Answer to the exercises

Answer to the exercises - Answer to the exercises

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Unformatted text preview: Answer to the exercises P164(B)求求求求求求求求 x5 4 3 − x + 4x + C 53 1 求3求 x − − 2 ln | x | +C x 求1求 求2求 x 3 − 求4求 ln | x | − 3 2 (5) x 2 − 2 x + C 3 1 (7) − − arctan x + C x (9) ln | x | + arctan x + C (11) − cot t − t + C (13) 2 sin x + C p186 求求求求1. ABC 2.BC p187 (B) 3 2 1(1) − (2 − 3 x) 2 + C 9 1 (3) − cos(4 x + 1) + C 4 1 (5) − cos x 2 + C 2 (7) ln | x − 2 | +C 52 x + 2x + C 2 2 3 + 2 +C x 2x 11 (6) 44 t +C 11 (8) x + arctan x + C (10) e x + x + C x sin x +C (12) + 2 2 (14) − cot x − 2 x + C 3.ACD 4.AD 1 (2) ln | 3 x + 1 | +C 3 1 (4) − sin(1 − 4 x) + C 4 (6) − 1 3 − 4x 2 + C 4 1 +C x (8) 2 sin x + C 1 (10) − ln | 1 − 2e x | +C 2 1 1+ ex (11) ln | | +C 2 1− ex 1 +C (13) 1 − sin x cos 3 x +C 3 1 4x +C (14) arctan 12 3 (9) cos (15) 2 2 + x3 + C 3 1 (17) ln 3 x + C 3 2(1) 2 x − ln | 1 + x | +C (12) − 1 (16) ln | 1 + 2 ln x | +C 2 1 (18) (1 + tan x) 3 + C 3 (2) 2 ln | 1 + x | +C 3 (3) 2e x +C (4) 4( x − 6) 2 + 5 2 ( x − 6) 2 +C 5 1 3 (5) 2 arctan x − 1 + C (6) 1 −2 x 1 −2 x xe − e + C 2 4 3(1) − 2 4 (3x + 1) 2 + 3x + 1 + C 27 9 (2) − xe − x + C 1 1 (3) − x cos 3 x + sin 3 x + C 3 9 1 1 (4) x sin 2 x + cos 2 x + C 2 4 x2 1 arctan x − ( x − arctan x) + C (6) x arcsin x + 1 − x 2 + C 2 2 1 1 x2 1 x2 (7) ln( x − 1) − [ + x + ln( x − 1)] + C (8) − ln x − + C x x 2 22 (5) 1 x −1 | +C 4(1) ln | 3 x+2 (3) ln | 1 2x − 1 | +C (2) ln | 4 2x + 1 x+2 | +C x+3 7 13 (4) − ln | x − 2 | + ln | x − 5 | +C 3 3 1 (5) ln | x 2 − 4 x + 3 | +C 2 (6) ln | x 3 + x 2 + 1 | +C 1 x+2 +C (7) arctan 2 2 p212(求)求求求 1. C 1 x+2 +C (8) ln( x 2 + 4 x + 8) + arctan 2 2 2.D 3.C 4. AD 5.C (B)1(1) 9 (3) 40 3 (2) (4) 2(1) (6) (8) (2) 3 16 (9) 2(1 − (6) 1 (8) π ) 4 3 8 3 2 (4) 1 1 (5) arctan 3 3 (7) π 6 π1 − 42 1 3 + ln 2 2 (3) 0 49 3 5 6 (5) e 2 + 2 ln 2 − e (7)0 6.C 5 2 (10)2(2-arctan2) 2 π 2 (4) 2 ln 2 − 1 1 (6) 4 3(1) 1 (2) (3) 0 (5) − 1 2e 2 3 32 4(1) A = ∫ (2 x + 3 − x )dx = 3 −1 2 1 1 (2) A = ∫ (2 − x 2 − x 2 )dx = −1 8 3 2 1 3 A = ∫ (2 − )dx + ∫ (2 − x)dx = − ln 2 (3) x 2 1 1 2 2 1 3 (4) A = ∫ ( x − )dx = − ln 2 x 2 1 1 4 0 1 (5) A = 2 ∫ x dx + ∫ ( x − x + 2)dx = 3 9 2 ...
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This note was uploaded on 03/29/2012 for the course MATH 120 taught by Professor Folh during the Spring '08 term at Eastern Michigan University.

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