Chapter 5 The Definite Integral

(1)

Chapter 5 The Definite Integral

Section 5.1

Estimating with Finite Sums

(pp. 263-273)

Exploration 1 Which RAM is the

Biggest?

1.

LRAM > MRAM > RRAM

2.

MRAM > RRAM > LRAM

3. RRAM > MRAM > LRAM, because the heights of the rectangles increase as you move toward the right under an increasing function.

4. LRAM > MRAM > RRAM, because the heights of the rectangles decrease as you move toward the right under a decreasing function.

Quick Review 5.1

1. 80mphi 5hr=400mi

2. 48mphi 3hr=144mi

3. 10ft/sec2 i10sec=100ft/sec 100 ft/sec mi

ft h mph

i 1 i

5280 3600

1 68 18

sec . ≈

4. 300 000 3600 1

24 1

365

, km/ sec sec

hr

hr day

days

i i i

1

1yr i1yr ≈9 46 10. × 12km

5. (6mph)(3h) (+ 5mph)(2h)=18mi+10mi=28mi

6. 20gal/min 1 h 60 min 1200

1 h gal

i i =

7. (− °1 C/h)(12 h)+( .1 5°C)(6h)= − °3 C

8. 300

1

24

1 25 920

ft /sec 3600 sec h

h 1 day day

3 _i _i _i ₌

, ,0000 ft3

9. 350people/mi2i50mi2=17 500, people

10. 70 3600

1 1 0 7 176 400

times/sec

h h times

i seci i . = ,

Section 5.1

Exercises

1. Since v(t) = 5 is a strait line, compute the area under the curve.

x=( ) ( )t v t =( )( )4 5 =20

2. Sincev t( )= +2t 1 creates a trapezoid with the x-axis, compute the area of the curve under the trapezoid.

A h a b

a t v b t v

= +

= = = = + =

= = = =

2

0 0 2 0 1 1

4 4 2 4

( )

( ) ( ) ( ) ( )++ = =

= + =

1 9 4

4

2 9 1 20

h

A ( )

3. Each rectangle has base 1. The height of each rectangle is Found by using the pointst=( . , . , . , . )0 5 1 5 2 5 3 5 in the equationv t t( ) 2+₁._{The area under the curve is}

approximately1 5 4

13 4

29 4

53

4 25

+ + +

⎛ ⎝⎜

⎞

⎠⎟= , so the particle is close to x = 25.

4. Each rectangle has base 1. The height of each rectangle is found by using the pointsy=( . , . , . , . , . )0 5 1 5 2 5 3 5 4 5 in the equationv t( )= +t2 1 The area under the curve is.

approximately1 5 4

13 4

29 4

53 4

85

4 46 25

+ + + +

⎛

⎝⎜ ⎞⎠⎟= . , so the

(2)

5. (a) y 2

x 2 R

(b) y

x

2

Δx=

− ⎛

⎝⎜ ⎞⎠⎟+ ⎛⎝⎜ ⎞⎠⎟− 1

2

0 1

2 2

1 2

2

LRAM: [2(0) ( ) ] ⎛⎛

⎝⎜ ⎞⎠⎟ ⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥ ⎛ ⎝⎜ ⎞⎠⎟

+ − ⎛

⎝⎜ ⎞⎠

2

1 2

2 1 1 1

2 [ ( ) ( ) ] _{⎟⎟ +} ⎛

⎝⎜ ⎞⎠⎟− ⎛⎝⎜ ⎞⎠⎟ ⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥ ⎛

⎝⎜ ⎞⎠⎟= = 2 3

2 3 2

1 2

5

4 1

2

..25

6. (a) y

x

2

RRAM: 2 1

2 1 2

1

2 2

2

⎛

⎝⎜ ⎞⎠⎟− ⎛⎝⎜ ⎞⎠⎟ ⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥ ⎛

⎝⎜ ⎞⎠⎟+[ (( ) ( ) ]1 1 1 2

2 3 2

3 2

2

− ⎛

⎝⎜ ⎞⎠⎟

+ ⎛

⎝⎜ ⎞⎠⎟− ⎛⎝⎜ ⎞⎠⎟ ⎡

⎣ ⎢ ⎢

⎤

⎦⎦ ⎥ ⎥ ⎛

⎝⎜ ⎞⎠⎟+ − ⎛⎝⎜ ⎞⎠⎟= = 1

2 2 2 2

1 2

5 4 1 25

2

[ ( ) ( ) ] .

(b) y

x

2

MRAM: 2 1

4 1 4

1

2 2

3

2

⎛ ⎝⎜

⎞ ⎠⎟− ⎛⎝⎜

⎞ ⎠⎟ ⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥ ⎛ ⎝⎜

⎞ ⎠⎟+ 44

3 4

1 2

2 5 4

2

⎛ ⎝⎜

⎞ ⎠⎟− ⎛⎝⎜

⎞ ⎠⎟ ⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥ ⎛ ⎝⎜

⎞ ⎠⎟

+ ⎛ ⎝⎜

⎞ ⎠⎟⎟− ⎛⎝⎜

⎞ ⎠⎟ ⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥ ⎛ ⎝⎜

⎞ ⎠⎟+ ⎛⎝⎜

⎞ ⎠⎟− ⎛⎝ 5

4 1

2 2

7 4

2

⎜⎜ ⎞_⎠⎟ ⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥ ⎛ ⎝⎜

⎞ ⎠⎟= =

2

1 2

11 8 1 375.

7.

n LRAM_n MRAM_n RRAM_n

10 1.32 1.34 1.32

50 1.3328 1.3336 1.3328

100 1.3332 1.3334 1.3332

500 1.333328 1.333336 1.333328

8. The area is 1 333 4 3

. = .

9.

10 12.645 13.4775 14.445

50 13.3218 13.4991 13.6818

100 13.41045 13.499775 13.59045

500 13.482018 13.499991 13.518018

Estimate the area to be 13.5.

10.

10 1.16823 1.09714 1.03490

50 1.11206 1.09855 1.08540

100 1.10531 1.09860 1.09198

500 1.09995 1.09861 1.09728

1000 1.09928 1.09861 1.09795

(3)

11.

10 0.98001 0.88220 0.78367

50 0.90171 0.88209 0.86244

100 0.89190 0.88208 0.87226

500 0.88404 0.88208 0.88012

1000 0.88306 0.88208 0.88110

Estimate the area to be 0.8821.

12. _n

LRAM_n MRAM_n RRAM_n

10 1.98352 2.00825 1.98352

50 1.99934 2.00033 1.99934

100 1.99984 2.00008 1.99984

500 1.99999 2.00000 1.99999

Estimate the area to be 2.

13.Usef x( )= 25−x2and approximate the volume using πr h2 =π( 25−n_i2)2Δx,so for the MRAM program, use π(25−x2) on the interval⎡⎣−5 5, ⎤⎦.

n MRAM

10 526.21677

20 524.25327

40 523.76240

80 523.63968

160 523.60900

14.V=4 = ≈

3 5

500

3 523 59878

3

π( ) π .

n error % error

10 2.61799 0.5

20 0.65450 0.125

40 0.16362 0.0312

80 0.04091 0.0078

160 0.01023 0.0020

15.LRAM: Area

≈ + + + +

= +

f( ) f( ) f( ) f( )

( .

2 2 4 2 6 2 22 2

2 0 0

i i i i

i 66 1 4 0 5

44 8

4

+ + +

=

≈

. . )

. ( ) sec

( )

i

mg/L RRAM: Area

f 22 6 2 8 2 24 2 2 0 6 1 4 2 7

+ + + +

= + + + +

f( ) f( ) f( )

( . . .

i i i

00 44 8

)

. ( sec

= mg/L)

Patient's cardiac output: 5 m

i

g g

44.8 (mg/L) L /min

Note th

iseci sec

min .

60

1 ≈6 7

aat estimates for the area may vary.

16. (a) LRAM:1i(0 12 22 10 5 13 11 6+ + + + + + + + + =2 6) 87in.=7 25. ft

(b) RRAM:1 12 22 10 5 13 11 6 2i( + + + + + + + + + =6 0) 87in.=7 25. ft

17.5 min = 300 sec

(a)LRAM:300i(1 1 2 1 7+ . + . + + 1 2. )=5220m

(b) RRAM: 300i( .1 2 1 7 2 0+ . + . + + = 0) 4920m

18. LRAM:10i(0+44 15+ + + 30)=3490ft

RRAM:10 ft

Average f

i(44 15 35 35) 3840 3490

+ + + + =

= tt+3840ft= ft

2 3665

19. (a) LRAM: 0.001(0+40 62+ + + 137)=0 898. mi

RRAM: 0.001(40 62 82 142) 1.04 mi

Average

+ + + + =

=0.9969 mi

(b)The halfway point is 0.4845 mi. The average of LRAM and RRAM is 0.4460 at 0.006 h and 0.5665 at 0.007 h. Estimate that it took 0.006 h = 21.6 sec. The car was going 116 mph.

20. (a) Use LRAM withπ(16−x2).

S₈≈146 08406.

S8 is an overestimate because each rectangle is below

the curve.

(b) V S

V

−

≈ =

8

0 09. 9%

21. (a) Use RRAM withπ(16−x2).

S₈≈120 95132.

S8 is an underestimate because each rectangle is below

the curve.

(b) V S

V

−

≈ =

8

0 10. 10%

22. (a) Use LRAM with π(64−x2)on the interval [4, 8], n = 8.

S≈372 27873. m3

(b) V S

V

−

≈ =

(4)

23. (a) ( )( .5 6 0 8 2 9 1+ . + . + + 12 7 30. )( )≈15 465, ft3

(b) ( )( .5 8 2 9 1 9 9+ . + . + + 13 0 30. )( )≈16 515, ft3

24.Use LRAM with πxon the interval [ 0 5, ],n=5. 1(0+ +π 2π π+3 +4π)=10π≈31 41593.

25. Use MRAM withπxon the interval [0, 5].n=5. 1 1

2 3 2

5 2

7 2

9 2

25

2 39 26991

π+ π+ π+ π+ π π

⎛ ⎝⎜

⎞

⎠⎟= ≈ .

26. (a) LRAM₅:

32 00 19 41 11 77 7 14. + . + . + . +4 33. =74 65. ft/sec

(b) RRAM₅:

19 41 11 77 7 14. + . + . +4 33 2 63. + . =45 28. ft/sec

(c)The upper estimates for speed are 32.00 ft/sec for the first sec, 32.00 + 19.41 = 51.41 ft/sec for the second sec, and 32.00 + 19.41 + 11.77 = 63.18 ft/sec for the third sec. Therefore, an upper estimate for the distance fallen is 32.00 + 51.41 + 63.18 = 146.59 ft.

27. (a) 400ft /sec– (5sec)(32ft/sec2)=240ft/sec

(b)Use RRAM with 400 – 32x on [0, 5], n = 5. 368 + 336 + 304 + 272 + 240 = 1520 ft

28. (a) Upper = 70 + 97 + 136 + 190 + 265 = 758 gal Lower = 50 + 70 + 97 + 136 + 190 = 543 gal

(b)Upper = 70 + 97 + 136 + 190 + 265 + 369 + 516 + 720 = 2363 gal

Lower = 50 + 70 + 97 + 136 + 190 + 265 + 369 + 516 = 1693 gal

(c)25,000 – 2363 = 22,637 gal 22 637

720 31 44

1693 2 ,

. ≈

− =

h (worst case)

25,000 33 307

32 37 , .

gal 23,307

720 ≈ h (best case)

29. (a) Since the release rate of pollutants is increasing, an upper estimate is given by using the data for the end of each month (right rectangles), assuming that new scrubbers were installed before the beginning of January. Upper estimate:

30 0 20 0 25 0 27 0 34 0 45 0 52 60 9

( . . . )

.

+ + + + +

≈ tons off pollutants

A lower estimate is given by using the data for the end of the previous month (left rectangles). We have no data for the beginning of January, but we know that pollutants were released at the new-scrubber rate of 0.05 ton/day, so we may use this value.

Lower Estimate:

30 0 05 0 20 0 25 0 27 0 34 0 45 46 8

( . . . )

.

+ + + + +

≈ tons off pollutants

(b) Using left rectangles, the amount of pollutants released by the end of October is

30 0 05 0 20 0 25 0 27 0 34 0 45 0 52 0

( . . . .

. .

+ + + + +

+ + 663 0 70+ . +0 81. )≈126 6. tons

Therefore, a total of 125 tons will have been released into the atmosphere by the end of October.

30.The area of the region is the total number of units sold, in millions, over the 10-year period. The area units are (millions of units per year)(years) = (millions of units).

31.True. Because the graph rises from left to right, the left-hand rectangles will all lie under the curve.

32. False. For example, all three approximations are the same if the function is constant.

33. E.y=4x−x2=0 4

0 4

2 x x x

= = ,

Use MRAM on the interval [0, 4], n= 4. 1(1.75 + 3.75 + 3.75 + 1.75) = 11

34.D.

35.C.

π π π π

4 0 4

3 4

sin( )+sin⎛ sin

⎝⎜ ⎞ ⎠⎟+

⎛ ⎝⎜

⎞ ⎠⎟+

⎛ ⎝⎜

⎞ sin

2 ⎠⎠⎟

⎛ ⎝⎜

⎞ ⎠⎟

+ + +

⎛ ⎝

⎜ ⎞_⎠⎟

π 4 0

2

2 1

2 2

36.D.

37. (a) The diagonal of the square has length 2, so the side length is 2.Area=( 2)2=2

(b)Think of the octagon as a collection of 16 right triangles with a hypotenuse of length 1 and an acute angle

measuring2

16 8

π π₌ _.

Area= ⎛ ⎝⎜

⎞ ⎠⎟ ⎛ ⎝⎜

⎞ ⎠⎟ =

16 1

2 8 8

4

sin cos

sin

π π

π 4 4 2 2 2 828

= ≈ .

(c)Think of the 16-gon as a collection of 32 right triangles with a hypotenuse of length 1 and an acute angle

measuring2

32 16

π ₌ π_.

Area= ⎛ ⎝⎜

⎞ ⎠⎟ ⎛ ⎝⎜

⎞ ⎠⎟ =

32 1

2 16 16

8

sin cos

si

π π

n

nπ .

8≈3 061

(d)Each area is less than the area of the circle,π. As n

increases, the area approachesπ.

38.The statement is false. We disprove it by presenting a counterexample, the functionf x( )=x2over the interval 0≤ ≤x 1,withn=1.MRAM₁=1 0 5f( . )=0 25.

LRAM RRAM

MRAM

1 1

1

2

1 0 1 1

2 0 1

2 0 5

+

= +

= + = ≠

f( ) f( )

(5)

39.RRAM_nf=(Δx f x)[ ( )₁ +f x( ₂)+ + f x( _n₋₁)+f x( _n)]

=(Δx)[ (f x0)+f x( )1 + f x( 2)+ + f x( n−1)]

LRAM

+ −

= + −

( )[ ( ) ( )]

( )[ ( ) (

Δ Δ

x f x f x f x f x f x

n

n n

0 0 0)]

But f(a) = f(b) by symmetry, so f(xn) – f(x0) = 0.

Therefore, RRAMnf= LRAMnf.

40. (a) Each of the isosceles triangles is made up of two right triangles having hypotenuse 1 and an acute angle

measuring2 2

π π

n= n. The area of each isosceles triangle

isA

n n n

T = ⎛_⎝⎜

⎞ ⎠⎟ ⎛ ⎝⎜

⎞ ⎠⎟= 2 1

2

1 2

2 sinπ cosπ sin π..

(b) The area of the polygon is

A nA n n

A n

n

P T

n P n

= =

→∞ →∞

2 2

2 2 sin ,

lim lim sin

π

π π so

(c)Multiply each area byr2_:

A r

n

A nr n

A r

T

P

n P

=

= = →∞

1 2

2

2 2

2

sin π π

π sin lim

Section 5.2

Definite Integrals (274–284)

Exploration 1 Finding Integrals by

Signed Areas

1.−2. (This is the same area as sinx dx,

0

π

∫

but below the

x-axis.)

2. 0. (The equal areas above and below the x- axis sum to zero.)

3. 1. (This is half the area of sinx dx.)

0

π

∫

4.2π +2. The same area as sinx dx 0

π

∫

sits above a rectangle of area π ×2.)

5. 4. (Each rectangle in a typical Riemann sum is twice as tall as in sinx dx.)

0

π

∫

6. 2. (This is the same region as in sinx dx,

0

π

∫

translated 2

units to the right.)

7. 0. (The equal areas above and below the x-axis sum to zero.)

8. 4. (Each rectangle in a typical Riemann sum is twice as wide as in sinx dx.)

0

π

∫

(6)

10. 0. (The equal areas above and below the x-axis sum to zero, since sin x is an odd function.)

Exploration 2 More Discontinuous

Integrands

1. The function has a removable discontinuity at x = 2.

2. The thin strip above x = 2 has zero area, so the area under the curve is the same as

∫

(x+2),

0 3

which is 10.5.

3. The graph has jump discontinuities at all integer values, but the Riemann sums tend to the area of the shaded region shown. The area is the sum of the areas of 5 rectangles (one of them with height 0):

int( )x dx= + + + + = .

∫

0 1 2 3 4 10

0 5

Quick Review 5.2

1. n n

2 1 5

2 2 2 2 2

1 2 3 4 5 55

=

∑

=( ) +( ) +( ) +( ) +( ) =

2. (3 2) [ ( )3 0 2] [ ( )3 1 2] [ ( )3 2 2]

0 4

k k

− = − + − + −

=

∑

+[ ( )3 3 − +2] [ ( )3 4 − =2] 20

3. 100 1 100 1 2 3 4 5

2 0

4

2 2 2 2 2

(j ) [( ) ( ) ( ) ( ) ( )

j

+ = + + + +

=

∑

]]

=5500

4. k

k=

∑

1 99

5. 2

0 25

k

k=

∑

6. 3 2

1 500 k k=

∑

7.2 2 3 2 3

1 50

2

1 50

x x x x

x= x= x=

∑

+

∑

=

∑

( + )

8. xK x x

k

= = =

∑

+

∑

=

∑

0 8

9 20

0 20

9. (− ) = =

∑

1 0

0

k

k n

if n is odd.

10. (− ) = =

∑

1 1

0

k

k n

if n is even.

Section 5.2

Exercises

1. lim (

n _k k

n

k

c x x dx

→∞ =

∑

=

∫

2

1

2 0

2

Δ where n is any partition of [0, 2].

2. lim ( ) ( )

n k k

k n

k

c c x x x dx

→∞ =

∑

− =

∫

− −

2

1

2 7 5

3 Δ 3 where n is any

partition of [−7 5 ., ]

3. lim

n _k _k

n

k

c x xdx

→∞ =

∑

=

∫

1 1

4

Δ where n is any partition of [1, 4].

4. lim

n _k _k

n

k

c x xdx

→∞ =

∑

− =

∫

−

1 1

1 2

3

Δ where n is any partition of

[2, 3].

5. lim

n k

k n

k

c x x dx

→∞ =

∑

4− =

∫

4−

2

1

2 0 1

Δ where n is any partition

of [0, 1].

6. lim (sin ) sin

n _k k

n

k

c x x dx

→∞ =

∑

=

∫

−

3

1

3

Δ π_π where n is any partition of

[−π π, ].

7. 5 5 1 2 15

2 1

dx −

∫

= [ − − =( )]

8.

∫

(−20) = −( 20 7 3)( − = −) 80

3 7

dx

9.

∫

(−160) = −( 160 3 0)( − = −) 480

0 3

dt

10. π θ π π

2 2 1 4

3 2

4 1

d

− −

∫

= [− − − =( )]

11. 0 5 0 5 3 4 2 1 2 75

2 1 3 4

. . [ . ( . )] .

. .

−

∫

ds= − − =

12. 2 2 18 2 4

2 18

(7)

13. Graph the region undery= +x − ≤ ≤x

2 3 for 2 4.

x dx

2 3

1

2 6 2 5 21

2 4

+ ⎛

⎝⎜ ⎞⎠⎟ = + =

−

∫

( )( )

14. Graph the region undery= − +2x 4 1≤ ≤x

2 3 2

for .

( ) ( )( )

/ /

− + = + =

∫

2 4 1

21 3 1 2

1 2 3 2

x dx

15. Graph the region undery= 9−x2 for− ≤ ≤3 x 3.

This region is half of a circle radius 3.

9 1

2 3

9 2

2 3

3 ₂

− = =

−

∫

x dx π( ) π

16. Graph the region undery= 16−x2 for− ≤ ≤4 x 0.

The region is one quarter of a circle of radius 4.

16 1

4 4

2 4 0

− = =

−

∫

x dx π(4)2 π

17.Graph the region undery= x for− ≤ ≤2 x 1.

x dx

−

∫

1₂ =1 + =

2 2 2 1 2 1 1

5 2 ( )( ) ( )( )

18. Graph the region under y= −1 x for− ≤ ≤1 x 1.

(1 ) 1( )( )

22 1 1

1 1

− = =

−

(8)

19. Graph the region undery= −2 x for− ≤ ≤1 x 1.

(2 ) 1( )( ) ( )( )

2 1 1 2 1

21 1 2 3

1 1

− = + + + =

−

∫

x dx

20.Graph the region undery= +1 1−x2 for− ≤ ≤1 x 1.

(1 1 ) ( )( )2 1 1 ( )

2 1 2 2

2 1

1 ₂

+ − = + = +

−

∫

x dx π π

21.Graph the region undery=θforπ θ≤ ≤2π

θ θ π π π π π

π π

d = − + =

∫

122 2

3 2

2 2

( )( )

22.Graph the region undery=rfor 2≤ ≤r 5 2.

r dr= − + =

∫

2 125 2 2 2 5 2 24

5 2

( )( )

23.

∫

bx dx=1 b b = b

2

1 2

2

0 ( )( )

24. 4 1

2 4 2

2 0 x dx b b b

b

= =

∫

( )( )

25. 2 1

2 2 2

2 2 s ds b a b a b a a

b

= − + = −

∫

( )( )

26. 3 1

2 3 3

3 2

2 2 t dt b a b a b a a

b

= − + = −

∫

( )( ) ( )

27. x dx a a a a a

a a

= − + =

∫

12 2 2

3 2

2 2

( )( )

28. x dx a a a a a a a

a a

= − + = − =

∫

12 3 3

1 2 3

2 2 2 3

( )( ) ( )

29. 87 87

8 11

8 11 dt= t

∫

87 11( )−87 8( )=261m esil

30. 25 25

0 60

0 60 dt= t

∫

25 60( )−25 0( )=1500gallons

31. 300 300

6 7 5 6

7 5

dt= t

∫

. .

calories 300 7 5( . )−300 6( )=450

32. 0 4 0 4

8 5 11 8 5

11

. .

. . dt= t

∫

0 4 11. ( )−0 4 8 5. ( . )=1liter

33. NINT x

x2+4 x 0 5 0 9905 ⎛

⎝⎜

⎞ ⎠⎟≈

, , , .

34.3 2+ ⋅NINT(tan 0 ≈ 3) 4.3863

x x, , ,

35. NINT(4−x2,x,−2 2. ) 10.6667≈

(9)

37. (a)The function hasa discontinuity atx=0.

(b)

x

x dx= − + =

−

∫

2 3 1

2 3

38. (a) The function has discontinuities at x= − − − − −5 4 3 2 1 0 1 2 3 4 5, , , , , , , , , , .

(b)

2 3 18 16 14

12 1

int( ) ( ) ( ) ( )

( ) (

x− dx= − + − + −

+ −

∫

−6 + −

5

0

0 8 6 4 2

0 2 88

) (+ − + − + − + −) ( ) ( ) ( ) + + = −

39. (a) The function hasa discontinuity atx= −1.

(b)

x x dx

2

3

4 1

1 1 2 4 4

1 2 3 3

7 2 −

+ = − + = −

−

∫

( )( ) ( )( )

40. (a) The function hasa discontinuity atx=3.

(b)

9 3

1 2 2 2

1 2 9 9

77 2

2

−

− = − = −

−

∫

5 x x dx 6

( )( ) ( )( )

41.False. Consider the function in the graph below.

x y

a b

42.True. All the products in the Riemann sums are positive.

43.E.

∫

( ( )f x +4)dx 2

5

= +

=

∫

+ =

∫

f x dx dx

x

( ) 4

18 4 30

2 5 2

5

2 5

44.D. (4 )

4 4

− −

∫

x dx

= + + −

= + −

∫

−

∫

−

− −

4

2 2

0 4 4 4

4 0

4 4

0 4 2

4 0 dx x dx x dx

x x x

2

==16

45.C.

46.A.

47.Observe that the graph off x( )=x3is symmetric with respect to the origin. Hence the area above and below the

x-axis is equal for− ≤ ≤1 x 1.

x dx3 x x

1 = − +

− (area below -axis) (area above -axis)

1 1

0

∫

=

48.The graph off x( )=x3+3 is three units higher than the graph ofg x( )=x3.The extra area is ( )( )3 1 =3.

(x3 )dx 0

1

3 1

4 3

13 4

+ = + =

∫

49.Observe that the region under the graph off x( )= −(x 2)3

for 2≤ ≤x 3is just the region under the graph of

g x( )=x3_{for 0}≤ ≤x ₁_{translated two units to the right.}

(x− ) dx= x dx=

∫

2

∫

1

4

3 2

3 ₃

0 1

50.Observe that the graph off x( )= x3is symmetric with respect to the y-axis and the right half is the graph of

g x( )=x3.

x dx3 x dx 1

1 ₃

0 1

2 1

2 −

∫

=

∫

=

51. Observe from the graph below that the region under the graphf x( )= −1 x3for0≤ ≤x 1cuts out a region Rfrom the square identical to the region under the graph of

g x( )=x3for 0≤x≤1.

R y

x 1

y = f(x)

1

(1 ) 1 1 1

4 3 4

3 3

0 1 0

1

− = −

_∫

= − =

(10)

52.Observe from the graph off x( )=(x−1)3_for− ≤ ≤1 x 2 that there are two regions below the x-axis and one region above the axis, each of whose area is equal to the area of the region under the graph of g x( )=x3for 0≤ ≤x 1.

(x− ) dx= ⎛−

⎝⎜ ⎞ ⎠⎟+ ⎛⎝⎜

⎞ ⎠⎟= − −

∫

1 2 1

4 1 4

1 4

3 1 2

53.Observe that the graph off x( )= ⎛x x

⎝⎜2⎞⎠⎟ ≤ ≤2

3

for 0 is just a

horizontal stretch of the graph ofg x( )=x3for 0≤ ≤x 1 by

a factor of 2. Thus the area underf x( )= ⎛x ⎝⎜2⎞⎠⎟

3

for 0≤ ≤x 2

is twice the area under the graph ofg x( )=x3for 0≤ ≤x 1.

x

dx x dx

2 2

1 2

3

0

2 ₃

0 1

⎛

⎝⎜ ⎞⎠⎟ = =

∫

54. Observe that the graph off x( )=x3_{is symmetric with}

respect to the orgin. Hence the area above and below the

x-axis is equal for − ≤ ≤8 x 8.

x dx3 x x

8 8

−

∫

= −(area below -axis) (+ area above -axiis)=0

55.Observe from the graph below that the region between the graph off x( )=x3−_{1 and the}_x_{-axis for 0}_{≤ ≤}_x ₁_{cuts out a}

region R from the square identical to the region under the graph ofg x( )=x3_{for 0}≤ ≤x ₁.

y = f(x)

R y

x

1

–1

(x3 )dx 0

1

1 1 1

4 3 4

− = − + = −

∫

56.Observe from the graph below that the region between the graph off x( )=3xand the x-axis for 0≤ ≤x 1cuts out a region Rfrom the square identical to the region under the graph ofg x( )=x3for 0≤ ≤x 1.

R y

x 1

y = f(x)

1

x dx 0

1

1 1

4 3 4

∫

= − =

57. (a) As xapproaches 0 from the right, f(x) goes to ∞.

(b)Using right endpoints we have

1 1 1

2 0

1 ₂

x dx k

n n

n

∫

= ⎛

⎝⎜ ⎞⎠⎟ ⎛

⎝ ⎜ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ⎟ ⎛ ⎝⎜

⎞ ⎠⎟ →∞

lim

kk n

n _k n

n n k

n

=

→∞ =

→∞

∑

=

= ⎛_⎝⎜ + + +

1

2 1

2 2

1 1

2

1 lim

lim ⎞⎞

⎠⎟.

Note thatn and so

n n n

n

1 1

2

1

1 1

2 2

+ + +

⎛ ⎝⎜

⎞

⎠⎟> → ∞

+

,

2 2

1

2+ + 2

⎛

⎝⎜ n ⎞⎠⎟→∞.

58. (a) Δx n x

k n k

=1, =

RRAM= ⎛

⎝⎜ ⎞⎠⎟ + ⎛⎝⎜ ⎞⎠⎟ + + ⎛⎝⎜ ⎞⎠⎟

1 2 1 2 2 1

n n n n

n n

i i

2 2

2

1

i

n

k n n k

n

= ⎛

⎝⎜ ⎞ ⎠⎟ ⎛ ⎝

⎜⎜ ⎞_⎠⎟⎟ =

∑

(b) k

n n

k

n n k

k n _⎛

⎝⎜ ⎞ ⎠⎟ ⎛ ⎝

⎜⎜ ⎞_⎠⎟⎟ = ⎛_⎝⎜ ⎞_⎠⎟ = =

∑

2

1

2

3 3

2

1 1

i

kk n

k n

=

∑

1 1

(c) 1 1 1 2 1

6

1 2 1

6

3 2

1 3

n k n

n n n n n n

n k

n

=

∑

= i ( + )( + )= ( + )(₃₃ + )

(d) lim lim (

n _k n

k n n

n n

→∞ ₌ →

∞ _⎛

⎝⎜ ⎞⎠⎟ ⎛ ⎝

⎜⎜ ⎞_⎠⎟⎟ = +

∑

2

1

i

∞ 1

1 2 1

6 3 )( n )

n

+

= + +

→∞ lim

n

n n n

n

2 3

6

3 2 3

= =2 6

(11)

58. Continued (e)Since x dx2

0 1

∫

equals the limit of any Riemann sum over the interval [0, 1]as n approaches∞, part (d) proves that

x dx2 0

1 1

3

∫

= .

Section 5.3

Definite Integrals and

Antiderivatives (pp. 285–293)

Exploration 1 How Long is the Average

Chord of a Circle?

1. The chord is twice as long as the leg of the right triangle in the first quadrant, which has length r2−x2by the Pythagorean Theorem.

2. Average value= 1

− −

∫

− −

r r r r x dx r

( ) 2 .

2 2

3. Average value= − −

∫

2 2

2 2 r r r x dx

r

=1

ri(area of semicircle of radiusr)

=1 2

2

r r iπ

= πr 2

4. Although we only computed the average length of chords perpendicular to a particular diameter, the same

computation applies to any diameter. The average length of a chord of a circle of radius r isπr

2.

5. The function y= 2 r2−x2 is continuous on [−r r, ],so the Mean Value Theorem applies and there is a cin [ , ]a b so that y(c) is the average value πr

2.

Exploration 2 Finding the Derivative of

an Integral

Pictures will vary according to the value of x chosen. (Indeed, this is the point of the exploration.) We show a typical solution here.

1. We have chosen an arbitray xbetween a and b.

2. We have shaded the region using vertical line segments.

3. The shaded region can be written as f t dt a

x

( )

∫

using the

definition of the definite integral in Section 5.2. We use tas a dummy variable because x cannot vary between a and itself.

4. The area of the shaded region is our value of F(x).

5. We have drawn one more vertical shading segment to representΔF.

(12)

7. Now the (signed) height of the newly-added vertical segment is f(x).

8. The (signed) area of the segment is ΔF=Δx f xi ( ), so

′ = =

→

F x F

x f x

( )

0

lim ( )

Δ Δ Δ

x

Quick Review 5.3

1. dy

dx=sinx

2. dy

dx=cosx

3. dy

dx

x x

=sec tan =

sec tan

4. dy

dx x

x x

=cos =

sin cot

5. dy

dx

x x x

= +

+ =

sec tan sec

2

6. dy

dx= ⎛x⎝⎜x x x

⎞

⎠⎟+ − =

1

1n 1 1n

7. dy

dx

n x

n n

= +

+ =

( 1) 1

8. dy

dx x

x x

x

= −

+ = − +

1

2 1 1 2 2

2 1 2

2 1

2 2

( ) i(n ) ( )

n

9. dy

dx xe e x x

= +

10. dy

dx= x +

1 1

2

Section 5.3

Exercises

1. (a)

∫

g x dx( ) =0

2 2

(b)

∫

g x dx( ) = −

∫

g x dx( ) = −8

1 5 5

1

(c) 3 3 3 4 12

1 2 1

2

f x dx( ) =

∫

f x dx( ) = − = −( )

∫

(d)

∫

f x dx( ) =

∫

f x dx( ) +

∫

f x dx( )

1 5 2

1 2

5

= −

∫

f x dx( ) +

∫

f x dx( )

1 5 1

2

= + =4 6 10

(e)

∫

[ ( )f x −g x dx( )] =

∫

f x dx( ) −

∫

g x dx( )

1 5 1

5 1

5

= − = −6 8 2

(f) [4 ( ) ( )] 4 ( ) ( )

1 5 1

5 1

5

f x −g x dx=

∫

f x dx−

∫

g x dx

∫

=4

∫

−

∫

1 5 1

5

f x dx( ) g x dx( ) =4 6( )− =8 16

2. (a)

∫

−2 = −2

∫

= − − =2 1 2

1 9

f x dx( ) f x dx( ) ( )

(b)

∫

[ ( )f x +h x dx( )] =

∫

f x dx( ) +

∫

h x dx( )

7 9 7

9 7

9

= + =5 4 9

(c) [2 ( ) 3 ( )] 2 ( ) 3 ( )

7 9 7

9 7

9

f x − h x dx=

∫

f x dx+

∫

h x dx

∫

=2

∫

−3

∫

7 9 7

9

f x dx( ) h x dx( ) =2 5( )−3 4( )= −2

(d)

∫

f x dx( ) = −

∫

f x dx( ) =1

1 9 9

1

(e)

∫

f x dx( ) =

∫

f x dx( ) +

∫

f x dx( )

9 7 1

9 1

7

=

∫

f x dx( ) −

∫

f x dx( )

7 9 1

9

= − − = −1 5 6

(f)

∫

[ ( )h x −f x dx( )] =

∫

h x dx( ) −

∫

f x dx( )

9 7 9

7 9

7

= −

∫

h x dx( ) +

∫

f x dx( )

7 9 7

9

= − + =4 5 1

3. (a)

∫

f u du( ) =5

1 2

(b) 3 3 5 3

1 2 1

2

f z dz( ) =

∫

f z dz( ) =

∫

(c)

∫

f t dt( ) = −

∫

f t dt( ) = −5

1 2 2

1

(d)

∫

[−f x dx( )] = −

∫

f x dx( ) = −5

1 2 1

2

4. (a) g t dt( ) = − g t dt( ) = − −

−

∫

2

3 0 0

3

(b) g u du( ) = −

∫

2

3 0

(c) [− ( )] = − ( ) = − −

−

∫

g x dx g x dx 3 0 3

0

2

(d) g r( )dr g r dr( ) 2

1

2 3 1

0 3

0

= ₋ =

−

∫

5. (a)

∫

f z dz( ) =

∫

f z dz( ) +

∫

f z dz( )

0 4 3

0 3

4

=

∫

f z dz( ) +

∫

f z dz( )

0 4 0

3

= − + =3 7 4

(b)

∫

f t dt( ) =

∫

f t dt( ) +

∫

f t dt( )

0 3 4

0 4

3

= −

∫

f t dt( ) +

∫

f t dt( )

0 3 0

4

= − + = −7 3 4

6. (a) h r dr( ) = h r dr( ) + h r dr( ) − −

∫

1

3 1

1 1

3

= − + =

− −

∫

h r dr( )

∫

h r dr( )

1 1

1 3

(13)

6. Continued

(b) − = − −

− −

∫

h u du( ) h u du( ) h u du( )

1 1 3

1 3

1

= − =

−

∫

h u du( ) h u du( ) 6

1 1 1

3

7. max sin (x2

) = sin (1) on [0, 1] sin(x2)dx sin ( )

0 1

1 1

≤ <

∫

8. max x+ =8 3andmin x+ =8 2 2on[ , ]0 1

2 2 8 3

0 1

≤

_∫

x+ dx≤

9. (b−a) min ( )f x ≥0 on [ , b]a

0≤ −(b a) min ( )f x ≤

∫

f x dx( )

a b

10. (b−a) max ( )f x ≤0 on [ , ]a b f x dx b a f x a

b

( ) ≤ −( ) max ( )≤

∫

0

11.An antiderivative of x2 1 F x 1x3 x

3

− is ( )= − .

av x dx

F F

= −

= ⎡_⎣ − ⎤_⎦

= − =

∫

1

3 1

1 3

3 0

1 3

0 0 0

2 0

3

( )

( ) ( )

( )

Findx=cin[ ,0 3]such thatc2− =1 0

c c

2 ₁

1 = = ±

Since is in1 [ ,0 3],x=1.

12.An antiderivative of −x F x = −x

2 3

2 is ( ) 6.

av= ⎛−x dx F F

⎝

⎜ ⎞_⎠⎟ = − = ⎛_⎝−

∫

1

3 2

1

3 3 0

1 3

9 2

2

0 3

[ ( )] ( )] _⎜⎜ ⎞

⎠⎟= − 3 2

Findx=cin[ , ]0 3 such that−c = − 2

3 2

2

.

c c

x 2 ₃

3

3 0 3 3

= = ±

= Since is in [ , ], .

13.An antiderivative of−3x2−1isF x( )= − −x3 x.

av x dx F F

x c

= − − = − = −

=

∫

1

1 3 1 1 0 2

0

2 0

1

( ) ( ) ( )

[ ,

Find in such that

Since

1 3 1 2

3 1

1 3 1

3

2 2

2

] − − = −

− = −

=

c c

c

c ±

1 1 3

0 1 1

3 is in [ , ],x= .

14.An antiderivative of (x−1) F x( )=1(x− ) .

3 1

2 3

is

av= x− dx= F −F = ⎛ +

⎝⎜ ⎞ ⎠⎟ 1

3 1

1

3 3 0

1 3

8 3

1 3

2 0

3

( ) [ ( ) ( )]

∫∫

=

= − =

− = ± =

1

0 3 1 1

1 1

2

Findx cin such that c c

c

[ , ] ( ) .

2

2 0

0 3 0 2

or

Since both are in or

c

x x

=

= =

.

[ , ], .

15.The region between the graph and the x-axis is a triangle of height 3 and base 6, so the area of the region

is1

2( )( )3 6 =9.

av f( )= f x dx( ) = = −

∫

1 6

9 6

3 2

4 2

.

16. The region between the graph and the x-axis is a rectangle with a half circle of radius 1 cut out. The area of the region

is 2(1)−1 = −

2 1

4 2

2

π( ) π.

av f( )= f t dt( ) = ⎛ − . ⎝⎜ ⎞⎠⎟= − −

∫

1 2

4 2

4 4

1

1 π π

17.There are equal areas above and below the x-axis.

av f( )= 1

∫

f t dt( ) = = 2

1

2 0 0

0 2

π π

π

i

18.Since tan θ is an odd function, there are equal areas above and below the x-axis.

av f( ) f d

/ / ( )

/

= 1

_∫

₋ = =

2

0 0

4 4

π π θ θ π

π

i

19.

∫

2 sinx dx= −cos(2π) cos( )+ π π

π

= −2

20. / cosx dx=sin⎛ ( )

⎝⎜ ⎞⎠⎟− =

∫

0π π2 0 1

2

sin

21.

∫

e dxx = −e1 e = −e 0

1 ₀

1 π/

22. / sec2 tan tan

0 4

4 0 1

x dx= ⎛

⎝⎜ ⎞

⎠⎟− =

∫

π π

23. 2 2 4 1 15

1 4 1

4 ₂ ₂

x dx=x = − =

∫

24. 3 2 3 2 1 9

1 2 1

2 ₃ ₃

x dx=x = − − =

− −

∫

( )

25. 5 5 5 6 5 2 40

2 6

dx= x = − − =

− −

∫

( ) ( )

26. 8 8 8 7 8 3 32

3 7

dx= x = − =

∫

( ) ( )

27. 1

1 2 1 1 2

1

1 ₁ ₁

+ = − − =

−

− −

∫

_x dx tan ( ) tan ( ) π

28. 1

1

2 0 6

2

1 0

1 2 ₁

− =

⎛

⎝⎜ ⎞⎠⎟− =

− −

(14)

29. 1 1

1 xdx e

= − =

∫

ln e ln1

30.

∫

−x dx− = = − = −

x 2 1 4 1 4 1 1 4 1 1 3 4

31. av f( )= sinx dx

−

∫

1 0 0

π π

= 1 − − − 0 = 2

π( cosπ ( cos ) π

32. av f

e e e xdx e e e e

( )= (ln ln )

−

∫

= − 1 2 1 1 2 2 =ln2 e

33. av f( )= sec x dx tan tan( )

− = ⎛ ⎝⎜ ⎞ ⎠⎟−

∫

1 4 0 4 0 2 0 4 π π π = 4 π

34. av f

x dx

( )= tan tan ( )

− + =

( )

− − −

∫

1 1 0 1

1 2 1 0

1 1

0 1

= π 4

35. av f( ) x x dx x x

( ) =

− −

∫

− + =

(

+

)

−

1

2 1 3 2

1 3

2 3 2

1 2 1

2

=4

36.av f( )= sec tanx x dx sec sec

− = ⎛ ⎝⎜ ⎞⎠⎟− ⎛ ⎝⎜ 1 3 0 3 3 0 π π π ⎞⎞ ⎠⎟

∫

0 3 π = 3 π

37. minf=1 max =

2and f 1

1 2 1 1 1 4 0 1 ≤ + ≤

∫

_x dx

38. f( . )0 5 16 17 = 1 2 16 17 1 1 1 2 1 4 0 0 5 ⎛ ⎝⎜ ⎞ ⎠⎟ ⎛ ⎝⎜ ⎞ ⎠⎟≤ + ≤ ⎛⎝⎜ ⎞ ⎠⎟

∫

. _x dx ( )

8 8 17 1 1 1 2 1 2 1 2 1 1 4 0 0 5 4 ≤ + ≤ ⎛ ⎝⎜ ⎞ ⎠⎟ ⎛ ⎝⎜ ⎞ ⎠⎟≤ +

∫

_x dx

x d . xx x dx ≤ ⎛ ⎝⎜ ⎞ ⎠⎟ ⎛ ⎝⎜ ⎞ ⎠⎟ ≤ + ≤

∫

12

16 17 1 4 1 1 8 17 0 5 1 4 0 5 . . 1 1 4 0 1 4 0 1 8 17 1 4 1 1 1 2 8 17 49 68 1 1

∫

+ ≤ + ≤ + ≤ + ≤ x dx x

dx 333

34

39.Yes, bav f dx f x dx a

b

a ( ) =

∫

( ) .

∫

This is because av(f) is a constant, so

av f dx av f x

av f b av f a b a b a b ( ) ( ) ( ) ( ) ( = ⎡⎣ ⎤⎦ = − =

∫

i i i −− = − − ⎡ ⎣⎢ ⎤ ⎦⎥ =

∫

a av f

b a

b a f x dx

f x dx a b a b ) ( ) ( ) ( ) ( ) 1

∫∫

40. (a) 300 mi

(b) 150mi 150 8 30 mph

mi

50 mph h

+ =

(c) 300mi 37 5

8 h = . mph

(d) The average speed is the total distance divided by the total time. Algebraically, d d

t t 1 2 1 2

+

+ . The driver computed 1 2 1 1 2 2 d t d t + ⎛ ⎝⎜ ⎞

⎠⎟. The two expressions are not equal.

41.Time for first release m 10 m3

= 1000 =100

3

/min min

Time for second release m m

Ave

3

= 100 =

20 3/min 50min

rrage rate total released total time

m

= =2000

150

3

miin =13 /min 1 3

3

m

42. sinx dx≤ x dx= ⎡ x

⎣⎢ ⎤ ⎦⎥ =

∫

0 1 ₂ 0 1 0 1 1 2 1 2

43. secx dx≥ ⎛ +x dx x x

⎝ ⎜ ⎞_⎠⎟ =⎡ + ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ =

∫

1 2 6 7 6 2 0 1 3 0 1 0 1 1

∫

44.Let L(x) = cx + d. Then the average value of f on [a, b] is

av f

b a cx d dx

b a cb db ca a b ( )= ( ) − + = − + ⎛ ⎝ ⎜ ⎞_⎠⎟ −

∫

1 1 2 2 2 2 2 1 2 2 + ⎛ ⎝ ⎜ ⎞_⎠⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = − − ₊ ₋ ⎡ ⎣ ⎢ da b a

c b a

d b a

( ) ( ) 2 ⎢⎢ ⎤ ⎦ ⎥ ⎥ = + + = + + + = +

c b a d

ca d cb d

L a L b

( ) ( ) ( ) ( ) ( ) 2 2 2 2

45.False. For example, sin 0 = sin π = 0, but the average value of sin x on [0, π] is greater than 0.

46. False. For example, 2 0 2 3 2 3

3 3

x dx but −

(15)

47.A. There is no rule for the multiplication of functions.

48.D. There is no rule for the negation of the bounds.

49.B. av f( )= cosxdx (sin sin )

−

∫

= −

1 5 1

1

4 5 1

1 5

⫽ – 0.450.

50.C. 10= 1

−

∫

b a aF x dx b

( )

10(b a) f x dx( )

a b

− =

_∫

51. (a) Area=1 2bh

(b) h

bx C

2

2₊

(c) y x dx h bx

hb b bh

b b

( )

0

2

0 2

2 2

1 2

∫

= ⎡

⎣⎢ ⎤

⎦⎥ = =

52.av x

k x dx k k x

k k k

k k k

k

k k

( )

(

= =

+ ⎡ ⎣⎢

⎤ ⎦⎥ =

+ +

∫

1 1 1

1

0 0

1

++1)

Graphy x

x x y x

x

1 1

2

1 =

+ =

+

( ) and on a graphing calculator and find the point of intersection for x> 1.

Thus,k≈2 39838.

53. An antiderivative of F′ (x) is F(x) and an antiderivative of

G′(x) is G(x).

F x dx F b F a

G x dx G b G a a

b

a b

( ) ( ) ( )

∫

= −

′ = −

Sincee ′ = ′ ′ = ′ so

−

∫

F x G x F x dx G x dx

F b

a b

( ) ( ), ( ) ( ) ,

( ) FF a( )=G b( )−G a( ).

Quick Quiz Sections 5.1–5.3

1. D. ( ( )F x )dx a b dx a

b

a b

+ = + +

_∫

∫

3 2 3

a+2b+3b−3a=5b−2a

2. B.

3. C. x dx2 x 3

2

2 ₃ ₃

2 2

3 2

3 0

= = − =

∫

.

4. (a) f′′( )x =6x+12

′′ = + +

= −

= ′ =

+ +

∫

f x dx x x c

y x m f

( )

( ) ( )

3 12

4 5

4

3 0 12 0

2

2 _cc

c

= =

4 4

′ = + +

= +

∫

+ +

∫

f x dx x x dx f x x x x c f

( ) ( )

3 12 4

6 4

0

2

3 2

== + + + = −

= −

= + + −

( ) ( ) ( )

( )

0 6 0 4 0 5

5

6 4 5

3 2

c c

f x x x x

(b) av f( ) x x x dx

( ) ( )

=

− −

∫

− + + −

1

1 1 6 4 5

3 2 1 1

1

2 4 2 2 5 3

4

3 2

1 1 x

x x x

+ + −

⎛ ⎝

⎜ ⎞_⎠⎟ = −

−

Section 5.4

Fundamental Theorem of

Calculus (pp. 294–305)

Exploration 1 Graphing NINT

f

2. The function y⫽ tan x has vertical asymptotes at all odd multiples of π

2. There are six of these between –10 and 10.

3. In attempting to find F( – 10) ⫽ tan( )t 3

10

−

∫

dt + 5, the calculator must find a limit of Riemann sums for the integral, using values of tan t for t between – 10 and 3. The large positive and negative values of tan t found near the asymptotes cause the sums to fluctuate erratically so that no limit is approached. (We will see in Section 8.3 that the “areas” near the asymptotes are infinite, although NINT is not designed to determine this.)

4.y⫽ tan x

5. The domain of this continuous function is the open interval π π

2 3

2

, .

⎛ ⎝⎜ ⎞⎠⎟

6. The domain of F is the same as the domain of the continuous function in step 4, namely π π

2 3

2

, .

(16)

7. We need to choose a closed window narrower than π π

2 3

2 , ⎛ ⎝⎜

⎞

⎠⎟ to avoid the asymptotes.

8. The graph of F looks the graph in step 7. It would be decreasing on π π

2, ⎛

⎝⎜ ⎤_⎦⎥ and increasing on π π, 3

2 ⎡ ⎣⎢

⎞ ⎠⎟, with

vertical asymptotes at x⫽ π

2and x⫽ 3

2 π_.

Exploration 2 The Effect of Changing

a

in

f (t) dt

a x

∫

1.

2.

3. Since NINT (x2

, x, 0, 0) ⫽ 0, the x-intercept is 0.

4. Since NINT (x2_{, x,}_{5, 5)}_⫽_{0, the}_x_{-intercept is 5.}

5. Changing a has no effect on the graph ofy d

dx a f t dt x

=

_∫

( ) .

It will always be the same as the graph of y⫽f(x).

6. Changing a shifts the graph of y⫽ f t dt a

x

( )

∫

vertically in such a way that a is always the x-intercept. If we change from a1 to a2, the distance of the vertical shift is f t dt

a a

( ) .

2 1

∫

Quick Review 5.4

1. dy

dx=cos(x ) x= xcos(x ) 2 _i₂ ₂ 2

2. dy

dx=2(sin )(cos )x x =2sin cosx x

3. dy

dx= x x x − x x

=

2 2

2

(sec )(sec tan ) (tan )(sec ) sec22xtanx−2 tanxsec2x=0

4. dy

dx= x− x=

3 3

7

7 0

5. dy

dx x

=2 ln2

6. dy

dx= x = x

− 1 2

1 2

1 2/

7. dy

dx

x x x

x

x x x

x

=( sin )( ) (cos )( )− −₂ 1 = − sin +₂cos

8. dy

dt t dy dt t

=cos , = −sin

dy dx

dy dt dx dt

t

t t

= =

− = −

/ /

cos

sin cot

9. Implicitly differentiate:

xdy

dx y y

dy dx dy

dx x y y dy

dx y

+ + =

− = − +

= − ( )

( ) ( )

1 1 2

2 1

++

− =

+ − 1 2

1 2

x y y

y x

10. dy

dx=dx dy= x

1 1

3 /

Section 5.4 Exercises

1. dy

dx d

dx t dt x

x

=

_∫

( )

sin2 =

0 sin

2

2. dy

dx d

dx t t dt x x x

=

_∫

(

3+ 2

)

=3 + 2

2 cos cos

3. dy

dx d

dx t t dt x x x

=

_∫

( )

3− 5 =

( )

−

0

3 5

4. dy

dx d

dx dt

t

x _x

=

_∫

₋ 1+ = 1+

2 e

5 _e5

5. dy

dx d

dx u dt x

x

=

_∫

( )

tan3 =tan

0

3

6. dy

dx d

dx e u du e x

u x

x

=

_∫

sec = sec

4

7. dy

dx d dx

t

t dt x

x x

= +

+ = ++

∫

7₁1 2 ₁1 2

8. dy

dx d dx

t tdt

x x x

= −

+ = −+

−

∫

32

2 3 sin cos

sin cos

9. dy

dx d

dx dt du dx x t

x _x _x

=

_∫

e3 =e = e

0

2 ₂ ₂

2

10. dy

dx d

dx t dt x

du

dx x x x

=

_∫

cot3 =cot 2 =2 3

6 2

2

cot

11. dy

dx d dx

u u du

x x x

(17)

12. dy dx d dx u udu x = + + = + − + 1 1 1 1 2 2 2 2 sin cos

sin ( )

cos ( − −

∫

x _x

)

13. dy

dx d

dx t dt

d

dx t dt

x

=

_∫

6ln

( )

1+ 2 = −

_∫

ln

( )

1+ 2

6

=ln 1

( )

+x2

14. dy

dx d

dx t t dt d

dx t t dt x

x

=

_∫

24+ +1 = −

_∫

24+ +1

7 7

= − 2x4+ +x 1

15. dy dx d dx t t dt d dx t t dt x x = − = − − =

∫

3 cos2

∫

cos c

5 2 5 3 2 2 o osx x du dx 3

6₋₂

= − + 3 2 2 3 6 x x x cos 16. dy dx d dx t t t dt d dx t t t d x = − + + = − − + +

∫

5 2 2 3

25 3 2 9 6 2 9 6 tt x 25 5 2

∫

= − + + = − +

25 10 9

125 6

250 100 90

4 2 6 5 3 x x x du dx

x x xx

x

125 6+6

17. dy

dx d

dx r dr d

dx r dr

x

=

_∫

0 sin

( )

2 = −

_∫

sin

( )

2

0

= −sinxdu= −sin dx x x 2 18. dy dx d

dx p dp

d

dx p

x

=

_∫

ln

(

+

)

= −

_∫

ln

(

+

)

3 2

10 ₂

10

3 2 ₂

2 2 ddp

= −ln

(

2+p2

)

du= −6 ln

(

2 9+ 4

)

dx x x

19. dy

dx d

dx t dt x

du dx x du d x x

=

_∫

₂cos = +cos

3

3 2

2 cos2 2

xx

=3x2cos2x3+2xcos2x2

20. dy

dx d

dx t dt x du dx x du dx x x

=

_∫

2 = 2 − 2

sin cos

cos sin

= −sinx cos2x−cosxsin2x

21. y=

∫

xsin3t dt 5

22. y=

∫

xe−ttant dt 8 23. _Es n Esn 10 4 10 = −

24. y= x − t dt+

−

∫

3 4

3 cos

25. y=

∫

xcos2 t dt−

7 5 2

26. y=

∫

xe dtt +1

0

27. 2 1 2

1 2 3 1 2 3 − ⎛ ⎝⎜ ⎞⎠⎟ =⎡⎣ − ⎤⎦

∫

/ x dx x Inx /

= − − −⎛ ⎝⎜ ⎞ ⎠⎟ = − + = − − =

( ln ln

ln ln

6 1 1

2

5 3 1

2

5 3 2

3)

5

5−ln6≈3 208.

28. 3 1

3 3

2 1

x_dx x

− −

∫

= ⎛ ⎝⎜ ⎞ ⎠⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ ln =⎛ ⎝⎜ ⎞ ⎠⎟ − ⎛ ⎝⎜ ⎞ ⎠⎟ = − ≈ − 1 3 1 3 9 26

3 3 7 889

ln

ln .

29. (x2 x dx) x3 x3 2/ 0 1 0 1 1 3 2 3 1 3 2 3 + =⎡ + ⎣⎢ ⎤ ⎦⎥ = + ⎛ ⎝⎜ ⎞⎠⎟

∫

−− + =(0 0) 1

30. x3 2dx x5 2 0 5 0 5 2 5 2

5 25 5 0 10 5 22 3

/ / ( ) . = ⎡ ⎣⎢ ⎤ ⎦⎥ = − = ≈

∫

661

31. x− dx= −⎡ x−

⎣ ⎤⎦ = − ⎛_⎝⎜ − ⎞_⎠⎟=

∫

6 5 1 5

1 32 1

32

5 5 1

2 1

5 2

/ /

32. 2 2 2 2 1 1

2 2 2 1 ₂ 2 1 ₁ 2 1

x dx x dx x

− − ₋ − − ₋ − −

∫

=

∫

= ⎡_⎣− ⎤_{⎦ =} ⎡ − ⎣⎣⎢ ⎤ ⎦⎥=1

33.

∫

sinx dx= −⎡⎣ cosx⎤⎦ = − − =( )

0 0 1 1 2

π π

34. (1 cos ) sin ₀

0 + =⎡⎣ + ⎤⎦

∫

π x dx x x π

= + − + = ≈ ( ) ( ) . π π

0 0 0

3 142

35. 2 2 2

0 3 0

3

sec tan /

/

θ θ θ π

π

d = ⎡⎣ ⎤⎦

∫

= −

= ≈

2 3 0

2 3 3 464

( )

.

36. csc cot _//

/ / ₂ 6 5 6 6 5 6

θ θ θ _ππ

π π

d = −⎡⎣ ⎤⎦

∫

= − −

= ≈

3 3

2 3 3 464

( )

.

37. csc cot csc ( ) ( )

/ /

/ x x dx= −⎡⎣ x⎤⎦π = − − − =

π

π 4

3 4 4

3

2 2 0

ππ/4

∫

38. 4 4 ₀3 4 2 1 4

0 3

sec tan sec / ( )

/

x x dx= ⎡⎣ x⎤⎦ = − =

∫

π π

39.

( )

r+ dr=⎡

( )

r+

⎣⎢

⎤

⎦⎥− = − = −

∫

1 1