CMII_CE1S

# CMII_CE1S - CC: AD [ 2011-12 ] 243: College Mathematics II...

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Unformatted text preview: CC: AD [ 2011-12 ] 243: College Mathematics II — Class. Exercises [ Sol ] — Chapter 1: Integration and its Applications ª 1.1 Indefinite Integration ª 1.2 Definite Integration ª 1.3 Applications of Integration ª 1.4 Improper Integration Ch1: Integration and its Applications [ CE1S-1 ] q 1.1 Indefinite Integration Class Ex. 1 Evaluate (a) ∫ ( 1 − p x ) 2 3 x dx . [ 3 ] (b) ∫ e 2 x 3 p e x − 1 dx . [ 4 ] (c) ∫ ln ( 2 + x ) dx . [ 4 ] (d) ∫ x 2 + 1 x 2 + 3 x + 2 dx . [ 4 ] (e) ∫ x 2 ( x − 1 )( x 2 − 2 x + 2 ) dx . [ 4 ] (f) ∫ 3 x 2 − 2 x + 1 ( x + 1 )( x 2 − 2 x + 3 ) dx [ 4 ] Solution (a) ∫ ( 1 − p x ) 2 3 x dx = 1 3 ∫ 1 − 2 p x + x x dx = 1 3 ∫ 1 x − 2 x − 1 2 + 1 dx = 1 3 ( ln | x |− 4 x 1 2 + x ) + C (b) Let u = e x − 1. Then du dx = e x . ∴ dx = 1 e x du . Hence, ∫ e 2 x 3 p e x − 1 dx = ∫ e 2 x u − 1 3 1 e x du = ∫ ( u + 1 ) u − 1 3 du = ∫ u 2 3 du + ∫ u − 1 3 du = 3 5 u 5 3 + 3 2 u 2 3 + C = 3 5 ( e x − 1 ) 5 3 + 3 2 ( e x − 1 ) 2 3 + C (c) Let f ( x ) = ln ( 2 + x ) and g ′ ( x ) = 1. Then f ′ ( x ) = 1 2 + x and g ( x ) = x . By parts, we have ∫ ln ( 2 + x ) dx = x ln ( 2 + x ) − ∫ x 2 + x dx = x ln ( 2 + x ) − ∫ 1 − 2 2 + x dx = x ln ( 2 + x ) − x + 2ln | 2 + x | + C AD: College Mathematics II [ 2011-12 ] Louis Lau 1 1.1 Indefinite Integration [ CE1S-1 ] Solution (d) ∫ x 2 + 1 x 2 + 3 x + 2 dx = ∫ 1 − 3 x + 1 ( x + 1 )( x + 2 ) dx = x − ∫ 5 x + 2 − 2 x + 1 dx = x − 5ln | x + 2 | + 2ln | x + 1 | + C (e) Let x 2 ( x − 1 )( x 2 − 2 x + 2 ) = A x − 1 + Bx + C x 2 − 2 x + 2 = A ( x 2 − 2 x + 2 ) + ( Bx + C )( x − 1 ) ( x − 1 )( x 2 − 2 x + 2 ) Then x 2 = A ( x 2 − 2 x + 2 ) + ( Bx + C )( x − 1 ) = ( A + B ) x 2 + ( C − B − 2 A ) x + ( 2 A − C ) Solving    A + B = 1 C − B − 2 A = 2 A − C = , we have A = 1, B = 0, C = 2. Hence x 2 ( x − 1 )( x 2 − 2 x + 2 ) = 1 x − 1 + 2 x 2 − 2 x + 2 ∫ x 2 ( x − 1 )( x 2 − 2 x + 2 ) = ∫ 1 x − 1 dx + 2 ∫ 1 ( x − 1 ) 2 + 1 dx = ln | x − 1 | + 2tan − 1 ( x − 1 ) + C (f) 3 x 2 − 2 x + 1 ( x + 1 )( x 2 − 2 x + 3 ) = 1 x + 1 + 2 x − 2 x 2 − 2 x + 3 ∫ 3 x 2 − 2 x + 1 ( x + 1 )( x 2 − 2 x + 3 ) dx = ∫ 1 x + 1 dx + ∫ 2 x − 2 x 2 − 2 x + 3 dx = ln | x + 1 | + ln | x 2 − 2 x + 3 | + C AD: College Mathematics II [ 2011-12 ] Louis Lau 2 1.1 Indefinite Integration [ CE1S-1 ] Class Ex. 2 Evaluate the following integrals: (a) ∫ ( 1 + x ) 2 p x dx [ 3 ] (b) ∫ x 3 p 1 + x 2 dx [ 4 ] (c) ∫ 3 x + 7 ( 2 x − 1 )( x 2 + 2 x + 3 ) dx [ 4 ] Solution (a) ∫ 1 + 2 x + x 2 p x dx = ∫ x − 1 2 + 2 x 1 2 + x 3 2 dx = 2 x 1 2 + 4 3 x 3 2 + 2 5 x 5 2 + C ....
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## This note was uploaded on 02/22/2012 for the course CHEM yscn0027 taught by Professor Drtong during the Fall '10 term at HKU.

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CMII_CE1S - CC: AD [ 2011-12 ] 243: College Mathematics II...

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