57318-0136147054_05

57318-0136147054_05 - Section 5.2 5.2.1 5.2.2 5.2.3 5.2.4...

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Section 5.2 5.2.1: Z (3 x 2 +2 x +1) dx = x 3 + x 2 + x + C . 5.2.2: Z (3 t 4 +5 t 6) dt = 3 5 t 5 + 5 2 t 2 6 t + C . 5.2.3: Z (1 2 x 2 +3 x 3 ) dx = 3 4 x 4 2 3 x 3 + x + C . 5.2.4: Z ± 1 t 2 ² dt = 1 t + C . 5.2.5: Z (3 x 3 x 3 / 2 1) dx = 3 2 x 2 + 4 5 x 5 / 2 x + C . 5.2.6: Z ± x 5 / 2 5 x 4 x ² dx = Z ( x 5 / 2 5 x 4 x 1 / 2 ) dx = 2 7 x 7 / 2 + 5 3 x 3 2 3 x 3 / 2 + C . 5.2.7: Z ³ 3 2 t 1 / 2 +7 ´ dt = t 3 / 2 t + C . 5.2.8: Z ± 2 x 3 / 4 3 x 2 / 3 ² dx = Z (2 x 3 / 4 3 x 2 / 3 ) dx =8 x 1 / 4 9 x 1 / 3 + C . 5.2.9: Z ( x 2 / 3 +4 x 5 / 4 ) dx = 3 5 x 5 / 3 16 x 1 / 4 + C . 5.2.10: Z ± 2 x x 1 x ² dx = Z (2 x 3 / 2 x 1 / 2 ) dx = 4 5 x 5 / 2 2 x 1 / 2 + C . 5.2.11: Z (4 x 3 4 x +6) dx = x 4 2 x 2 +6 x + C . 5.2.12: Z ± 1 4 t 5 5 t 2 ² dt = Z ( 1 4 t 5 5 t 2 ) dt = 1 24 t 6 t 1 + C . 5.2.13: Z 7 e x/ 7 dx =49 e x/ 7 + C . 5.2.14: Z 1 7 x dx = Z 1 7 · 1 x dx = 1 7 ln | x | + C . Another correct answer is Z 1 7 x dx = 1 7 ln | 7 x | + C 1 = 1 7 ln | x | + 1 7 ln 7 + C 1 = 1 7 ln | x | + C 2 , the last step because the constant C 2 is simply the constant C 1 + 1 7 ln 7. 5.2.15: Z ( x 4 dx = 1 5 ( x 5 + C . Note that many computer algebra systems give the answer C + x x 2 x 3 + x 4 + x 5 5 . 429
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5.2.16: Z ( t +1) 10 dt = 1 11 ( t 11 + C . 5.2.17: Z 1 ( x 10) 7 dx = Z ( x 10) 7 dx = 1 6 ( x 10) 6 + C = 1 6( x 10) 6 + C . 5.2.18: Z z +1 dz = Z ( z 1 / 2 dz = 2 3 ( z 3 / 2 + C . 5.2.19: Z x (1 x ) 2 dx = Z ( x 1 / 2 2 x 3 / 2 + x 5 / 2 ) dx = 2 3 x 3 / 2 4 5 x 5 / 2 + 2 7 x 7 / 2 + C . 5.2.20: Z 3 x ( x 3 dx = Z ( x 10 / 3 +3 x 7 / 3 x 4 / 3 + x 1 / 3 ) dx = 3 13 x 13 / 3 + 9 10 x 10 / 3 + 9 7 x 7 / 3 + 3 4 x 4 / 3 + C . 5.2.21: Z 2 x 4 3 x 3 +5 7 x 2 dx = Z ( 2 7 x 2 3 7 x + 5 7 x 2 ) dx = 2 21 x 3 3 14 x 2 5 7 x 1 + C . 5.2.22: Z (3 x +4) 2 x dx = Z x 1 / 2 ( 9 x 2 +24 x +16 ) dx = Z ( 9 x 3 / 2 x 1 / 2 x 1 / 2 ) dx = 18 5 x 5 / 2 x 3 / 2 +32 x 1 / 2 + C . 5.2.23: Z (9 t + 11) 5 dt = 1 54 (9 t + 11) 6 + C . Mathematica gives the answer C + 161051 t + 658845 t 2 2 + 359370 t 3 + 441045 t 4 2 + 72171 t 5 + 19683 t 6 2 . 5.2.24: Z 1 (3 z + 10) 7 dz = Z (3 z + 10) 7 dz = 1 18 (3 z + 10) 6 + C . 5.2.25: Z ( e 2 x + e 2 x ) dx = 1 2 e 2 x 1 2 e 2 x + C . 5.2.26: Z ( e x + e x ) 2 dx = Z ( e 2 x +2+ e 2 x ) dx = 1 2 e 2 x +2 x 1 2 e 2 x + C . 5.2.27: Z (5 cos10 x 10 sin5 x ) dx = 1 2 sin 10 x +2cos5 x + C . 5.2.28: Z (2 cos πx +3sin ) dx = 2 π sin 3 π cos + C . 5.2.29: Z (3 cos πt +cos3 ) dt = 3 π sin + 1 3 π sin3 + C . 5.2.30: Z (4 sin2 2sin4 ) dt = 2 π cos2 + 1 2 π cos4 + C . 5.2.31: D x ( 1 2 sin 2 x + C 1 ) =s in x cos x = D x ( 1 2 cos 2 x + C 2 ) . Because 1 2 sin 2 x + C 1 = 1 2 cos 2 x + C 2 , it follows that C 2 C 1 = 1 2 sin 2 x + 1 2 cos 2 x = 1 2 . 430
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5.2.32: F 0 1 ( x )= 1 (1 x ) 2 , F 0 2 ( x 1 x + x (1 x ) 2 = 1 (1 x ) 2 . F 1 ( x ) F 2 ( x C 1 for some constant C 1 on ( −∞ , 1); F 1 ( x ) F 2 ( x C 2 for some constant C 2 on (1 , + ). On either interval, F 1 ( x ) F 2 ( x 1 x 1 x =1. 5.2.33: Z sin 2 xdx = Z ( 1 2 1 2 cos2 x ) dx = 1 2 x 1 4 sin 2 x + C and Z cos 2 = Z ( 1 2 + 1 2 x ) dx = 1 2 x + 1 4 sin 2 x + C . 5.2.34: (a): D x tan x =sec 2 x ;( b ) : Z tan 2 = Z ( sec 2 x 1 ) dx = (tan x ) x + C . 5.2.35: y ( x x 2 + x + C ; y (0) = 3, so y ( x x 2 + x +3.
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This document was uploaded on 11/28/2010.

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57318-0136147054_05 - Section 5.2 5.2.1 5.2.2 5.2.3 5.2.4...

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