Calculus with Analytic Geometry by edwards & Penney soln ch5

# Calculus with Analytic Geometry

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Section 5.2 C05S02.001: (3 x 2 + 2 x + 1) dx = x 3 + x 2 + x + C . C05S02.002: (3 t 4 + 5 t 6) dt = 3 5 t 5 + 5 2 t 2 6 t + C . C05S02.003: (1 2 x 2 + 3 x 3 ) dx = 3 4 x 4 2 3 x 3 + x + C . C05S02.004: 1 t 2 dt = 1 t + C . C05S02.005: (3 x 3 + 2 x 3 / 2 1) dx = 3 2 x 2 + 4 5 x 5 / 2 x + C . C05S02.006: x 5 / 2 5 x 4 x dx = ( 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: 3 2 t 1 / 2 + 7 dt = t 3 / 2 + 7 t + C . C05S02.008: 2 x 3 / 4 3 x 2 / 3 dx = (2 x 3 / 4 3 x 2 / 3 ) dx = 8 x 1 / 4 9 x 1 / 3 + C . C05S02.009: ( x 2 / 3 + 4 x 5 / 4 ) dx = 3 5 x 5 / 3 16 x 1 / 4 + C . C05S02.010: 2 x x 1 x dx = (2 x 3 / 2 x 1 / 2 ) dx = 4 5 x 5 / 2 2 x 1 / 2 + C . C05S02.011: (4 x 3 4 x + 6) dx = x 4 2 x 2 + 6 x + C . C05S02.012: 1 4 t 5 5 t 2 dt = ( 1 4 t 5 5 t 2 ) dt = 1 24 t 6 + 5 t 1 + C . C05S02.013: 7 dx = 7 x + C . C05S02.014: 4 3 x 2 6 3 x dx = 4 x 2 / 3 6 x 1 / 3 dx = 12 5 x 5 / 3 9 x 2 / 3 + C . C05S02.015: ( x + 1) 4 dx = 1 5 ( x + 1) 5 + C . Note that many computer algebra systems give the answer C + x + 2 x 2 + 2 x 3 + x 4 + x 5 5 . C05S02.016: ( t + 1) 10 dt = 1 11 ( t + 1) 11 + C . C05S02.017: 1 ( x 10) 7 dx = ( x 10) 7 dx = 1 6 ( x 10) 6 + C = 1 6( x 10) 6 + C . 1

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C05S02.018: z + 1 dz = ( z + 1) 1 / 2 dz = 2 3 ( z + 1) 3 / 2 + C . C05S02.019: x (1 x ) 2 dx = ( 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: 3 x ( x + 1) 3 dx = ( 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: 2 x 4 3 x 3 + 5 7 x 2 dx = ( 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: (3 x + 4) 2 x dx = x 1 / 2 ( 9 x 2 + 24 x + 16 ) dx = ( 9 x 3 / 2 + 24 x 1 / 2 + 16 x 1 / 2 ) dx = 18 5 x 5 / 2 + 16 x 3 / 2 + 32 x 1 / 2 + C . C05S02.023: (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: 1 (3 z + 10) 7 dz = (3 z + 10) 7 dz = 1 18 (3 z + 10) 6 + C . C05S02.025: 7 ( x + 77) 2 dx = 7 ( x + 77) 2 dx = 7( x + 77) 1 + C = 7 x + 77 + C . C05S02.026: 3 ( x 1) 3 dx = 3( x 1) 3 / 2 dx = 6( x 1) 1 / 2 + C = 6 x 1 + C . C05S02.027: (5cos10 x 10sin5 x ) dx = 1 2 sin10 x + 2cos5 x + C . C05S02.028: (2cos πx + 3sin πx ) dx = 2 π sin πx 3 π cos πx + C . C05S02.029: (3cos πt + cos3 πt ) dt = 3 π sin πt + 1 3 π sin3 πt + C . C05S02.030: (4sin2 πt 2sin4 πt ) dt = 2 π cos2 πt + 1 2 π cos4 πt + 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 1 ( x ) = 1 (1 x ) 2 , F 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
C05S02.033: sin 2 x dx = ( 1 2 1 2 cos2 x ) dx = 1 2 x 1 4 sin2 x + C and cos 2 x dx = ( 1 2 + 1 2 cos2 x ) dx = 1 2 x + 1 4 sin2 x + C . C05S02.034: (a): D x tan x = sec 2 x ; (b): tan 2 x dx = ( 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. C05S02.037: y ( x ) = 2 3 x 3 / 2 + C and y (4) = 0, so y ( x ) = 2 3 x 3 / 2 16 3 . C05S02.038: y ( x ) = 1 x + C and y (1) = 5, so y ( x ) = 1 x + 6.

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