dx x f 12 ) 10 4 ( 3 ) 2 8 ( 5 = + = 9.4 Since + + dx x x ) 1 4 ( 3 C x x x + + + = 2 4 2 4 , p.68 x x x + + 2 4 2 4 is a primitive function of 1 4 3 + + x x . By the Fundamental Theorem of Calculus, + + 2 0 3 ) 1 4 ( dx x x 14 0 ) 0 ( 2 4 0 2 ) 2 ( 2 4 2 2 4 2 4 2 4 2 0 2 4 = + + + + = + + = x x x 9.5 + 4 2 4 2 1 5 dx e e x x p.68 dx e dx e x x x + = = 4 2 5 3 4 2 ) 4 2 ( 1 5 4 2 5 3 3 1 = x e 5 6 5 12 3 1 3 1 = e e ) ( 3 1 7 e e = 9.6 2 0 3 4 dx x x p.69 = 2 0 3 4 dx x dx e x = 2 0 3 4 ln ) ( dx e x = 2 0 3 4 ln 2 0 3 4 ln 3 4 ln 1 = x e 3 4 ln 1 3 4 2 = 3 4 ln 9 7 = 9.7 π 4 0 2 sec xdx p.69 4 0 ] [tan π = x 1 0 1 0 tan 4 tan = = π = 9.8 π 6 0 7 sin 2 sin xdx x p.70 π = 6 0 ) 5 cos 9 (cos 2 1 dx x x π = 6 0 ) 5 cos 9 (cos 2 1 dx x x 6 0 5 5 sin 9 9 sin 2 1 π
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