Calculus with Analytic Geometry by edwards & Penney soln ch5

Calculus with Analytic Geometry

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Section 5.2 C05S02.001: Z (3 x 2 +2 x +1) dx = x 3 + x 2 + x + C . C05S02.002: Z (3 t 4 +5 t 6) dt = 3 5 t 5 + 5 2 t 2 6 t + C . C05S02.003: Z (1 2 x 2 +3 x 3 ) dx = 3 4 x 4 2 3 x 3 + x + C . C05S02.004: Z µ 1 t 2 dt = 1 t + C . C05S02.005: Z (3 x 3 x 3 / 2 1) dx = 3 2 x 2 + 4 5 x 5 / 2 x + C . C05S02.006: Z µ x 5 / 2 5 x 4 x 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 . C05S02.007: Z ³ 3 2 t 1 / 2 +7 ´ dt = t 3 / 2 t + C . C05S02.008: Z µ 2 x 3 / 4 3 x 2 / 3 dx = Z (2 x 3 / 4 3 x 2 / 3 ) dx =8 x 1 / 4 9 x 1 / 3 + C . C05S02.009: Z ( x 2 / 3 +4 x 5 / 4 ) dx = 3 5 x 5 / 3 16 x 1 / 4 + C . C05S02.010: Z µ 2 x x 1 x dx = Z (2 x 3 / 2 x 1 / 2 ) dx = 4 5 x 5 / 2 2 x 1 / 2 + C . C05S02.011: Z (4 x 3 4 x +6) dx = x 4 2 x 2 +6 x + C . C05S02.012: Z µ 1 4 t 5 5 t 2 dt = Z ( 1 4 t 5 5 t 2 ) dt = 1 24 t 6 t 1 + C . C05S02.013: Z 7 dx =7 x + C . C05S02.014: Z µ 4 3 x 2 6 3 x dx = Z ³ 4 x 2 / 3 6 x 1 / 3 ´ dx = 12 5 x 5 / 3 9 x 2 / 3 + C . C05S02.015: 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 . C05S02.016: Z ( t 10 dt = 1 11 ( t 11 + C . C05S02.017: Z 1 ( x 10) 7 dx = Z ( x 10) 7 dx = 1 6 ( x 10) 6 + C = 1 6( x 10) 6 + C . 1
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C05S02.018: Z z +1 dz = Z ( z +1) 1 / 2 dz = 2 3 ( z 3 / 2 + C . C05S02.019: 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 . C05S02.020: Z 3 x ( x 3 dx = Z ( x 10 / 3 +3 x 7 / 3 +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 . C05S02.021: 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 . C05S02.022: 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 . C05S02.023: 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 . C05S02.024: Z 1 (3 z + 10) 7 dz = Z (3 z + 10) 7 dz = 1 18 (3 z + 10) 6 + C . C05S02.025: Z 7 ( x + 77) 2 dx =7 Z ( x + 77) 2 dx = 7( x + 77) 1 + C = 7 x +77 + C . C05S02.026: Z 3 p ( x 1) 3 dx = Z 3( x 1) 3 / 2 dx = 6( x 1) 1 / 2 + C = 6 x 1 + C . C05S02.027: Z (5cos10 x 10sin5 x ) dx = 1 2 sin10 x + 2cos5 x + C . C05S02.028: Z (2cos πx + 3sin ) dx = 2 π sin 3 π cos + C . C05S02.029: Z (3cos πt + cos3 ) dt = 3 π sin + 1 3 π sin3 + C . C05S02.030: Z (4sin2 2sin4 ) dt = 2 π cos2 + 1 2 π cos4 + C . C05S02.031: D x ( 1 2 sin 2 x + C 1 ) =sin 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 . C05S02.032: 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. 2
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C05S02.033: Z sin 2 xdx = Z ( 1 2 1 2 cos2 x ) dx = 1 2 x 1 4 sin2 x + C and Z cos 2 = Z ( 1 2 + 1 2 x ) dx = 1 2 x + 1 4 x + C . C05S02.034: (a): D x tan x =sec 2 x ; (b): Z tan 2 = Z ( sec 2 x 1 ) dx = (tan x ) x + C . C05S02.035: y ( x )= x 2 + x + C ; y (0) = 3, so y ( x x 2 + x +3. C05S02.036: y ( x 1 4 ( x 2) 4 + C and y (2) = 1, so y ( x 1 4 ( x 2) 4 +1.
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Calculus with Analytic Geometry by edwards & Penney...

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