Solucionario - Problemas Pares, C.04 ,Vol1 ,Cálculo, Larson, Hostetler

8/2/2019 Solucionario - Problemas Pares, C.04 ,Vol1 ,Clculo, Larson, Hostetler

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C H A P T E R 4

Integration

Section 4.1 Antiderivatives and Indefinite Integration . . . . . . . . . 450

Section 4.2 Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456

Section 4.3 Riemann Sums and Definite Integrals . . . . . . . . . . . 462

Section 4.4 The Fundamental Theorem of Calculus . . . . . . . . . . 466

Section 4.5 Integration by Substitution . . . . . . . . . . . . . . . . . 472

Section 4.6 Numerical Integration . . . . . . . . . . . . . . . . . . . 479

Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483

Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 488


2/34

C H A P T E R 4

Integration

Section 4.1 Antiderivatives and Indefinite Integration

Solutions to Even-Numbered Exercises

450

2.d

dxx4 1

x C 4x3 1x2 4.

x12 x32 x2 1

x32

d

dx2x2 3

3x C ddx

2

3x32 2x12 C

6.

Check:d

d

C

r Cdr

d 8.

Check: ddx

1

x2 C 2x

3

y 2x2

2 C

1

x2 c

dy

dx 2x3

Given Rewrite Integrate Simplify

10. 1

x C

x1

1 Cx2dx 1

x2dx

12.1

4x 4

3

2x2 C

x4

4 3x

2

2 Cx3 3xdxxx2 3dx

14.1

9x C

1

9

x1

1 C

1

9x2dx

1

3x2

dx

16.

Check:d

dx5x x2

2 C 5 x

5 xdx 5x x22 C 18.Check:

d

dxx 4 2x 3x C 4x 3 6x 2 1

4x3 6x2 1dx x 4 2x3 x C

20.

Check:d

dxx4

4 2x2 2x C x3 4x 2

x3 4x 2dx x 44 2x2 2x C

22.

Check: x 1

2x

d

dx2

3x32 x12 C x12 12x12

2

3x32 x12 C

x32

32

1

2x12

12 Cx 12xdx x12 1

2x12dx


3/34

Section 4.1 Antiderivatives and Indefinite Integration 451

24.

Check:d

dx4

7x74 x C x34 1 4x3 1

4x3 1dx x34 1dx 47x74 x C 26.Check:

d

dx1

3x3 C 1x4

1x4

dx x4dx x33

C 1

3x3 C

28.

Check:

x2 2x 3

x4

d

dx1

x

1

x2

1

x3 C x2 2x3 3x4

1

x

1

x2

1

x3 C

x1

1

2x2

2

3x3

3 C

x

2

2x 3x4

dx x2 2x3 3x4dx 30.

Check:

2t2 12

d

dt4

5t5

4

3t3 t C 4t4 4t2 1

4

5t5

4

3t3 t C

2t2 12dt 4t

4 4t2 1dt

32.

Check:d

dt1

3t3

3

4t4 C t2 3t3 1 3tt2

1 3tt2dt t2 3t3dt 13t3 34t4 C 34.Check:

d

dt3t C 3

3 dt 3t C

36.

Check:d

dt1

3t3 cos t C t2 sin t

t2 sin tdt 13t3 cos t C 38.Check:

d

d1

33 tan C 2 sec2

2 sec2d 133 tan C

40.

Check:

secytany secy

d

dysecy tany C secy tany sec2y

secy tany Csecytany

secydy

secy tany

sec

2

ydy 42.

cosx

1 cos2x

Check:d

dxcscx C cscx cotx

1

sinx

cosx

sinx

cscx cotxdx cscx C

cosx

1 cos2xdx

cosx

sin2xdx

1

sinxcosx

sinxdx

44.

x4 6 82

2

4

2

4

6

C= 0

C= 3

C= 2

y

fx x 46.

x

y

4

6

8

24 2 4

f x( ) 2= +x

2

2

f x( ) =x

2

2

f

fx x2

2

C

fx x 48.

x24

2

4

4

f x( ) =

f x( ) 1= +

1

1

x

x

f

y

fx 1

x C

fx 1

x2


4/34

452 Chapter 4 Integration

50.

y

x

2

2x

1

2 32 23 C C 1

y 2x 1dx x2 2x C

dy

dx 2x 1 2x 2, 3, 2 52.

y 1

x 2, x > 0

3 1

1 C C 2

y x2dx 1x C

dy

dx

1

x2 x2, 1,3

54. (a)

x 4

5

5

4

y

56.

gx 2x3 1

g0 1 203 CC 1

gx 6x2 dx 2x3 Cgx 6x2, g0 1

(b)

y x3

3x

7

3

C7

3

3 1

3 1 C

3 13

3 1 C

y x3

3x C 4 4

5

( 1, 3)

5dy

dxx2 1, 1, 3

58.

fs 3s2 2s4 23

f2 3 322 224 C 12 32 CC 23

fs 6s 8s3ds 3s 2 2s4 Cfs 6s 8s3, f2 3

60.

fx 1

12x4 6x 3

f0 0 0 C2 3 C2 3

fx

1

3

x3 6

dx

1

12

x4 6x C2

fx 13x3 6

f0 0 C1 6 C1 6

fx x2dx 13x3 C1f0 3

f0 6

fx x2 62.

fx sinx 2x 6

f0 0 0 C2 6 C2 6

fx cosx 2dx sinx 2x C2fx cosx 2

f0 1 C1 1 C1 2

fx sinxdx cosx C1f0 6

f0 1

fx sinx


5/34


64.

P7 100732 500 2352 bacteria

Pt 2

3150t32 500 100t32 500

P1 23

k 500 600 k 150

P0 0 C 500 C 500

Pt kt12dt 23kt32 C

dP

dt kt, 0 t 10 66. Since is negative on is decreasing on

Since is positive on is increasing

on has a relative minimum at Since is

positive on fis increasing on

x1 2 323

1

2

3

2

3f

f

f

y

,.,,f0, 0.f0,.

f0,,f, 0.f, 0,f

68.

ft 16t2 v0t s0

f0 0 0 C2 s0 C2 s0

ft st 32t v0dt 16t2 v0t C2ft 32t v0

f0 0 C1 v0 C1 v0

ft vt

32 dt 32t C

1

f0 s0

f0 v0

ft at 32 ftsec2 70.

(a)

Choosing the positive value,

(b)

1617 65.970 ftsec

v1 172 321 17

2 16

vt st 32t 16

t1 17

2 2.562 seconds.

t1 17

2

16t2 t 4 0

st 16t2 16t 64 0

s0 64 ft

v0 16 ftsec

72. From Exercise 71, (Using the

canyon floor as position 0.)

t2 1600

4.9 t 326.53 18.1 sec

4.9t2 1600

ft 0 4.9t2 1600

ft 4.9t2 1600. 74. From Exercise 71, If

then

for this tvalue. Hence, and we solve

v02 3880.8 v0 62.3 msec.

4.9 v02 9.82 198

4.9 v02 9.8 v0

2 9.82 198

4.9 v0

2

9.82 v0

2

9.8 198

4.9 v09.82

v0 v09.8 2 200

t v09.8

vt 9.8t v0 0

ft 200 4.9t2 v0t 2,

ft 4.9t2 v0t 2.


6/34


76.

When

v2 v02 2GM1y

1

R

v2 2GM

y v0

2

2GM

R

1

2v2

GM

y

1

2v0

2

GM

R

C1

2v0

2

GM

R

1

2v02 GM

R C

y R, v v0.

1

2v2

GM

y C

vdv GM 1y2

dy 78.

(a)

(b) when and

(c) when

v73 37

3 57

3 3 223

4

3

t7

3.at 6t 14 0

3 < t < 5.0 < t 0

at vt 6t 14

vt xt 3t2 14t 15 3t 5t 3

t3 7t2 15t 9

0 t 5xt t 1t 32

80. (a)

(b) for k 0, 1, 2, . . .t k,vt 0 sin t

ft cos t 4

f0 3 cos0 C2 1 C2 C2 4

ft vtdt sin tdt cos t C2vt atdt cos tdt sin t C1 sin t since v0 0at cos t

82.

st 8.25t2 66t

vt 16.5t 66

at 16.5

132 when a 33

2 16.5.

s66a

a2

66a

2

6666a

at 66 0 when t66

a.

vt 0 after car moves 132 ft.

st a

2t2 66t Let s0 0.

vt at 66

at a

15 mph 22 ftsec

30 mph 44 ftsec

v0 45 mph 66 ftsec (a)

(b)

(c)

It takes 1.333 seconds to reduce the speed from 45 mph to

30 mph, 1.333 seconds to reduce the speed from 30 mph

to 15 mph, and 1.333 seconds to reduce the speed from

15 mph to 0 mph. Each time, less distance is needed to

reach the next speed reduction.

0

73.33 117.33

132

feet feet

45mph

=66ft/s

ec

30mph

=44ft/s

ec

15mph

=22ft/s

ec

0mph

s 4416.5 117.33 ft

t44

16.5 2.667

16.5t 66 22

s 2216.5 73.33 ftt

22

16.5 1.333

16.5t 66 44


7/34


84. No, car 2 will be ahead of car 1. If and are the respective velocities, then 300

v2tdt > 30

0v1tdt.v2tv1t

86. (a) (b)

s6 196.1 feet

Note: Assume s0 0 is initial position

st vtdt 0.6139t44 5.525t3

3

0.0492t2

2 65.9881tv 0.6139t3 5.525t2 0.0492t 65.9881

88. Let the aircrafts be located 10 and 17 miles away from the airport, as indicated in the figure.

(a) When aircraft A lands at time you have

kA 50

tA 5012510 625SAt S1t

625

2t2 150t 10

tA 10

125.

125tA 10

1

250

tAt2A 150tA 10

sAtA 1

2

kAt2A 150tA 10 0

vAtA kAtA 150 100 kA 50

tA

tA

sAt 1

2kAt

2 150t 10 sB

1

2kBt

2 250t 17

10 170

A BirportvAt kAt 150 vB kBt 250

Similarly, when aircraft B lands at time you have

(b)

0

0

0.1

20

sA

sB

kB

135

tB 135

365

34

49,275

34S

B

t S2t 49,275

68t2 250t 17

tB 34

365.

365

2tB 17

1

2135

tBt2B 250tB 17

sBtB 1

2kBt

2B 250tB 17 0

vBtB

kBtB

250

115

kB

135

tB

tB

(c)

Yes, for t > 0.0505.d < 3

0

0

0.1

20

d3

d sBt sAt

90. True 92. True


8/34


Section 4.2 Area

96.

Thus, for some constant k. Since,

and

Therefore,

[Note that and satisfy these

properties.]

cx cosxsx sinx

sx2 cx2 1.

k 1.c0 1,s0 0

sx2 cx2 k

0

2sxcx 2cxsx

d

dxsx2 cx2 2sxsx 2cxcx94. False. has an infinite number of antiderivatives, each

differing by a constant.

f

2. 6

k3

kk 2 31 42 53 64 50 4. 5

j3

1

j

1

3

1

4

1

5

47

60

6. 4

i1

i 12 i 13 0 8 1 27 4 64 9 125 238

8. 15

i1

5

1 i10.

4

j11 j4

2

12. 2nn

i11 2in 1

2

14. 1nn1

i01 in

2

16.

215162 45 19515

i1

2i 3 215

i1

i 315 18.

1011216 10 37510

i1

i2 1 10

i1

i2 10

i1

1

20.

102112

4 10112 3080

10

i1

ii2 1 10

i1

i3 10

i1

i 22. sum seq (TI-82)

152162

4 1516 14,160

15

i1

i3 2i 15 215 1 2

4 2

1515 1

2

3 2x,x, 1, 15, 1 14,160x

24.

s 4 4 2 01 10

S 5 5 4 21 16 26.

s 2 1 23 1

2

1

3 9

2 4.5

S 5 2 1 23 1

2 55

6

28.

s8 0 21

4 14 214 12 214 . . . 74 214 5.685

1

416 1

22

23

2 1

5

26

27

2 2 6.038

54 214 32 214 74 214 2 214

S8 14 214 12 214 34 214 1 214

>


9/34

Section 4.2 Area 457

30.

s5 1 152

15 1 2

52

15 1 3

52

15 1 4

52

15 0 0.659

1

51 24

521

516

59

5 0.859

S5 115 1 1

52

15 1 2

52

15 1 3

52

15 1 4

52

15

32. limn

64n3nn 12n 1

6 64

6limn

2n3 3n2 n

n3 64

62

64

3

34. limn

1n2nn 1

2 1

2limn

n2 n

n2 1

21

1

2

36.

S10,000 2.0005

S1000 2.005

S100 2.05

S10 25

10 2.5

n

j1

4j 3

n2

1

n2

n

j1

4j 3 1

n24nn 1

2 3n 2n 5n Sn

38.

S10,000 1.000066657

S1000 1.00066567

S100 1.006566

S10 1.056

1

3n

33n3 2n2 3n 2 Sn

1

3n33n3 6n2 3n 4n2 6n 2

4

n3n3 2n2 n

4

2n2 3n 1

6

n

i1

4i2i 1

n4

4

n4

n

i1

i3 i2 4

n4n2n 12

4

nn 12n 1

6

40. limn

4

21 1

n 2 limn4

n2nn 1

2 limnn

i12in

2

n limn4

n2n

i1

i

42.

21 2 43 26

3

2 limn

1 2 2n 4

3

2

n

2

3n2

limn

2

n3n3 4nnn 1

2 4nn 12n 1

6

limn

2

n3n

i1

n2 4nn

i1

i 4 n

i1

i 2

limn

n

i11 2in

2

2n limn2

n3n

i1

n 2i2


10/34


44.

2 limn

10 13n 4

n2 20

2 limn

1 3 3n 4 6

n

2

n2 2

4

n

2

n2

2 limn

1

n4n4 6n2nn 12 12n

nn 12n 1

6 8n2n 12

4

2 limn

1

n4

n

i1

n3 6n2i 12ni2 8i 3

limn

n

i11 2in

3

2n 2 limn1

n4n

i1

n 2i3

46. (a)

(c) Since is increasing, on

(d) on

(e)

Sn n

i1fxix

n

i1f1 i2n

2

n n

i11 i2n

2

n

[xi1,xifMi fxi

n

i1

f1 i 12n2

n n

i11 i 12n

2

n

sn n

i1fxi1x

xi1,xi.fmi fxi1y x

x2 4

1

2

3

4

y

x 5 10 50 100

3.6 3.8 3.96 3.98

4.4 4.2 4.04 4.02Sn

sn

(b)

Endpoints:

1 < 1 12n < 1 22n < . . . < 1 n 12n < 1 n2n

1 < 1 2

n< 1

4

n< . . . < 1

2n

n 3

x 3 1

n

2

n

(f )

limn

2 2n 1n limn4 2n 4

limn

n

i11 i2n

2

n limn2

nn 2

nnn 1

2

limn

2 2n 2n 4

n limn4 2

n 4

limn

n

i11 i 12n

2

n limn2

nn 2

nnn 1

2 n


11/34


48. on

Area limn

Sn 6 27

2

39

2

6 27

n2n 1n

2 6 27

2 1 1

n

n

i132 3in 4

3

n 18 33

n2

n

i1

i 12

Sn n

i1

f2 3in 3

n

x4 6 8 10 1 2

2

4

6

8

10

12

yNote: x 5 2n 3

n2, 5.y 3x 4

50. on

Area limn

Sn 9

22 3 12

27

n3nn 12n 1

6

3

nn

9

22n2 3n 1

n2 3

27

n3

n

l1

l 23

n

n

l1

1

Sn n

l1f3in

3

n n

l13in

2

13n

y

x1 2

2

4

6

8

10

12

3

0, 3. Note:x 3 0n 3

ny x2 1

52. on Find area of region over the interval

Area 4

3

1

2Area lim

n

sn 1 1

3

2

3

1 1

n3

n

i1

i 2 1 nn 12n 1

6n3 1

1

62 3

n

1

n2

sn n

i1f in

1

n n

i11 in

2

1n

x

2

3

2 2

1

yNote: x 1n0, 1.1, 1.y 1 x2

54. on

Sincey both increases and decreases on is neither an upper nor lower sum.

Area limn

Tn 1 1

4

3

4

1 1

n

1

4

2

4n

1

4n2

2

n2

n

i1

i 1

n4

n

i1

i3 nn 1

n2

1

n4n2n 12

4

Tn n

i1f in

1

n n

i12 in

i

n3

1n

Tn0, 1,

x0.5 1.0 2.0

0.5

1.0

1.5

2.0

yNote: x 1 0n 1

n0, 1.y 2x x3


12/34


56. on

Area limn

sn 2 5

2

4

3

1

4

7

12

2 52

52n

43

2n

23n3

14

12n

14n2

n

i12 5in

4i 2

n2

i 3

n31

n 2 5

n2

n

i1

i 4

n3

n

i1

i2 1

n4

n

i1

i3

sn n

i1f1 in

1

n n

i11 in

2

1 in3

1n

x1 212

1

1

2

3

yNote: x 0 1n 1

n1, 0.y x2 x3

58.

y

x1

1

2

3

4

5

2 3 4 5

Area limn

Sn 2 1 3

2

nn 1

nnn 1

2 2 n 1

n

n

i1

1

22 2i

n 2

n 2

n

n

i11 in

Sn n

i1

g2 2in 2

n

gy 1

2y, 2 y 4. Note:y 4 2n

2

n 60.

1

1

2

3

5

2 3 4 5

y

x

Area limn

Sn 3 1 1

3

11

3

3 n 1

n

n 12n 1

6

1

n3n 2n

nn 1

2

1

n2

nn 12n 1

6

1

n

n

i1

3 2in i2

n2

1

n

n

i1

4 4in 1 2i

n

i2

n2

1

n

n

i1

41 in 1 i

n2

Sn n

i1

f1 in1

n

fy 4y y2, 1 y 2. Note:y 2 1n 1

n

62.

Area limn

Sn 2 1

4 1

3

2

19

4

2 n 12

n24

1

2n 12n 1

n2

3n 1

2n

1

n2n 1

n3

n2n 12

4

3

n2nn 12n 1

6

3

nnn 1

2

1

n

n

i12 i

3

n3

3i2

n2

3i

n

n

i11 in

3

11n

2

1

2

3

4

5

4 6 8 10

y

x

Sn n

i1

h1 in1

n

hy y3 1, 1 y 2 Note:y 1n


13/34


64.

Let

53

14 2 9

4 6 254 10

49

4 14

Area

n

i1fci

x

4

i1ci2

4ci1

c4 7

2c3

5

2,c2

3

2,c1

1

2,x 1,

ci xi xi1

2.

n 40 x 4,fx x2 4x, 66.

Let

8sin

16 sin

3

16 sin

5

16 sin

7

16 1.006

Area

n

i1fcix

4

i1sin ci

8

c4 7

16c3

5

16,c2

3

16,c1

16,x

8,

ci xi xi1

2.

n 40 x

2,fx sinx,

68. on 2, 6.fx 8

x2 1

n 4 8 12 16 20

2.3397 2.3755 2.3824 2.3848 2.3860Approximate area

70. on 0, 2.fx cosx

n 4 8 12 16 20

1.1041 1.1053 1.1055 1.1056 1.1056Approximate area

72. See the Definition of Area. Page 259.

74.

(Note: exact answer is 12.)

0 x 8fx 3x,

n 10 20 50 100 200

10.998 11.519 11.816 11.910 11.956

12.598 12.319 12.136 12.070 12.036

12.040 12.016 12.005 12.002 12.001Mn

Sn

sn

76.

a. square unitsA 3

x1 2 3 4

1

2

3

4

y 78. True. (Theorem 4.3)


14/34


Section 4.3 Riemann Sums and Definite Integrals

2.

limn

1

n43n4

4

n3

2

n2

4 limn3

4

1

2n

1

4n2 3

4

limn

1

n43n4 6n3 3n2

4

2n3 3n2 n

2

n2 n

2

limn

1

n43n2n 12

4 3nn 12n 1

6 nn 1

2

limn

1

n4

n

i1

3i 3 3i 2 i

limn

n

i1

fcixi limn

n

i1

3i

3

n3

3i 2 3i 1

n3

xi i 3

n3

i 13

n3

3i 2 3i 1

n3

ci i 3

n3x 1,x 0,y 0,fx 3x,

80. (a)

(b)

(c)

Let As

limn

An limx0

r2sinxx r21 r2x0.n,x 2n.

An n12r2 sin2

n r2n

2sin

2

n r2sin2n2n

A 1

2

bh 1

2

rrsin 1

2

r2 sin

h rsin

sin h

r

r

r

h

2

n

82. (a)

The formula is true for

Assume that the formula is true for

Then we have

Which shows that the formula is true for n k 1.

k 1k 2

kk 1 2k 1

k1

i1

2i k

i1

2i 2k 1

k

i1

2i kk 1.

n k:

n 1: 2 11 1 2

n

i1

2i nn 1 (b)

The formula is true for because

Assume that the formula is true for

Then we have

which shows that the formula is true for n k 1.

k 12

4 k 22

k 12

4k2 4k 1

k2k 12

4 k 13

k1

i1

i3 k

i1

i3 k 13

k

i1

i3 k2k 12

4

n k:

13 121 12

4

4

4

1

n 1

n

i1

i3 n2n 12

4


15/34

Section 4.3 Riemann Sums and Definite Integrals 463

4. on

3

2

xdx limn

52 252n 52

10 25n2nn 1

2 10

25

2 1 1

n 5

2

25

2n

n

i1fcixi

n

i1

f2 5in 5

n n

i12 5in

5

n 10 25

n2

n

i1

i

Note:x 3 2n 5

n, 0 as n2, 3.y x

6. on

6 12 8 26

31

3x2 dx limn

6 12n 1n 4n 12n 1

n2

6 12n 1

n 4

n 12n 1

n2

6

nn 4

nnn 1

2

4

n2nn 12n 1

6

6

nn

i11 4in

4i2

n2

n

i1

31 2in 2

2n

n

i1fcixi

n

i1f1 2in

2

n

1, 3. Note: x 3 1n 2

n, 0 as ny 3x2

8. on

15 27 27 15

2

1

3x2 2dx limn

15 27

n 1

n

27

2

n 12n 1

n2

15 27n 1

n

27

2n 12n 1

n2

3

n3n 18

nnn 1

2

27

n2nn 12n 1

6 2n

3

n

n

i131 6in

9i2

n2 2

3

nn

i131 3in

2

2

n

i1fcixi

n

i1f1 3in 3n

1, 2. Note:x 2 1n 3

n; 0 as ny 3x2 2

10.

on the interval 0, 4.

lim0

n

i1

6ci4 ci2xi 4

0

6x4 x2dx 12.

on the interval 1, 3.

lim0

n

i1 3ci2xi

3

1

3

x2dx

14. 20

4 2xdx 16. 20

x2dx 18. 11

1

x2 1dx 20. 4

0

tanxdx


16/34


22. 20

y 22dy 24. Rectangle

x

1

2

3

5

a a

Rectangle

y

A aa

4 dx 8a

A bh 24a

26. Triangle

x1 2 3 4

1

1

2

3

Triangle

y

A 40

x

2dx 4

A 1

2bh

1

242

28. Triangle

2

2

4

6

8

10

4 6 8 10

y

x

A 8

0

8 xdx 32

A 1

2bh

1

288 32

30. Triangle

x

Triangle

a a

a

y

A

a

aa xdx a

2

A 1

2bh

1

22aa

32. Semicircle

xr

r

r

r Semicircle

y

A

r

rr2 x2dx 12

r2

A 1

2r2

In Exercises 3440,

4

2x

3

dx

60,

4

2xdx

6,

4

2dx

2.

34. 22

x3 dx 0 36. 42

15 dx 1542

dx 152 30

38. 42

x3 4dx 42

x3 dx 442

dx 60 42 68 40.

62 26 60 36

42

6 2x x3dx 642

dx 242

xdx 42

x3 dx

42. (a)

(b)

(c)

(d) 63

5fxdx 563

fxdx 51 5

33

fxdx 0

3

6

fxdx 6

3

fxdx 1 1

60

fxdx 30

fxdx 63

fxdx 4 1 3 44. (a)

(b)

(c)

(d) 10

3fxdx 310

fxdx 35 15

11

3fxdx 311

fxdx 30 0

1

0

fxdx 0

1

fxdx 5 5 10

01

fxdx 11

fxdx 10

fxdx 0 5 5


17/34

Section 4.3 Riemann Sums and Definite Integrals 465

46. (a)

(c) f even55

fxdx 250

fxdx 24 8

50

fx 2dx 50

fxdx 50

2 dx 4 10 14 (b)

(d) f odd55

fxdx 0

Let u x 2.32

fx 2dx 50

fxdx 4

48. The right endpoint approximation will be less than the

actual area:

52. is integrable on but is not contin-

uous on There is discontinuity at To see

that

is integrable, sketch a graph of the region bounded by

and thex-axis for You see that

the integral equals 0.

x1 22

2

1

2

y

1 x 1.fx xx

11

xx

dx

x 0.1, 1.1, 1,fx xx 54.

b. square unitsA 43

x1 1314 42

1

2

3

4

y

56.

c. Area 27.

y

x1

1

2

3

4

5

6

7

8

9

2 3 4 5 6 7 8 9

58. 30

5

x2 1dx

4 8 12 16 20

7.9224 7.0855 6.8062 6.6662 6.5822

6.2485 6.2470 7.2460 6.2457 6.2455

4.5474 5.3980 5.6812 5.8225 5.9072Rn

Mn

Ln

n

60. 30

x sinxdx

4 8 12 16 20

2.8186 2.9985 3.0434 3.0631 3.0740

3.1784 3.1277 3.1185 3.1152 3.1138

3.1361 3.1573 3.1493 3.1425 3.1375Rn

Mn

Ln

n

62. False

10

xxdx 1

0

xdx1

0

xdx64. True 66. False

42

xdx 6


18/34


68.

12

4 3

2

12 3

2 2

3 1 0.7084

i1fcixi f6x1 f

3x2 f2

3 x3 f3

2 x4c4

3

2c3

2

3 ,c2

3 ,c1

6 ,

x4 x3 2

3,x2

12,x1

4,

x4 2x3 ,x2

3,x1

4,x0 0,

0, 2fx sinx,

70. To find use a geometric approach.

Thus,

20

xdx 12 1 1.

x1 2 31

1

1

2

3

y

2

0 xdx,

Section 4.4 The Fundamental Theorem of Calculus

2.

0

cosxdx 0 0

2

2

fx cosx

6. 72

3 dv 3v7

2

37 32 15

8. 52

3v 4dv 32 v2 4v5

2

752 20 6 8 39

2

10.

38

3

1

3x2 5x 4dx x3 5x2

2 4x

3

1

27 452 12 1 5

2 4

12. 11

t3 9tdt 14t4 9

2t2

1

1

14 9

2 1

4

9

2 0

14. 12u 1u2du

u2

2

1

u1

2

12 1 2 1

2 2

4.

is negative.22

x2 xdx 2 2

5

5

fx

x

2

x


19/34

Section 4.4 The Fundamental Theorem of Calculus 467

16. 33

v13dv 34v433

3

3

433 4 33 4 0

18. 812xdx 2

8

1

x12dx 22x128

1

22x8

1

8 22

20.

2

0

2 ttdt

2

0

2t12 t32dt

4

3

t32 2

5

t52

2

0

tt

15

20 6t

2

0

22

15

20 12 162

15

22.

1

23

5x53

3

8x83

1

8

x53

8024 15x

1

8

1

8039

32

80144

4569

80

18

x x2

2 3xdx

1

21

8

x23 x53dx

24.

4 16 18 9

2

13

2

92 1

2 24 8 18 9

2

x2

2

3

1

6x

x2

2

4

3

31

xdx 43

6 xdx

41

3 1x 31dx 31

3 x 3dx 43

3 x 3dx

26.

4

3 0

4

3

4

3 0 4

1

3 2 3

9 18 9

1

3 2 3

64

3 32 12

9 18 9

x3

3 2x2 3x

1

0

x3

3 2x2 3x

3

1

x3

3 2x2 3x

4

3

40x2 4x 3dx

1

0

x2 4x 3dx 31

x2 4x 3dx 43

x2 4x 3dx

28. 40

1 sin2

cos2d 4

0

d 4

0

4

30. 24

2 csc2xdx 2x cotx2

4 0 2 1

2 1

2

2

32. 22

2t cos tdt t2 sin t2

2

2

4 1

2

4 1 2

34. P 2

2

0

sin d 2cos 2

0

2

0 1

2

63.7%

36. A 11

1 x4dx x 15x51

1

8

538. A 2

1

1

x2dx 1x

2

1

1

2 1

1

2

40. A 0

x sinxdx x2

2 cosx

0

2

2 2

2 4

2

split up the integral at the zeros

x 1, 3


20/34


44. Since on

A

3

0 3x

x2

dx

3

2x2

x3

3 3

0

9

2.

0, 3,y 0 46.

c 392 1.6510

c3 9

2

9

c3 2

fc3

1

4

31

9

x3dx 92x2

3

1

1

2

9

2 4

48.

c 0.5971

cos c 33

2

fc

3

3 3

33

cosxdx sinx3

3 3 50.

2

3

1

3

16

3

1

3 13

1

4x2 1

x2dx 23

1

1 x2dx 2x 1x3

1

42. Since on

Area 80

1 x13dx x 34x 438

0

8 3

416 20

0. 8,y 0

52.

x 0.881

cosx 2

Average value 2

0.5

0 2.71

0.881, 2((

1.51

2 02

0

cosxdx 2sinx2

0

2

54. (a)

x1 2 3 4 5 6 7

1

2

3

4

A

A A A

1

2 3 4

y

8

1

23 1

1

21 2

1

22 1 31

A1 A2 A3 A4

71

fxdx Sum of the areas (b) Average value

(c)

Average value

x1 2 3 4 5 6 7

1

2

34

5

6

7

y

20

6

10

3

A 8 62 20

71

fxdx

7 1

8

6

4

3


21/34


56.

3.5 1.5 5.0

62

fxdx area or regionB 60

fxdx 20

fxdx 58. 20

2fxdx 220

fxdx 21.5 3.0

60. Average value 1

66

0

fxdx 1

63.5 0.5833 62.

1

R 0R0

kR2 r

2dr

k

RR2r

r3

3R

0

2kR

2

3

64.

(a)

Average profit

(b)

(c) The definite integral yields a better approximation.

1

66.5

0.5

5t 30dt1

652

3t32 30t

6.5

5.0

954.061

6 159.010

1

6155 157.071 158.660 160 161.180 162.247

954.158

6 159.026

P 5t 30

t 1 2 3 4 5 6

P 155 157.071 158.660 160 161.180 162.247

66. (a)

(b)

1 5

20

100

R 2.33t4 14.67t3 3.67t2 70.67t (c)

181.957

4

0

Rtdt 2.33t5

5 14.67t

4

4 3.67t

3

3 70.67t

2

2 4

0

68. (a) histogram

t

2

21

4

43

68

10

12

14

16

18

65 8 97

N

(b) customers

(c) Using a graphing utility, you obtain

(d)

(e)

The estimated number of customers is

(f) Between 3 P.M. and 7 P.M., the number of customers is approximately

Hence, per minute.3017240 12.6

7

3

Ntdt60 50.2860 3017.85.16260 5110.

90

Ntdt 85.162

2

2 10

16

Nt 0.084175t3 0.63492t2 0.79052 4.10317.

49806 7 9 12 15 14 11 7 260 8360


22/34


70.

F8 84

4 64 16 4 1068

F5 625

4 25 10 4 167.25

Note: F2 2

2

t3 2t 2dt 0F2 4 4 4 4 0

x4

4x2 2x 4

x4

4x2 2x 4 4 4

Fx x2

t3 2t 2dt t4

4 t2 2t

x

2

72.

F8 1

64

1

4

15

64

F5

1

25

1

4

21

100 0.21

F2 1

4

1

4 0

Fx x2

2

t3dt x

2

2t3 dt1

t2x

2

1

x2

1

4

74.

F8 1 cos 8 1.1455

F5 1 cos 5 0.7163

F2 1 cos 2 1.4161

Fx x0

sin d cos x

0

cosx cos 0 1 cosx

76. (a)

(b)d

dx1

4x4

1

2x2 x3 x xx2 1

x

0 tt2

1dt

x

0 t3

tdt

1

4t4

1

2t2

x

0

1

4x4

1

2x2

x2

4 x2

2

78. (a)

(b)d

dx2

3x32

16

3 x12 x

x4

tdt 23t32x

4

2

3x32

16

3

2

3x32 8 80. (a)

(b)d

dxsecx 2 secx tanx

x3

sec ttan tdt sec tx

3 secx 2

82.

Fx x2

x2 1

Fx x1

t2

t2 1dt 84.

Fx 4x

Fx x1

4tdt 86.

Fx sec3x

Fx x0

sec3tdt


23/34


88.

Fx 0

Fx xx

t3dt t4

4x

x 0 Alternate solution

Fx x31 x3 0

x

0t3

dt x

0t3

dt

0x

t3dt x0

t3dt

Fx xx

t3dt

90.

Alternate solution: Fx x232x 2x5

Fx x2

2

t3 dt t2

2x2

2

12t2x2

2

1

2x4

1

8Fx 2x5

92.

Fx sin x222x 2x sinx 4

Fx x2

0

sin 2d

94. (a) (b)

(c) Minimum ofg at

(d) Minimum at Relative maximum at

(e) On g increases at a rate of

(f) Zeros ofg: x 3,x 8.

12

4 3.6, 10

2, 2.10, 6.

6, 6.

2

2 4 6 10 122

4

6

4

6

y

x

x 1 2 3 4 5 6 7 8 9 10

1 2 0 0 3 63642gx

96. (a)

Horizontal asymptote: y 4

limt

gt 4

gt 4 4t2

(b)

The graph of does not have a horizontal asymptote.Ax

limx

Ax limx

4x 4x 8 0 8

4x2 8x 4

x

4x 12

x

4t 4tx

1

4x 4

x 8

Ax x

1 4 4t2dt

98. True 100. Let be an antiderivative of Then,

.fvxvx fuxux

Fvxvx Fuxux

d

dxvx

uxftdt ddxFvx Fux

vx

uxftdt Ft

vx

ux Fvx Fux

ft.Ft


24/34

Section 4.5 Integration by Substitution

du gxdxu gxfgxgxdx

2.

4. 2x 2 dx

6. sinx cosx dxcosxsin2x dxsec 2x tan 2xdx

3x2dxx3 1x2x3 1 dx

12.

Check:d

dxx3 55

15 C 5x

3 543x2

15 x3 54x2

x2x3 54dx 13x3 543x2dx 13x3 55

5 C

x3 55

15 C

14.

Check:d

dx4x2 34

32 C 44x

2 338x

32x4x2 33

x4x2 33dx 184x2 338xdx 184x2 34

4 C4x2 34

32 C

102.

(a)

(c)

(d) G0 0 f0 0

0

ftdt 0

Gx x fx x0

ftdt

G0 00s

s

0

ftdtds 0Gx x

0s

s

0

ftdtds(b) Let

G0 000

ftdt 0

Gx Fx xx

0

ftdt

Gx x0

Fsds

Fs ss0

ftdt.

104.

Using a graphing utility,

Total distance 50xtdt 27.37 units

xt 3t2 14t 15

xt t 1t 32 t3 7t2 15t 9


10.

Check:d

dx3

41 2x243 C 34

4

31 2x2134x 1 2x2134x

1 2x2134xdx 341 2x243 C

8.

Check:d

dxx2 94

4 C 4x

2 93

42x x2 932x

x2 932xdx x2 944 C


25/34

16.

Check:d

dt1

6t4 532 C 16

3

2t4 5124t3 t4 512t3

t3t4 5 dt 14t4 5124t3dt 14t4 5 32

32 C

1

6t4 532 C

18.

Check:d

du2u3 232

9 C 29

3

2u3 2123u2 u3 212u2

u2u3 2 du

1

3u3 2123u2du

1

3

u3 232

32 C

2u3 232

9

C

20.

Check:d

dx1

41 x4 C 141 x424x3

x3

1 x42

x31 x42dx 141 x424x3dx 141 x41 C 141 x4 C

22.

Check:d

dx1

316 x3 C 13116 x323x2

x2

16 x32

x216 x32dx 1316 x323x2dx 1316 x31

1 C1

316 x3 C

24.

Check:d

dx1 x4

2 C 12 121 x4124x3 x

3

1 x4

x31 x4

dx 1

41 x4124x3dx 1

41 x412

12 C

1 x4

2 C

26.

Check:

d

dx1

3x3

1

9x1

C x2

1

9x2

x2

1

3x2

x2 13x2dx x2 19x2dx x3

3

1

9x1

1 Cx3

3

1

9x C

3x4 1

9x C

28.

Check:d

dxx C

1

2x

12x

dx 1

2x12dx 12x12

12 Cx C

30.

Check:d

dt2

3t32

4

5t52 C t12 2t32 t 2t

2

t

t 2t2t

dt t12 2t32dt 23t32 45t52 C 215t325 6t C

32.

Check:d

dt1

12t4

1

4t C 13t3

1

4t2

t33 14t2dt 13t3 14t2dt 13t4

4 1

4t1

1 C1

12t4

1

4t C

Section 4.5 Integration by Substitution 473


26/34

34.

Check:d

dy4y2

714 y32 C ddy24y2

2

7y72 C 16y 2y52 2y8 y32

2y8 y32dy 28y y52dy 24y2 27y72 C 4y2

714 y32 C

36.

20

31 x3 C

10

3 1 x312

12 C

10

3 1 x3123x2dxy

10x2

1 x3dx 38.

x2 8x 1 C

1

2x2 8x 112

12 C

1

2x2 8x 1122x 8dxy

x 4

x2

8x 1

dx

42. 4x3 sinx4dx sinx44x3dx cosx4 C 44. cos 6xdx 16cos 6x6dx 16 sin 6x C

46. x sinx2dx 12sinx22xdx 12 cosx2 C

48. sec1 x tan1 xdx sec1 x tan1 x1dx sec1 x C

50. tanx sec2xdx tanx3232 C 23tanx32 C

52. sinxcos3xdx cosx3sinxdx cosx2

2 C

1

2 cos2x C

1

2sec2x C

54. csc2x2dx 2csc2x212dx 2 cotx2 C 56.Since Thus

fx sec x 1.

C 1.f13 1 sec3 C,

fx sec x tan xdx sec x C

40. (a)

x55

2

3

3

y (b)

y 1

2sinx2 1

1 1

2sin0 C C 10, 1:

1

2sin x2 C

y x cosx2dx 12cosx22xdx

dy

dxx cosx2, 0, 1



27/34

58.

1

152x 1323x 1 C

1

302x 1326x 2 C

1

302x 13232x 1 5C

u32

303u 5 C

1

42

5u52

2

3u32 C

1

4u32 u12dux2x 1 dx 12u 1u12du

dx 1

2dux

1

2u 1,u 2x 1, 60.

2

52 x32x 3C

2

52x325 2 xC

2u32

55 uC

2u32 2

5u52 C

3u12 u32dux 12 xdx 3 uudu

dx dux 2 u,u 2 x,

62. Let

2

3x 42x 13 C

2

3

x 42x 4 21 C

2

3u122u 21 C

4

3u32 14u12 C

2u12 7u12du 2x 1x 4

dx 2u 4 1u

du

u x 4,x u 4, du dx. 64.

3

7t 443t 3 C

3

7

t 443t 4 7 C

3u43

7u 7 C

3

7u73 3u43 C

u43 4u13dut3t 4 dt u 4u13du

dt dut u 4,u t 4,

66. Let

1

964 83 8 83 41,472

42

x2x3 82dx 1

34

2

x3 823x2dx 13x3 83

3 4

2

u x3 8, du 3x2dx.

68. Let

1

0

x1 x2dx 1

21

0

1 x2122xdx 131 x2321

0

0 1

3

1

3

du 2xdx.u 1 x2,

70. Let

20

x

1 2x2dx

1

42

0

1 2x2124xdx 121 2x22

0

3

2

1

2 1

du 4xdx.u 1 2x2,



28/34

74. Let

When When

16

3

1

42

327 23 23 2

1

42

3u32 2u12

9

1

51

x

2x 1dx 9

1

12u 1

u1

2du

1

491

u12 u12du

x 5, u 9.x 1, u 1.

u 2x 1, du 2 dx,x 1

2u 1.

76.

2

3

x cosxdx x2

2 sinx

2

3

2

8 1

2

183

2 52

72

2 3

2

78.

When When

80

u73 4u43 4u13du 310u103 12

7u73 3u43

8

0

4752

35Area 6

2

x23x 2 dx 80

u 223udu

u 8.x 6,u 0.x 2,

dx dux u 2,u x 2,

80. A 0

sinx cos 2xdx cosx 12 sin 2x

0

2

82. Let

Area 412

csc 2x cot 2xdx 1

24

12csc 2x cot 2x2dx 12 csc 2x

4

12

1

2

du

2 dx.u

2x,

84.

0

0

2

20

20

x3x 2 dx 7.581 86.

1 5

0

50

51

x2x 1 dx 67.505 88.

0

1

2

2

20

sin 2xdx 1.0

90.

They differ by a constant: .C2 C1 1

2

cos2x

2 C2

1 sin2x

2 C2

sin2x

2

1

2 C2

sinx cosxdx cosx1sinxdx cos2x2 C2sinx cosxdx sinx1cosxdx sin2x2 C1

72. Let

20

x34 x2dx 1

22

0

4 x2132xdx 384 x2432

0

3

8843 443 6

3

234 3.619

du 2xdx.u 4 x2,



29/34

92. is even.

2

3

2sin3x

3 2

0

22

sin2x cosxdx 20

sin2xcosxdx

fx sin2x cosx 94. is odd.

22

sinx cosxdx 0

fx sinx cosx

96. (a) since sinx is symmetric to the origin.

(b) since cosx is symmetric to they-axis.

(c)

(d) since and hence, is symmetric to the origin.sinx cosx sinx cosx22

sinx cosxdx 0

22

cosxdx 220

cosxdx 2 sinx2

0

2

44

cosxdx 240

cosxdx 2 sinx4

0

2

44

sinxdx 0

98.

sin 3x cos 3xdx

sin 3xdx

cos 3xdx 0 2

0

cos 3xdx 23 sin 3x

0

0

100. If then and x5 x23dx 125 x232xdx 12u3du.du 2xdxu 5 x2,

102.

Thus, When

$250,000.

Q50 t 50,Qt 2100 t3.

Q0 k

31003 2,000,000 k 6

Qt k

3100 t3

Q100 C 0

Qt k100 t2dt k3100 t3 CdQ

dt k100 t2 104.

(a)

Relative minimum: or June

Relative maximum: or January

(b) inches

(c) inches1

312

9

Rtdt1

313 4.33

120

Rtdt 37.47

0.4, 5.5

6.4, 0.7

0 12

0

8

R 3.121 2.399 sin0.524t 1.377

106. (a)

Maximum flow: at .

is a relative maximum.18.861, 61.178t 9.36R 61.713

0 24

0

70 (b) (5 thousand of gallons)Volume

24

0

Rtdt

1272



30/34

108. (a)

0 9.4

4

f

g

4

(e)

The graph ofh is that ofg shifted 2 units downward.

2

0

fxdx t

2fxdx 2 ht.

gt t0

fxdx

0 9.4

4

4

(b) g is nonnegative because the graph offis positive at the

beginning, and generally has more positive sections than

negative ones.

(d) No, some zeros off, like do not correspond toan extrema ofg. The graph ofg continues to increase

after becausefremains above thex-axis.x 2

x 2,(c) The points on g that correspond to the extrema offarepoints of inflection ofg.

110. False

xx2 12dx 12x2 12xdx 14x2 12 C

112. True

ba

sinxdx cosxb

a cos b cos a cosb 2 cos a b2

a

sinxdx

114. False

sin2 2x cos 2xdx 12sin 2x22 cos 2xdx 12sin 2x3

3 C

1

6sin3 2x C

116. Becausefis odd, Then

Let in the first integral.

When When

a0

fudu a0

fxdx 0

1a

fxdx a0

fudu a0

fxdx

x a, u a.x 0, u 0.

x u, dx du

a0

fxdx a0

fxdx.

aa

fxdx 0a

fxdx a0

fxdx

fx fx.



31/34

Section 4.6 Numerical Integration

Section 4.6 Numerical Integration 479

2. Exact:

Trapezoidal:

Simpsons: 10x

2

2 1dx 1121 4

142

2 1 212

2

2 1 434

2

2 1 1

2

2 1 76 1.1667

10x

2

2 1dx 181 2

142

2 1 212

2

2 1 234

2

2 1 1

2

2 1 7564 1.1719

10x

2

2 1dx x

3

6x

1

0

7

6 1.1667

4. Exact:

Trapezoidal:

Simpsons: 21

1

x2dx

1

121 44

52

2462

4472

1

4 0.5004

21

1

x2dx

1

81 24

52

2462

2472

1

4 0.5090

21

1

x2dx 1x

2

1

0.5000

6. Exact:

Trapezoidal:

Simpsons: 80

3xdx 1

30 4 2 32 4 33 2 34 4 35 2 36 4 37 2 11.8632

80

3xdx 1

20 2 2 32 2 33 2 34 2 35 2 36 2 37 2 11.7296

8

0

3xdx 34x438

0

12.0000

8. Exact:

Trapezoidal:

Simpsons: 3

1

4 x2dx 1

63 44 9

4 0 44 25

4 5 0.6667

31

4 x2dx 1

43 24 3

22

20 24 522

5 0.750031

4 x2dx 4x x3

3 3

1

3 11

3

2

3 0.6667

10. Exact:

Trapezoidal:

Simpsons: 20

xx2 1 dx 1

60 41

2122 1 2112 1 43

2322 1 222 1 3.392

20

xx2 1 dx 1

40 21

2122 1 2112 1 23

2322 1 222 1 3.457

20

xx2 1 dx 1

3x2 1322

0

1

3532 1 3.393

12. Trapezoidal:

Simpsons:

Graphing utility: 1.402

20

1

1 x3dx

1

61 41

1 123 21

1 13 4 11 323

1

3 1.405

2

0

11 x3

dx 141 2 11 123 2 11 13 2 11 323 13 1.397


32/34


14. Trapezoidal:

Simpsons:


2x sinxdx

242 458 sin58 234 sin34 478 sin78 0 1.4582

x sinxdx

162 1 258 sin58 234 sin34 278 sin78 0 1.430

16. Trapezoidal:

Simpsons:


0.257

40

tanx2dx 4

12 tan 0 4 tan44 2

2 tan42 2

4 tan344 2

tan42

0.2714

0

tanx2dx 48 tan 0 2 tan

44

2

2 tan42 2

2 tan344 2

tan

42

18. Trapezoidal:

Simpsons:


2

0

1 cos2xdx

242 41 cos28 21 cos24 41 cos238 1 1.910

20

1 cos2xdx

162 21 cos28 21 cos24 21 cos238 1 1.910

20. Trapezoidal:

Simpsons:


0

sinx

xdx

121 4 sin4

4

2 sin22

4 sin34

34 0 1.852

0

sinx

xdx

81 2 sin4

4

2 sin22

2 sin34

34 0 1.836

22. Trapezoidal: Linear polynomials

Simpsons: Quadratic polynomials

24.

(a) Trapezoidal: since

is maximum in when

(b) Simpsons:

since is maximum in whenx 0.0, 1f4x

Error 1 05

1804424

1

1920 0.0005

x 0.0, 1fx

Error 1 02

12422

1

96 0.01

f4x 24x 15

fx 6

x 14

fx 2

x 13

fx 1

x 12

fx 1

x 1


33/34

Section 4.6 Numerical Integration 481

26. in .

(a) is maximum when and

Trapezoidal: Error < 0.00001, > 16,666.67, n > 129.10; let

in

(b) is maximum when and

Simpsons: Error > 13,333.33, n > 10.75; let (In Simpsons Rule n must be even.)n 12.n41

180n424 < 0.00001,

f40 24.x 0f

4x

0, 1f4x 24

1 x5

n 130.n21

12n22

f0 2.x 0fx

0, 1fx 2

1 x3

28.

(a) in .

is maximum when and


(b) in

is maximum when and

Simpsons: Error < 0.00001, > 12,290.81, n > 10.53; let (In Simpsons Rule n must

be even.)

n 12.n432

180n456

81

f40

56

81.x 0f

4x

0, 2f4x 56

81x 1103

n 122.n28

12n4

2

9

f0 2

9.x 0fx

0, 2fx 2

9x 143

fx x 123

30.

(a) in .

is maximum when and


(b) in

is maximum when and

Simpsons: Error < 0.00001, > 15,793.61, n > 11.21; let n 12.n41 05

180n428.4285

f40.852 28.4285.x 0.852f

4x

0, 1f4x 16x4 12 sinx2 48x2 cosx2

n 139.n21 03

12n22.2853

f1 2.2853.x

1fx

0, 1fx 22x2 sinx2 cosx2

fx sinx2

32. The program will vary depending upon the computer or programmable calculator that you use.

34. on .0, 1fx 1 x2

n

4 0.8739 0.7960 0.6239 0.7489 0.7709

8 0.8350 0.7892 0.7100 0.7725 0.7803

10 0.8261 0.7881 0.7261 0.7761 0.7818

12 0.8200 0.7875 0.7367 0.7783 0.7826

16 0.8121 0.7867 0.7496 0.7808 0.7836

20 0.8071 0.7864 0.7571 0.7821 0.7841

SnTnRnMnLn


34/34


36. on .1, 2fx sinx

x

n

4 0.7070 0.6597 0.6103 0.6586 0.6593

8 0.6833 0.6594 0.6350 0.6592 0.6593

10 0.6786 0.6594 0.6399 0.6592 0.6593

12 0.6754 0.6594 0.6431 0.6593 0.6593

16 0.6714 0.6594 0.6472 0.6593 0.6593

20 0.6690 0.6593 0.6496 0.6593 0.6593

SnTnRnMnLn

38. Simpsons Rule:

17.476

83201 23 sin2d

3

6 1 23 sin2 0 41 23 sin2 16 21 23 sin28 . . . 1 23 sin22

n 8

40. (a) Trapezoidal:

Simpsons:

20

fxdx 2

384.32 44.36 24.58 45.79 26.14 47.25 27.64 48.08 8.14 12.592

20

fxdx 2

284.32 24.36 24.58 25.79 26.14 27.25 27.64 28.08 8.14 12.518

(b) Using a graphing utility,

Integrating,20

ydx 12.53

y 1.3727x3 4.0092x2 0.6202x 4.2844

42. Simpsons Rule:

3.14159

410

1

1 x2dx

4

361 4

1 162

2

1 262

4

1 362

2

1 462

4

1 562

1

2

n 6

44.

7435 sq m

Area 120

21275 281 284 276 267 268 269 272 268 256 242 223 0

46. The quadratic polynomial

passes through the three points.

px x x2x x3

x1 x2x1 x3y1

x x1x x3

x2 x1x2 x3y2

x x1x x2

x3 x1x3 x2y3

Documents

Solucionario - Problemas Pares, C.04 ,Vol1 ,Cálculo, Larson, Hostetler