Bioestadistica

Section 5.2
C05S02.001: (3x2 + 2x + 1) dx = x3 + x2 + x + C. (3t4 + 5t − 6) dt = 3 t5 + 5 t2 − 6t + C. 5 2 (1 − 2x2 + 3x3 ) dx = 3 x4 − 2 x3 + x + C. 4 3 − 1 t2 dt = 1 + C. t

C05S02.002:

C05S02.003:

C05S02.004:

C05S02.005:

(3x−3 + 2x3/2 − 1) dx = − 3 x−2 + 4 x5/2 − x + C. 2 5 x5/2 −
3 1/2 2t

C05S02.006:

√ 5 − x x4

dx =

(x5/2 − 5x−4 − x1/2 ) dx = 2 x7/2 + 5 x−3 − 2x3/2 + C. 7 3 3

C05S02.007:

+ 7 dt = t3/2 + 7t + C. (2x−3/4 − 3x−2/3 ) dx = 8x1/4 − 9x1/3 + C.

C05S02.008:

2 3 − 2/3 x3/4 x

dx =

C05S02.009:

(x2/3 + 4x−5/4 ) dx = 3 x5/3 − 16x−1/4 + C. 5 √ 1 2x x − √ x dx = (2x3/2 − x−1/2 ) dx = 4 x5/2 − 2x1/2 + C. 5

C05S02.010:

C05S02.011:

(4x3 − 4x + 6) dx = x4 − 2x2 + 6x + C. 1 5 5 t − 2 4 t
1 5 4t

C05S02.012:

dt =

− 5t−2 dt=

1 6 24 t

+ 5t−1 + C.

C05S02.013:

7 dx = 7x + C. √ 6 3 4 x2 − √ 3 x 4x2/3 − 6x−1/3 dx = 12 5/3 x − 9x2/3 + C. 5

C05S02.014:

dx =

C05S02.015:

(x + 1)4 dx = 1 (x + 1)5 + C. Note that many computer algebra systems give the answer 5 C + x + 2x2 + 2x3 + x4 + x5 . 5

C05S02.016:

(t + 1)

10

dt =

1 11 (t

+ 1)11 + C. (x − 10)−7 dx = − 1 (x − 10)−6 + C = − 6 1 1 +C. 6(x − 10)6

C05S02.017:

1 dx = (x − 10)7

C05S02.018:

√ √ √ 3

z + 1 dz =

(z + 1)1/2 dz = 2 (z + 1)3/2 + C. 3 (x1/2 − 2x3/2 + x5/2 ) dx = 2 x3/2 − 4 x5/2 + 2 x7/2 + C. 3 5 7 (x10/3 +3x7/3 +3x4/3 +x1/3 ) dx = − 3 x + 5 x−2 dx = 7 7
3 13/3 9 + 10 x10/3 + 9 x7/3 + 3 x4/3 +C. 13 x 7 4

C05S02.019:

x (1 − x)2 dx =
3

C05S02.020:

x (x + 1) dx =

C05S02.021:

2x4 − 3x3+ 5 dx = 7x2 (3x + 4) √ dx = x
2

2 2 7x

2 3 21 x

−

3 2 14 x

− 5 x−1 + C. 7

C05S02.022: =
18 5/2 5 x

x−1/2 9x2 + 24x + 16 dx =

9x3/2 + 24x1/2 + 16x−1/2 dx

+ 16x3/2 + 32x1/2 + C. (9t + 11)5 dt =
1 54 (9t

C05S02.023:

+ 11)6 + C. Mathematica gives the answer

C + 161051t + 1 (3z + 10)
7

658845t2 441045t4 19683t6 + 359370t3 + + 72171t5 + . 2 2 2
1 (3z + 10)−7dz = − 18 (3z + 10)−6 + C.

C05S02.024:

dz =

C05S02.025:

7 dx = 7 (x + 77)2 3 (x − 1)3 dx =

(x + 77)−2 dx = −7(x + 77)−1 + C = −

7 + C. x + 77 6 + C. x−1

C05S02.026:

3(x − 1)−3/2 dx = −6(x − 1)−1/2 + C = − √

C05S02.027:

(5 cos 10x − 10 sin 5x) dx = (2 cos πx + 3 sin πx) dx =

1 2

sin 10x + 2 cos 5x + C.

C05S02.028:

2 3 sin πx − cos πx + C. π π

C05S02.029:(3 cos πt + cos 3πt) dt =

3 1 sin πt + sin 3πt + C. π 3π 2 1 cos 2πt + cos 4πt + C. π 2π

C05S02.030:

(4 sin 2πt − 2 sin 4πt) dt = −
1 2

C05S02.031: Dx

sin2 x + C1 = sin x cos x = Dx − 1 cos2 x + C2 . Because 2 it follows that C2 − C1 = 1 1 1 sin2 x + cos2 x = . 2 2 2

1 1 sin2 x + C1 = − cos2 x + C2 , 2 2 C05S02.032: F1 (x) =

1 1−x+x 1 , F2 (x) = = . F1 (x) − F2 (x) = C1for some constant 2 2 (1 − x) (1 − x) (1 − x)2 C1 on (−∞, 1); F1 (x) − F2 (x) = C2 for some constant C2 on (1, +∞). On either interval, F1 (x) − F2 (x) = 1−x = 1. 1−x 2

C05S02.033:

sin2 x dx =
1 2

1 2

−

1 2

cos 2x dx = 1 x − 2
1 4

1 4

sin 2x + C and

cos2 x dx =

+

1 2

cos 2x dx = 1 x + 2 (b):

sin 2x + C. tan2 x dx = sec2 x − 1 dx = (tan x) − x + C.C05S02.034: (a): Dx tan x = sec2 x;

C05S02.035: y(x) = x2 + x + C; y(0) = 3, so y(x) = x2 + x + 3. C05S02.036: y(x) = 1 (x − 2)4 + C and y(2) = 1, so y(x) = 1 (x − 2)4 + 1. 4 4 C05S02.037: y(x) = 2 x3/2 + C and y(4) = 0, so y(x) = 2 x3/2 − 3 3 C05S02.038: y(x) = −
16 3 .

1 1 + C and y(1) = 5, so y(x) = − + 6. x x √ √ C05S02.039: y(x) = 2 x + 2 + C and y(2) = −1, so y(x) = 2 x + 2 − 5. C05S02.040:y = √ x + 9 dx = 2 (x + 9)3/2 + C; 0 = y(−4) = 2 (−4 + 9)3/2 + C = 3 3
10 3 2 3

√ · 5 5 + C;

y(x) = 2 (x + 9)3/2 − 3

√

5.

C05S02.041: y(x) = 3 x4 − 2x−1 + C; y(1) = 1, so y(x) = 3 x4 − 2x−1 + 9 . 4 4 4 C05S02.042: y(x) = 1 5 3 2 3 1 3 3 9 x − x − 2 + C = x5 − x2 − 2 + . 5 2 2x 5 2 2x 5
1 4

C05S02.043: y(x) = 1 (x − 1)4 + C; y(0) = 2 = 4

+ C, so C = 7 . 4

C05S02.044:...