M2_Bk2_Sol_Ch10_E.pdf

# M2_Bk2_Sol_Ch10_E.pdf - 10 New Progress in Senior...

This preview shows pages 1–3. Sign up to view the full content.

© Hong Kong Educational Publishing Co. 98 New Progress in Senior Mathematics Module 2 Book 2 (Extended Part) Solution Guide 10 10 10 10 10 p.119 p.119 h r 2 3 1 π pp.105 – 132 10.1 The required area = e xdx 1 ln p.105 1 ] [ 1 ) (ln ] ln [ 1 1 1 1 = = = = e e e e x e dx x x e x xd x x 10.2 The required area p.106 + + + = 3 1 2 1 0 2 ) 3 4 ( ) 3 4 ( dx x x x dx x x x + + = 3 1 2 3 1 0 2 3 ) 3 4 ( ) 3 4 ( dx x x x dx x x x 3 1 2 3 4 1 0 2 3 4 2 3 3 4 4 2 3 3 4 4 + + = x x x x x x = 3 8 12 5 12 37 = 10.3 Consider the points of intersection of the p.107 two curves. ) 2 ..... ( .......... 3 ) 1 ..( .......... 9 2 2 x y x y = = Substituting (1) into (2), we have 4 9 3 9 2 2 2 = = x x x 2 3 = x or 2 3 (rejected) Hence the curves intersect when . 2 3 = x The required area = 2 3 0 2 2 ) 3 9 ( dx x x 9 2 9 2 27 3 4 9 ) 4 9 ( 2 3 0 3 2 3 0 2 = = = = x x dx x 10.4 Solving 2 + = x y and , 3 x y = p.108 the point of intersection is (1, 3). Solving 2 + = x y and , 2 x y = the point of intersection is (2, 4). The required area + + = 2 1 2 1 0 2 ) 2 ( ) 3 ( dx x x dx x x 2 1 3 2 1 0 3 2 3 2 2 3 2 3 + + = x x x x x 6 7 6 7 + = 3 7 = 10.5 At the points of intersection, p.109 2 3 or 6 5 , 6 1 or 2 1 sin 0 ) 1 )(sin 1 sin 2 ( 0 1 sin sin 2 sin sin 2 1 sin 2 cos 2 2 π π π = = = + = + = = x x x x x x x x x x The points of intersection are ( 6 π , 2 1 ), ( 6 5 π , 2 1 ) and ( 2 3 π , 1). On the left hand side of ( 6 5 π , 2 1 ), sin x cos 2 x is positive. On the right hand side of ( 6 5 π , 2 1 ), sin x cos 2 x is negative. Applications of Definite Integrals

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
99 © Hong Kong Educational Publishing Co. Applications of Definite Integrals The required area π π π π + + = 2 3 6 5 6 5 6 ) 2 cos sin ( ) 2 cos (sin dx x x dx x x 2 3 6 5 6 5 6 2 2 sin cos 2 2 sin cos π π π π + + = x x x x 4 3 3 2 3 3 + = 4 3 9 = 10.6 The required area p.110 dy e y = 2 0 2 0 ] [ y e = 1 2 = e 10.7 The required area p.111 + + + = 3 1 2 3 1 1 2 3 ) 3 3 ( ) 3 3 ( dy y y y dy y y y 3 1 2 3 4 1 1 2 3 4 3 2 4 3 2 4 + + = y y y y y y y y ) 4 ( 4 = 8 = 10.8 (a) Slope of the tangent p.112 1 3 ) ( = = x x dx d 1 2 3 = = x x 3 = Equation of L : 2 3 3 3 1 ) 1 ( 3 1 = = = x y x y x y (b) For , 3 x y = 3 1 y x = For , 2 3 = x y 3 2 + = y x The required area + = 1 0 3 1 3 2 dy y y 1 0 3 4 2 4 3 3 2 6 + = y y y 12 1 = 10.9 The required volume p.120 π = 3 0 2 2 ) ( dx x 3 0 5 5 π = x 5 243 π = 10.10 The required volume p.121 π = 1 0 2 ) 1 ( dx x x + π = 1 0 2 ) 1 2 ( dx x x x + π = 1 0 2 3 ) 2 ( dx x x x 1 0 2 3 4 2 3 2 4 + π = x x x 12 π = 10.11 ) 1 ( . .......... 1 2 + = x y p.122 ) 2 ( .......... 5 2 x y = Substituting (1) into (2), 1 or 2 3 0 ) 1 )( 3 2 ( 0 3 2 5 ) 1 ( 2 2 2 = = + = + = + x x x x x x x The curves intersect at ( 4 13 , 2 3 ) and (1, 2).
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern