Calculus one and several variables 10E Salas solutions manual ch05

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU

JWDD027-05 JWDD027-Salas-v1 November 25, 2006 15:58

SECTION 5.2 229

CHAPTER 5

SECTION 5.2

1. Lf (P ) = 0( 14 ) + 1

2 ( 14 ) + 1( 1

2 ) = 58 , Uf (P ) = 1

2 ( 14 ) + 1( 1

4 ) + 2( 12 ) = 11

8

2. Lf (P ) = 23 ( 1

3 ) + 14 ( 5

12 ) + 0( 14 ) + (−1)(1) = − 97

144 ,

Uf (P ) = 1( 13 ) + 2

3 ( 512 ) + 1

4 ( 14 ) + 0(1) = 97

144

3. Lf (P ) = 14 ( 1

2 ) + 116 ( 1

4 ) + 0( 14 ) = 9

64 , Uf (P ) = 1( 12 ) + 1

4 ( 14 ) + 1

16 ( 14 ) = 37

64

4. Lf (P ) = 1516 ( 1

4 ) + 34 ( 1

4 ) + 0( 12 ) = 27

64 , Uf (P ) = 1( 14 ) + 15

16 ( 14 ) + 3

4 ( 12 ) = 55

64

5. Lf (P ) = 1( 12 ) + 9

8 ( 12 ) = 17

16 , Uf (P ) = 98 ( 1

2 ) + 2( 12 ) = 25

16

6. Lf (P ) = 0( 125 ) + 1

5 ( 325 ) + 2

5 ( 525 ) + 3

5 ( 725 ) + 4

5 ( 925 ) = 14

25 ,

Uf (P ) = 15 ( 1

25 ) + 25 ( 3

25 ) + 35 ( 5

25 ) + 45 ( 7

25 ) + 1( 925 ) = 19

25

7. Lf (P ) = 116 ( 3

4 ) + 0( 12 ) + 1

16 ( 14 ) + 1

4 ( 12 ) = 3

16 , Uf (P ) = 1( 34 ) + 1

16 ( 12 ) + 1

4 ( 14 ) + 1( 1

2 ) = 4332

8. Lf (P ) = 916 ( 1

4 ) + 116 ( 1

2 ) + 0( 12 ) + 1

16 ( 14 ) + 1

4 ( 12 ) = 5

16 ,

Uf (P ) = 1( 14 ) + 9

16 ( 12 ) + 1

16 ( 12 ) + 1

4 ( 14 ) + 1( 1

2 ) = 98

9. Lf (P ) = 0(π6

)+ 1

2

(π3

)+ 0(π2

)= π

6 , Uf (P ) = 12

(π6

)+ 1(π3

)+ 1(π2

)= 11π

12

10. Lf (P ) = 12 (π3 ) + 0(π6 ) + (−1)(π2 ) = −π

3 , Uf (P ) = 1(π3 ) + 12 (π6 ) + 0(π2 ) = 5π

12

11. (a) Lf (P ) ≤ Uf (P ) but 3 �≤ 2.

(b) Lf (P ) ≤∫ 1

−1

f(x) dx ≤ Uf (P ) but 3 �≤ 2 ≤ 6.

(c) Lf (P ) ≤∫ 1

−1

f(x) dx ≤ Uf (P ) but 3 ≤ 10 �≤ 6.

12. (a) Lf (P ) = (x0 + 3)(x1 − x0) + (x1 + 3)(x2 − x1) + · · · + (xn−1 + 3)(xn − xn−1),

Uf (P ) = (x1 + 3)(x1 − x0) + (x2 + 3)(x2 − x1) + · · · + (xn + 3)(xn − xn−1)

(b) For each index i

xi−1 + 3 ≤ 12(xi−1 + xi) + 3 ≤ xi + 3

Multiplying by Δxi = xi − xi−1 gives

(xi−1 + 3)Δxi ≤12(x2i − x2

i−1

)+ 3(xi − xi−1) ≤ (xi + 3)Δxi.

Summing from i = 1 to i = n, we find that

Lf (P ) ≤ 12(x2

1 − x20

)+ 3(x1 − x0) · · · +

12(x2n − xn−1

2)

+ 3(xn − xn−1

)≤ Uf (P )



230 SECTION 5.2

The middle sum collapses to12(xn

2 − x02)

+ 3(xn − x0) =12(b2 − a2) + 3(b− a)

Thus ∫ b

a

(x + 3)dx =12(b2 − a2) + 3(b− a)

13. (a) Lf (P ) = −3x1(x1 − x0) − 3x2(x2 − x1) − · · · − 3xn(xn − xn−1),

Uf (P ) = −3x0(x1 − x0) − 3x1(x2 − x1) − · · · − 3xn−1(xn − xn−1)


−3xi ≤ − 32

(xi + xi−1

)≤ −3xi−1.


−3xi Δxi ≤ − 32

(xi

2 − x2i−1

)≤ −3xi−1 Δxi.

Summing from i = 1 to i = n, we find that

Lf (P ) ≤ − 32

(x1

2 − x02)− · · · − 3

2

(xn

2 − x2n−1

)≤ Uf (P ).

The middle sum collapses to

− 32

(xn

2 − x02)

= − 32 (b2 − a2).

Thus

Lf (P ) ≤ −32(b2 − a2) ≤ Uf (P ) so that

∫ b

a

−3x dx = −32(b2 − a2).

14. (a) Lf (P ) = (1 + 2x0)(x1 − x0) + (1 + 2x1)(x2 − x1) + · · · + (1 + 2xn−1)(xn − xn−1),

Uf (P ) = (1 + 2x1)(x1 − x0) + (1 + 2x2)(x2 − x1) + · · · + (1 + 2xn)(xn − xn−1)


1 + 2xi−1 ≤ 1 + (xi−1 + xi) ≤ 1 + 2xi


(1 + 2xi−1) Δxi ≤ (xi − xi−1) +(xi

2 − x2i−1

)≤ (1 + 2xi) Δxi.

Proceeding as before, we get ∫ b

a

(1 + 2x) dx = (b− a) + (b2 − a2)

15.∫ 2

−1

(x2 + 2x− 3) dx 16.∫ 3

0

(x3 − 3x) dx

17.∫ 2π

0

t2 sin(2t + 1) dt 18.∫ 4

1

√t

t2 + 1dt



SECTION 5.2 231

19. 20.

21. Δx1 = Δx2 = 18 , Δx3 = Δx4 = Δx5 = 1

4

m1 = 0, m2 = 14 , m3 = 1

2 , m4 = 1, m5 = 32

f(x∗1) = 1

8 , f(x∗2) = 3

8 , f(x∗3) = 3

4 , f(x∗4) = 5

4 , f(x∗5) = 3

2

M1 = 14 , M2 = 1

2 , M3 = 1, M4 = 32 , M5 = 2

(a) Lf (P ) = 2532 (b) S∗(P ) = 15

16 (c) Uf (P ) = 3932

22.∫ 1

0

2x dx = 1.

23. Lf (P ) = x03(x1 − x0) + x1

3(x2 − x1) + · · · + x3n−1(xn − xn−1)

Uf (P ) = x13(x1 − x0) + x2

3(x2 − x1) + · · · + xn3(xn − xn−1)

For each index i

x3i−1 ≤ 1

4

(xi

3 + xi2xi−1 + xix

2i−1 + x3

i−1

)≤ xi

3

and thus by the hint

x3i−1(xi − xi−1) ≤ 1

4

(xi

4 − x4i−1

)≤ xi

3(xi − xi−1).

Adding up these inequalities, we find that

Lf (P ) ≤ 14

(xn

4 − x04)≤ Uf (P ).

Since xn = 1 and x0 = 0, the middle term is14:∫ 1

0

x3 dx =14.

24. (a) Lf (P ) = x04(x1 − x0) + x1

4(x2 − x1) + · · · + xn−14(xn − xn−1),

Uf (P ) = x14(x1 − x0) + x2

4(x2 − x1) + · · · + xn4(xn − xn−1)


xi−14 ≤ xi

4 + xi3xi−1 + xi

2xi−12 + xixi−1

3 + xi−14

5≤ xi

4


xi−14 Δxi ≤

15(xi

5 − xi−15)≤ xi

4 Δxi.



232 SECTION 5.2

Summing and collapsing the middle sum gives

Lf (P ) ≤ 15(xn

5 − x05)≤ Uf (P ),

Thus ∫ 1

0

x4 dx =15(15 − 05) =

15.

25. Necessarily holds: Lg(P ) ≤∫ b

a g(x) dx <∫ b

a f(x) dx ≤ Uf (P ).

26. Need not hold. Consider the partition {0, 2, 3} on [0, 3] where f(x) = x and g(x) = 1.

Then∫ b

a f(x) dx = 412

and∫ b

a g(x) dx = 3,but Lg(P ) = 3 and Lf (P ) = 2.

27. Necessarily holds: Lg(P ) ≤∫ b

a g(x) dx <∫ b

a f(x) dx

28. Need not hold. Consider the partition {0, 1, 3} on [0, 3] where f(x) = 2 and g(x) = 3 − x.

Then∫ b

a f(x) dx = 6 and∫ b

a g(x) dx = 412, but Ug(P ) = 7 and Uf (P ) = 6.

29. Necessarily holds: Uf (P ) ≥∫ b

a f(x) dx >∫ b

a g(x) dx

30. Need not hold. Use the same counter example as Exercise 30.

31. Let P = {x0, x1, x2, . . . , xn} be a regular partition of [a, b] and let Δx = (b− a)/n.

Since f is increasing on [a, b],

Lf (P ) = f(x0)Δx + f(x1)Δx + · · · + f(xn−1)Δx

and

Uf (P ) = f(x1)Δx + f(x2)Δx + · · · + f(xn)Δx.

Now,

Uf (P ) − Lf (P ) = [f(xn) − f(x0)]Δx = [f(b) − f(a)]Δx.

32. Proceed as in Exercise 31.

33. (a) f ′(x) =x√

1 + x2> 0 for x ∈ [0, 2]. Thus, f is increasing on [0, 2].

(b) Let P = {x0, x1, . . . , xn} be a regular partition of [0, 2] and let Δx = 2/n

By Exercise 30,∫ 2

0

f(x) dx− Lf (P ) ≤ |f(2) − f(0)| 2n

=2(√

5 − 1)n

∼= 2.47n

It now follows that∫ 2

0 f(x) dx− Lf (P ) < 0.1 if n > 25.

(c)∫ 2

0

f(x) dx ∼= 2.96

34. (a) f ′(x) =−2x

(1 + x2)2< 0 on (0, 1) ⇒ f is decreasing.



SECTION 5.2 233

(b) Uf (P ) −∫ 1

0 f(x) dx ≤ |f(1) − f(0)|Δx = | 12 − 1| 1n =12n

.

so need12n

= 0.05, or n = 10.

(c) Using Uf (P ) with n = 10, we have∫ 1

0

11 + x2

dx ∼= 0.78

35. Let S be the set of positive integers for which the statement is true. Since1(2)2

= 1, 1 ∈ S. Assume

that k ∈ S. Then

1 + 2 + · · · + k + k + 1 = (1 + 2 + · · · + k) + k + 1 =k(k + 1)

2+ k + 1

=(k + 1)(k + 2)

2Thus, k + 1 ∈ S and so S is the set of positive integers.

36. See Exercise 5 in section 1.8.

37. Let f(x) = x and let P = {x0, x1, x2, . . . , xn} be a regular partition of [0, b]. Then Δx = b/n and

xi =ib

n, i = 0, 1, 2, . . . , n.

(a) Since f is increasing on [0, b],

Lf (P ) =[f(0) + f

(b

n

)+ f

(2bn

)+ · · · + f

((n− 1)b

n

)]b

n

=[0 +

b

n+

2bn

+ · · · + (n− 1)bn

]b

n

=b2

n2[1 + 2 + · · · + (n− 1)]

(b) Uf (P ) =[f

(b

n

)+ f

(2bn

)+ · · · + f

((n− 1)b

n

)+ f(b)

]b

n

=[b

n+

2bn

+ · · · + (n− 1)bn

+ b

]b

n

=b2

n2[1 + 2 + · · · + (n− 1) + n]

(c) By Exercise 35,

Lf (P ) =b2

n2· (n− 1)n

2=

12b2(n2 − n

n2

)=

12b2(

1 − 1n

)=

12b2(1 − ||P ||)

Lf (P ) =b2

n2· n(n + 1)

2=

12b2(n2 + n

n2

)=

12b2(

1 +1n

)=

12b2(1 + ||P ||)

(d) For any partition P,Lf (P ) ≤ §∗(P ) ≤ Uf (P ). Since

lim||P ||→0

Lf (P ) = lim||P ||→0

Uf (P ) =12b2,

lim||P ||→0

S∗(P ) =12b2 by the pinching theorem.



234 SECTION 5.2

38. Let f(x) = x2 and let P = {x0, x1, x2, . . . , xn} be a regular partition of [0, b]. Then Δx = b/n and

xi =ib

n, i = 0, 1, 2, . . . , n.

(a) Since f is increasing on [0, b],

Lf (P ) =[f(0) + f

(b

n

)+ f

(2bn

)+ · · · + f

((n− 1)b

n

)]b

n

=[0 +

b2

n2+

4b2

n2+ · · · + (n− 1)2b2

n

]b

n

=b3

n3[1 + 22 + · · · + (n− 1)2]

(b) Uf (P ) =[f

(b

n

)+ f

(2bn

)+ · · · + f

((n− 1)b

n

)+ f(b)

]b

n

=[b2

n2+

4b2

n2+ · · · + n2b2

n2

]b

n

=b3

n3[1 + 22 + · · · + n2]

(c) By Exercise 36,

Lf (P ) =b3

n3· (n− 1)n(2n− 1)

6= b3

(2n3 − 3n2 + n

6n3

)=

16b3 = (2 − 3||P || + ||P ||2)

Uf (P ) =b3

n3· n(n + 1)(2n− 1)

6= b3

(2n3 + 3n2 + n

6n3

)=

16b3 = (2 + 3||P || + ||P ||2)

(d) For any partition P,Lf (P ) ≤ §∗(P ) ≤ Uf (P ). Since

lim||P ||→0

Lf (P ) = lim||P ||→0

Uf (P ) =13b3,

lim||P ||→0

S∗(P ) =13b3 by the pinching theorem.

39. Choose each x∗i so that f(x∗

i ) = mi. Then S∗i (P ) = Lf (P ).

Similarly, choosing each x∗i so that f(x∗

i ) = Mi gives S∗i (P ) = Uf (P ).

Also, choosing each x∗i so that f(x∗

i ) =12(mi + Mi) (they exist by the intermediate value theorem)

gives

S∗i (P ) =

12(m1 + M1)Δx1 + · · · + 1

2(mn + Mn)Δxn

=12[m1Δx1 + · · · + mnΔxn + M1Δx1 + · · · + MnΔxn]

=12[Lf (P ) + Uf (P )].

40. (a) Lf (P ) =18125

∼= 7.24, Uf (P ) =22125

∼= 8.84

(b)12[Lf (P ) + Uf (P )] =

40250

∼= 8.04 (c) S∗(P ) ∼= 7.98



SECTION 5.3 235

41. (a) Lf (P ) ∼= 0.6105, Uf (P ) ∼= 0.7105

(b)12[Lf (P ) + Uf (P )] ∼= 0.6605 (c) S∗(P ) ∼= 0.6684

42. (a) Lf (P ) ∼= 1.0224, Uf (P ) ∼= 1.1824

(b)12[Lf (P ) + Uf (P )] ∼= 1.1024 (c) S∗(P ) ∼= 1.1074

43. (a) Lf (P ) ∼= 0.53138, Uf (P ) ∼= 0.73138

(b)12[Lf (P ) + Uf (P )] ∼= 0.63138 (c) S∗(P ) ∼= 0.63926

SECTION 5.3

1. (a)∫ 5

0

f(x) dx =∫ 2

0

f(x) dx +∫ 5

2

f(x) dx = 4 + 1 = 5

(b)∫ 2

1

f(x) dx =∫ 2

0

f(x) dx−∫ 1

0

f(x) dx = 4 − 6 = −2

(c)∫ 5

1

f(x) dx =∫ 5

0

f(x) dx−∫ 1

0

f(x) dx = 5 − 6 = −1

(d) 0 (e)∫ 0

2

f(x) dx = −∫ 2

0

f(x) dx = −4

(f)∫ 1

5

f(x) dx = −∫ 5

1

f(x) dx = 1

2. (a)∫ 8

4

f(x) dx =∫ 8

1

f(x) dx−∫ 4

1

f(x) dx = 11 − 5 = 6

(b)∫ 3

4

f(x) dx = −∫ 4

3

f(x) dx = −7

(c)∫ 3

1

f(x) dx =∫ 4

1

f(x) dx−∫ 4

3

f(x) dx = 5 − 7 = −2

(d)∫ 8

3

f(x) dx =∫ 8

1

f(x) dx−∫ 3

1

f(x) dx = 11 − (−2) = 13

(e)∫ 4

8

f(x) dx = −∫ 8

4

f(x) dx = −6

(f)∫ 4

4

f(x) dx = 0

3. With P ={

1,32, 2}

and f(x) =1x

, we have

0.5 ≤ 712

= Lf (P ) ≤∫ 2

1

dx

x≤ Uf (P ) =

56< 1.



236 SECTION 5.3

4. Using P = {0, 12 , 1}, we have 0.6 < 0.65 = Lf (P ) ≤

∫ 1

0

11 + x2

dx ≤ Uf (P ) = 0.9 < 1.

5. (a) F (0) = 0 (b) F ′(x) = x√x + 1 (c) F ′(2) = 2

√3

(d) F (2) =∫ 2

0

t√t + 1 dt (e) −F (x) =

∫ 0

x

t√t + 1 dt

6. (a) F (π) =∫ π

π

t sin t dt = 0 (b) By Theorem 5.3.5, F ′(x) = x sinx.

(c) F ′(π2 ) = π2 sin π

2 = π2 (d) F (2π) =

∫ 2π

π

t sin t dt

(e) −F (x) =∫ π

x

t sin t dt.

7. F ′(x) =1

x2 + 9; (a)

110

(b)19

(c)437

(d)−2x

(x2 + 9)2

8. F ′(x) = −√x2 + 1 (a) −

√2 (b) −1 (c) − 1

2

√5 (d)

−x√x2 + 1

9. F ′(x) = −x√x2 + 1; (a)

√2 (b) 0 (c) − 1

4

√5 (d) −

(√x2 + 1 +

x2

√x2 + 1

)

10. F ′(x) = sinπx (a) 0 (b) 0 (c) 1 (d) π cosπx

11. F ′(x) = cosπx; (a) −1 (b) 1 (c) 0 (d) −π sinπx

12. F ′(x) = (x + 1)3 (a) 0 (b) 1 (c)278

(d) 3(x + 1)2

13. (a) Since P1 ⊆ P2, Uf (P2) ≤ Uf (P1) but 5 �≤ 4.

(b) Since P1 ⊆ P2, Lf (P1) ≤ Lf (P2) but 5 �≤ 4.

14. (a) constant functions. (b) constant functions.

15. constant functions

16. We know this is true for a < c < b. Assume a < b. If c = a or c = b, the equality becomes∫ b

a

f(x) dt =∫ b

a

f(x) dt, trivially true. If c < a, we get

∫ c

a

f(t) dt +∫ b

c

f(t) dt = −∫ a

c

f(t) dt +∫ b

c

f(t) dt =∫ b

a

f(t) dt, as desired

The other possible cases are proved in a similar manner.

17. F ′(x) =x− 11 + x2

= 0 =⇒ x = 1 is a critical number.

F ′′(x) =(1 + x2) − 2x(x− 1)

(1 + x2)2, soF ′′(1) =

12> 0 means x = 1 is a local minimum.



SECTION 5.3 237

18.F ′(x) =

x− 41 + x2

= 0 =⇒ x = 4 is a critical number.

F ′′(x) =(1 + x2) − 2x(x− 4)

(1 + x2)2, so F ′′(4) =

117

> 0 means x = 4

is a local minimum.

19. (a) F ′(x) =1x> 0 for x > 0.

Thus, F is increasing on (0,∞);

there are no critical numbers.

(b) F ′′(x) = − 1x2

< 0 for x > 0.

The graph of F is concave down on (0,∞);

there are no points of inflection.

(c)

20. (a) F ′(x) = x(x− 3)2, (b) F ′′(x) = (x− 3)2 + 2x(x− 3) = 3(x− 3)(x− 1).

F is increasing on [0,∞); The graph of F is concave up on (−∞, 1) ∪ (3,∞);

F is decreasing on (−∞, 0]; The graph of F is concave down on (1, 3);

critical numbers 0,−3. Inflection points at x = 1, x = 3.

(c)

− 1 1 2 3x

2

4

6

y

21. (a) F is differentiable, therefore continuous (b) F ′(x) = f(x) f is differentiable;F ′′(x) = f ′(x)

(c) F ′(1) = f(1) = 0 (d) F ′′(1) = f ′(1) > 0

(e) f(1) = 0 and f increasing (f ′ > 0) implies f < 0 on (0, 1) and f > 0 on (1,∞).

Since F ′ = f, F is decreasing on (0, 1) and increasing on (1,∞);

F (0) = 0 implies F (1) < 0.

22. (a) G is differentiable, therefore continuous (b) G′(x) = g(x) and g is differentiable; G′′(x) = g′(x)

(c) G′(1) = g(1) = 0 (d) G′′(x) = g′(x) < 0 for x < 1

G′′(x) = g′(x) > 0 for x > 1(e) G′(x) = g(x) > 0 for all x �= 0.



238 SECTION 5.3

23. (a)

− 1 1 2 3x

2

4

y (b)

− 1 1 2 3x

2

4

6

y

F (x) =

⎧⎨⎩2x− 1

2x2 + 5

2 −1 ≤ x ≤ 0

2x + 12x

2 + 52 0 < x ≤ 3

(c) f is discontinuous at x = 0, but not differentiable; F is continuous but differentiable at x = 0.

24. (a) (b)

F (x) =

⎧⎪⎪⎨⎪⎪⎩

x3

3+

x2

20 ≤ x ≤ 1

x2 − 16

1 < x ≤ 3

(c) f is continuous at x = 1, but not differentiable. F is continuous and differentiable at x = 1.

25. Let u = x3. Then F (u) =∫ u

1

t cos t dt and

dF

dx=

dF

du

du

dx= u cosu (3x2) = 3x5 cosx3.

26. Let u = cosx.dF

dx=

dF

du

du

dx=√

1 − u2 (−sinx) =√

1 − cos2 x (−sinx) = −| sinx| sinx

27. F (x) =∫ 1

x2(t− sin2 t) dt = −

∫ x2

1

(t− sin2 t) dt. Let u = x2. Then

dF

dx=

dF

du

du

dx= −(u− sin2 u)(2x) = 2x

[sin2(x2) − x2

].

28. Let u =√x.

dF

dx=

dF

du

du

dx=

u2

1 + u4

12√x

=x

1 + x2

12√x

29. (a) F (0) = 0

(b) F ′(0) = 2 +sin 2(0)1 + 02

= 2

(c) F ′′(0) =(1 + 0)22 cos 2(0) − sin 2(0)(2)(0)

(1 + 0)2= 2



SECTION 5.3 239

30. (a) F (0) = 0

(b) Let u = x2. Then f(u) = 2√u +

∫ u

0

sin 2t1 + t2

dt.

dF

dx=

dF

du

du

dx= 2 +

sin 2u1 + u2

(2x) = 2 +sin 2x2

1 + x4(2x)

31. f(x) =d

dx

(2x

4 + x2

)=

8 − 2x2

(4 + x2)2

(a) f(0) =12

(b) f(x) = 0 at x = −2, 2

32. (a) F (x) =∫ x

0

tf(t)dt = sinx− x cosx.

F ′(x) = xf(x) = cosx− cosx + x sinx = x sinx =⇒ f(π2 ) = sin π2 = 1

(b) f ′(x) = cosx

33. By the hintF (b) − F (a)

b− a= F ′(c) for some c in (a, b). The result follows by observing that

F (b) =∫ b

a

f(t) dt , F (a) = 0 , and F ′(c) = f(c).

33. Set G(x) =∫ x

a

f(t) dt. Then F (x) =∫ a

c

f(t) dt + G(x). First, note that∫ a

c

f(t) dt

is a constant. By (5.3.5) G, and thus F , is continuous on [a, b], is differentiable

on (a, b), and F ′(x) = G′(x) = f(x) for all x in (a, b).

34. Choose point c ∈ (a, b) and set F (x) =∫ x

c

f(t) dt. Since∫ x

c

f(t) dt =∫ x

a

f(t) dt−∫ c

a

f(t) dt, (Exercise 16)

it follows that

F ′(x) =d

dx

(∫ x

a

f(t) dt−∫ c

a

f(t) dt)

= f(x)

by Theorem 5.3.5.

35. (a) F ′(x) = f(x) = G′(x), on [a, b]. Therefore, by Theorem 4.2.4, F and G differ by a constant.

(b) F (x) = −∫ c

a f(t) dt +∫ x

a f(t) dt and G(x) = −∫ d

a f(t) dt +∫ x

a f(t) dt.

Thus F (x) −G(x) = −∫ c

a f(t) dt +∫ d

a f(t) dt =∫ d

c f(t) dt, a constant

36. (a) F ′(x) = x∫ x

1 f(u) du (b) F ′(1) = 0

(c) F ′′(x) = xf(x) +∫ x

1 f(u) du (d) F ′′(1) = f(1)



240 SECTION 5.4

37. (a) F ′(x) = 0 at x = −1, 4; F is increasing on (−∞,−1], [4,∞); F is decreasing on [−1, 4]

(b) F ′′(x) = 0 at x = 32 ; the graph of F is concave up on

(32 ,∞

); concave down on

(−∞, 3

2

)38. (a) F ′(x) = 2 − 3 cosx = 0 =⇒ cosx = 2

3 =⇒ x ∼= 0.8411, 5.4421;

F is decreasing on [0, 0.8411] ∪ [5.4421, 2π]; F is increasing on [0.8411, 5.4421].

(b) F ′′(x) = sinx = 0 =⇒ x = π;

the graph of F is concave up on (0, π) and concave down on (π, 2π).

39. (a) F ′(x) = 0 at x = 0, π2 , π,

3π2 , 2π

F is increasing on [π2 , π], [ 3π2 , 2π]; F is decreasing on [0, π2 ], [π, 3π

2 ]

(b) F ′′(x) = 0 at x = π4 ,

3π4 , 5π

4 , 7π4 ;

the graph of F is concave up on(π4 ,

3π4

),(

5π4 , 7π

4

); concave down on

(0, π

4

),(

3π4 , 5π

4

),(

7π4 , 2π

)40. (a) F ′(x) = −(2 − x)2 = 0 at x = 2;F ′(x) < 0 for all x �= 2 =⇒ F is decreasing on (−∞,∞).

(b) F ′′(x) = (2 − x) = 0 at x = 2; the graph of F is concave up on (−∞, 2) and concave down

on (2,∞).

SECTION 5.4

1.∫ 1

0

(2x− 3) dx = [x2 − 3x]10 = (−2) − (0) = −2

2.∫ 1

0

(3x + 2) dx =[3x2

2+ 2x

]10

=72

3.∫ 0

−1

5x4 dx = [x5]0−1 = (0) − (−1) = 1

4.∫ 2

1

(2x + x2) dx =[x2 +

13x3

]21

=163

5.∫ 4

1

2√x dx = 2

∫ 4

1

x1/2 dx = 2[23x3/2

]41

=43

[x3/2

]41

=43(8 − 1) =

283

6.∫ 4

0

3√x dx =

∫ 4

0

x13 dx =

[34x4/3

]40

=3444/3 = 3 3

√4

7.∫ 5

1

2√x− 1 dx =

∫ 5

1

2(x− 1)1/2 dx =[43(x− 1)3/2

]51

=43[43/2 − 0] =

323



SECTION 5.4 241

8.∫ 2

1

(3x3

+ 5x)

dx =[−3

2x−2 +

52x2

]21

=698

9.∫ 0

−2

(x + 1)(x− 2) dx =∫ 0

−2

(x2 − x− 2) dx =[x3

3− x2

2− 2x

]0−2

=[0 −(−8

3− 2 + 4

)]=

23

10.∫ 0

1

(t3 + t2) dt =[14t4 +

13t3]01

= − 712

11.∫ 2

1

(3t +

4t2

)dt =

∫ 2

1

(3t + 4t−2) dt =[32t2 − 4t−1

]21

=[(6 − 2) −

(32− 4)]

=132

12.∫ −1

−1

7x6 dx = 0

13.∫ 1

0

(x3/2 − x1/2) dx =[25x5/2 − 2

3x3/2

]10

=[(

25− 2

3

)− 0]

= − 415

14.∫ 1

0

(x3/4 − 2x1/2) dx =[47x7/4 − 4

3x3/2

]10

= −1621

15.∫ 1

0

(x + 1)17 dx =[

118

(x + 1)18]10

=118

(218 − 1)

16.∫ a

0

(a2x− x3) dx =[a2x2

2− x4

4

]a0

=a4

4

17.∫ a

0

(√a−

√x )2 dx =

∫ a

0

(a− 2√a x1/2 + x) dx =

[ax− 4

3√a x3/2 +

x2

2

]a0

= a2 − 43a2 +

a2

2=

16a2

18.∫ 1

−1

(x− 2)2 dx =[13(x− 2)3

]1−1

=263

19.∫ 2

1

6 − t

t3dt =

∫ 2

1

(6t−3 − t−2) dt =[−3t−2 + t−1

]21

=[−3

4+

12

]− [−3 + 1] =

74

20.∫ 3

1

(x2 − 1

x2

)dx =

[13x3 +

1x

]31

= 8

21.∫ 2

1

2x(x2 + 1) dx =∫ 2

1

(2x3 + 2x) dx =[x4

2+ x2

]21

= 12 − 32

=212

22.∫ 1

0

3x2(x3 + 1) dx =∫ 1

0

(3x5 + 3x2) dx =[12x6 + x3

]10

=32

23.∫ π/2

0

cosx dx = [sinx]π/20 = 1



242 SECTION 5.4

24.∫ π

0

3 sinx dx = [−3 cosx]π0 = 6

25.∫ π/4

0

2 sec2 x dx = 2 [tanx]π/40 = 2

26.∫ π/3

π/6

secx tanx dx = [secx]π/3π/6 = 2 − 2√

33

27.∫ π/4

π/6

cscu cotu dx = [− cscu]π/4π/6 = −√

2 − (−2) = 2 −√

2

28.∫ π/3

π/4

− csc2 u du = [cotu]π/3π/4 =√

33

− 1

29.∫ 2π

0

sinx dx = [− cosx]2π0 = −1 − (−1) = 0

30.∫ π

0

12

cosx dx =[12

sinx

]π0

= 0

31.∫ π/3

0

(2πx− 2 sec2 x

)dx =

[1πx2 − 2 tanx

]π/30

=π

9− 2

√3

32.∫ π/2

π/4

cscx(cotx− 3 cscx) dx =∫ π/2

π/4

(cscx cotx− 3 csc2 x) dx = [− cscx + 3 cotx]π/2π/4 =√

2 − 4

33.∫ 3

0

[d

dx

(√4 + x2

)]dx =

[√4 + x2

]30

=√

13 − 2

34.∫ π/2

0

[d

dx(sin3 x)

]=[sin3 x

]π/20

= 1

35. (a) F (x) =∫ x

1

(t + 2)2 dt =⇒ F ′(x) = (x + 2)2

(b)∫ x

1

(t + 2)2 dt =[t3

3+ 2t2 + 4t

]x1

=x3

3+ 2x2 + 4x− 6

13

=⇒ F ′(x) = x2 + 4x + 4 = (x + 2)2

36. (a) F (x) =∫ x

0

(cos t− sin t) dt =⇒ F ′(x) = cosx− sinx

(b)∫ x

0

(cos t− sin t) dt = [sin t + cos t]x0 = sinx− cosx− 1

=⇒ F ′(x) = cosx− sinx

37. (a) F (x) =∫ 2x+1

1

12 secu tanu du ⇒ F ′(x) = sec (2x + 1) tan (2x + 1)

(b)∫ 2x+1

1

12

secu tanu du =[12

secu]2x+1

1

=12

sec (2x + 1) − 12

sec 1

=⇒ F ′(x) = sec (2x + 1) tan (2x + 1)



SECTION 5.4 243

38. (a) F (x) =∫ 2

x2t(t− 1) dt ⇒ F ′(x) = −x2(x2 − 1)2x

(b)∫ 2

x2t(t− 1) dt =

[t3

3− t2

2

]2x2

=23− x6

3+

x4

2

=⇒ F ′(x) = −2x5 + 2x3 = −2x3(x2 − 1)

39. (a) F (x) =∫ x

2

dt

t(b) F (x) = −3 +

∫ x

2

dt

t

40. (a) F (x) =∫ x

3

√1 + t2 dt (b) F (x) = 1 +

∫ x

3

√1 + t2 dt

41. Area =∫ 4

0

(4x− x2) dx =[2x2 − x3

3

]40

=323

42. Area =∫ 9

1

(x√x + 1) dx =

∫ 9

1

(x3/2 + 1) dx =[25x5/2 + x

]91

=5245

43. Area =∫ π/4

−π/2

2 cosx dx = 2 [sinx]π/4−π/2 =√

2 + 2

44. Area =∫ π/3

0

(secx tanx) dx = [secx]π/30 = 2 − 1 = 1

45. (a)∫ 5

2

(x− 3) dx =[x2

2− 3x

]52

=32

(b)∫ 5

2

|x− 3| dx =∫ 3

2

(3 − x) dx +∫ 5

3

(x− 3) dx

=[3x− x2

2

]32

+[x2

2− 3x

]53

=52

46. (a)∫ 2

−4

(2x + 3) dx =[x2 + 3x

]2−4

= 6

(b)∫ 2

−4

|2x + 3| dx =∫ −3/2

−4

(−2x− 3) dx +∫ 2

−3/2

(2x + 3) dx =[−x2 − 3x

]−3/2

−4+[x2 + 3x

]2−3/2

=372

47. (a)∫ 2

−2

(x2 − 1) dx =[x3

3− x

]2−2

=43

(b)∫ 2

−2

|x2 − 1| dx =∫ −1

−2

(x2 − 1) dx +∫ 1

−1

(1 − x2) dx +∫ 2

1

(x2 − 1) dx

=[x3

3− x

]−1

−2

+[x− x3

3

]1−1

+[x3

3− x

]21

= 4

48. (a)∫ π

−π/2

cosx dx = [sinx]π−π/2 = 1

(b)∫ π

−π/2

| cosx| dx =∫ π/2

−π/2

cosx dx +∫ π

π/2

− cosx dx = [sinx]π/2−π/2 + [− sinx]ππ/2 = 3



244 SECTION 5.4

49. valid 50. not valid; sec2 x is not defined at x = 12π,

and x = 32π

51. not valid; 1/x3 is not defined at x = 0 52. valid

53. (a) x(t) =∫ t

0

(10u− u2) du =[5u2 − u3

3

]t0

= 5t2 − t3

3, 0 ≤ t ≤ 10

(b) v′(t) = 10 − 2t; v has an absolute maximum at t = 5. The object’s position at t = 5 is

x(5) =2503

.

54. (a) We need x(t) such that x′(t) = 3 sin t + 4 cos t and x(0) = 1.

Then x(t) = −3 cos t + 4 sin t+ C, x(0) = −3 + C = 1 =⇒ C = 4

=⇒ x(t) = −3 cos t + 4 sin t + 4.

(b) Maximum displacement when v(t) = 0 : 3 sin t + 4 cos t = 0

=⇒ tan t = − 43 =⇒ sin t = 4

5 , cos t = − 35

So xmax = −3(−35 ) + 4( 4

5 ) + 4 = 9

55.∫ 4

0

f(x) dx =∫ 1

0

(2x + 1) dx +∫ 4

1

(4 − x) dx =[x2 + x

]10

+[4x− x2

2

]41

=132

56.∫ 4

−2

f(x) dx =∫ 0

−2

(2 + x2) dx +∫ 4

0

(12x + 2) dx =

[2x +

x3

3

]0−2

+[x2

4+ 2x

]40

=563

57.∫ π

−π/2

f(x) dx =∫ π/3

−π/2

(1 + 2 cosx dx +∫ π

π/3

[3πx + 1

]dx = [x + 2 sinx]π/3−π/2 +

[3x2

2π+ x

]ππ/3

= 2 +√

3 + 176 π

58.∫ 3π/2

0

f(x) dx =∫ π/2

0

2 sinx dx +∫ 3π/2

π/2

(2 + cosx) dx = [−2 cosx]π/20 + [2x + sinx]3π/2π/2 = 2π

59. (a) f is continuous on [−2, 2].

For x ∈ [−2, 0], g(x) =∫ x

−2

(t + 2)dt =[12t2 + 2t

]x−2

=12x2 + 2x + 2.

For x ∈ [0, 1], g(x) =∫ 0

−2

(t + 2) dt +∫ x

0

2 dt = 2 + [2t]x0 = 2 + 2x.

For x ∈ [1, 2], g(x) =∫ 0

−2

(t + 2) dt +∫ 1

0

2 dt +∫ x

1

(4 − 2t)dt = 2 + 2 +[4t− t2

]x1

= 1 + 4x− x2.

Thus g(x) =

⎧⎪⎪⎨⎪⎪⎩

12 x

2 + 2x + 2, −2 ≤ x ≤ 0

2x + 2, 0 ≤ x ≤ 1

1 + 4x− x2, 1 ≤ x ≤ 2

⎤⎥⎥⎦

− 2 1 2x

1

2

y



SECTION 5.4 245

(b)

2 1 2x

2

4

y

(c) f is continuous on [−2, 2]; f is differentiable on (−2, 0), (0, 1), and (1, 2).g is differentiable on (−2, 2).

60. (a) g(x) =∫ x

−1

f(x) dt =∫ x

−1

(1 − t2) dt =[t− t3

3

]x−1

= 2x− x3

3+

53, for − 1 ≤ x ≤ 1

g(x) =∫ 1

−1

(1 − t2) dt +∫ x

1

1 dt =103

+ [t]x1 =73

+ x, for 1 < x < 3

g(x) =∫ 3

−1

f(t) dt +∫ x

3

(2t− 5) dt =163

+[t2 − 5t

]x3

=343

+ x2 − 5x, for 3 ≤ x ≤ 5

− 1 1 2 3 4 5x

1

2

y

(b)

− 1 1 2 3 4 5x

1

y

(c) f is continuous on [−1, 1) ∪ (1, 5], f is differentiable on (−1, 1) ∪ (1, 3) ∪ (3, 5).

g is differentiable on (−1, 1) ∪ (1, 5).

61. Follows from Theorem 5.3.2 since f(x) is an antiderivative of f ′(x).

62. Let F (x) = f2(x). Then F ′(x) = 2f(x)f ′(x).

Thus∫ b

a

f(t)f ′(t) dt =12

∫ b

a

F ′(t) dt =12[F (b) − F (a)] =

12[f2(b) − f2(a)].

63.d

dx

[∫ x

a

f(t) dt]

= f(x);∫ x

a

d

dt[f(t)] dt = f(x) − f(a)

64. F (x) =∫ x

0

xf(t) dt = x

∫ x

0

f(t) dt; F is a product.

F ′(x) = x f(x) +∫ x

0

f(t) dt



246 SECTION 5.5

SECTION 5.5

1. A =∫ 1

0

(2 + x3) dx =[2x +

x4

4

]10

=94

2. A =∫ 2

0

(x + 2)−2 dx =[ −1x + 2

]20

=14

3. A =∫ 8

3

√x + 1 dx =

∫ 8

3

(x + 1)1/2 dx =[23(x + 1)3/2

]83

=23[27 − 8] =

383

4. A =∫ 8

0

(3x2 + x3) dx =[x3 +

14x4

]80

= 1536

5. A =∫ 1

0

(2x2 + 1)2 dx =∫ 1

0

(4x4 + 4x2 + 1) dx =[45x5 +

43x3 + x

]10

=4715

6. A =∫ 8

0

12√x + 1

dx =[√

x + 1]80

= 2

7. A =∫ 2

1

[0 − (x2 − 4)] dx =∫ 2

1

(4 − x2) dx =[4x− x3

3

]21

=[8 − 8

3

]−[4 − 1

3

]=

53

8. A =∫ π/3

π/6

cosx dx = [sinx]π/3π/6 =√

3 − 12

9. A =∫ π/2

π/3

sinx dx = [− cosx]π/2π/3 = (0) −(−1

2

)=

12

10. A = −∫ −1

−2

(x3 + 1) dx = −[x4

4+ x

]−1

−2

=114

11.A =

∫ 1

0

[x1/2 − x2] dx

=[23x

3/2 − 13x

3]10

= 13

12.A =

∫ 4

0

(6x− x2 − 2x) dx =323



SECTION 5.5 247

13.A =

∫ 2

−1

[(5 − x2) − (3 − x)] dx

=∫ 2

−1

(2 + x− x2) dx

=[2x +

x2

2− x3

3

]2−1

=[4 + 2 − 8

3

]−[−2 + 1

2 + 13

]= 9

2

14.A =

∫ 2

−4

(8 − x2 − 2x) dx = 36

15.A =

∫ 2

−2

[(8 − x2) − (x2)] dx

=∫ 2

−2

(8 − 2x2) dx

=[8x− 2

3x3]2−2

=[16 − 16

3

]−[−16 + 16

3

]= 64

3

16. A =∫ 16

0

(√x− 1

4x

)dx =

323



248 SECTION 5.5

17.A =

∫ 10

0

[x− 1√

10x3/2

]dx

=

[x2

2− 2

√10

50x5/2

]10

0

= 50 − 2√

1050

(10)5/2 = 10

18. A =∫ 27

0

[−x−

(−√

27x)]

dx =2432

19.

A =∫ −2

−3

[(√

3 + x ) − (−√

3 + x )] dx +∫ 6

−2

[(√

3 + x ) −(

12x

)]dx

=∫ −2

−3

2(3 + x)1/2 dx +∫ 6

−2

[(3 + x)1/2 − 1

2x

]dx

=[43(3 + x)3/2

]−2

−3

+[23(3 + x)3/2 − x2

4

]6−2

=[43− 0]

+[(18 − 9) −

(23− 1)]

=323

20.A =

∫ 2

0

[√2x− (−

√2x)]dx +

∫ 8

2

[√2x− x + 4

]dx

= 18



SECTION 5.5 249

21.A =

∫ 2

0

[2x− x] dx +∫ 4

2

[4 − x] dx

=[12x

2]20

+[4x− 1

2x2]42

= 2 + [8 − 6] = 4

22. A =∫ 4

0

(x2 +

√x)dx =

803

23.A =

∫ π/2

−π/2

[cosx− (4x2 − π2)] dx

=[sinx− 4

3x3 + π2x

]π/2−π/2

= [1 − 16π

3 + 12π

3] − [−1 + 16π

3 − 12π

3]

= 2 + 23π

3

24.A =

∫ π

0

(πx− x2 − sinx) dx =π3

6− 2



250 SECTION 5.5

25.A =

∫ π/2

0

[x− sinx] dx

=[x2

2+ cosx

]π/20

=π2

8− 1

26. A =∫ π

0

(x + 1 − cosx) dx =π

2(π + 2)

27. (a)∫ 4

−3

(x2 − x− 6) dx =[13x3 − 1

2x2 − 6x

]4−3

= − 916

;

the area of the region bounded by the graph of f and the x-axis for x ∈ [−3,−2] ∪ [3, 4]

minus the area of the region bounded the graph of f and the x-axis for x ∈ [−2, 3].

(b) A =∫ −2

−3

(x2 − x− 6) dx +∫ 3

−2

(−x2 + x + 6) dx +∫ 4

3

(x2 − x− 6) dx

=[13 x

3 − 12 x

2 − 6x]−2

−3+[− 1

3 x3 + 1

2 x2 + 6x

]3−2

+[13 x

3 − 12 x

2 − 6x]43

=176

+1256

+176

=532

(c) A = −∫ 3

−2

(x2 − x− 6) dx =1256

28. (a)∫ 3π/4

−π/2

2 sinx dx = [−2 cosx]3π/4−π/2 =√

2 = area above − area below

(b) A =∫ 0

−π/2

−2 sinx dx +∫ 3π/4

0

2 sinx dx = [2 cosx]0−π/2 + [−2 cosx]3π/40 =√

2 + 4

(c) A =∫ 0

−π/2

−2 sinx dx = [2 cosx]0−π/2 = 2



SECTION 5.5 251

29. (a)∫ 2

−2

(x3 − x) dx =[14x4 − 1

2x2

]2−2

= 0

(b)A = 2

[−∫ 1

0

(x3 − x) dx +∫ 2

1

(x3 − x) dx]

= −2[ 14 x4 − 1

2 x2]10 + 2[ 14 x

4 − 12 x

2]21

=12

+92

= 5

30. (a)∫ π

−π

(cosx + sinx) dx = [sinx− cosx]π−π = 0

(b)A = −

∫ −π/4

−π

f(x) dx +∫ 3π/4

−π/4

f(x) dx−∫ π

3π/4

f(x) dx

= 4√

2

31. (a)∫ 3

−2

(x3 − 4x + 2) dx =[14x4 − 2x2 + 2x

]3−2

=654

(b)

A ∼=∫ 0.54

−2

(x3 − 4x + 2) dx−∫ 1.68

0.54

(x3 − 4x + 2) dx +∫ 3

1.68

(x3 − 4x + 2) dx

=[14 x

4 − 2x2 + 2x]0.54−2

−[14 x

4 − 2x2 + 2x]1.680.54

+[14 x

4 − 2x2 + 2x]31.68

= 8.52 + .81 + 8.54 = 17.87



252 SECTION 5.5

32. (a)∫ π/2

−π/2

(3x2 − 2 cosx) dx =[x3 − 2 sinx

]π/2−π/2

=π3

4− 4

(b)A ∼= −

∫ 0.71

−0.71

(3x2 − 2 cosx) dx + 2∫ π/2

0.71

(3x2 − 2 cosx) dx

∼= 7.53

33.A =

∫ 1

0

(x2 + 1) dx +∫ 3

1

(3 − x) dx

=[13 x

3 + x]10

+[3x− 1

2 x2]31

=43

+ 2 =103

34.A =

∫ 1

0

3√x dx +

∫ 2

1

(4 − x2) dx

= [2x3/2]10 +[4x− x3

3

]21

= 2 +53

=113

35.A =

∫ π/4

0

sinx dx +∫ π/2

π/4

cosx dx

= [− cosx]π/40 + [sinx]π/2π/4

= 2 −√

2

36.A = 2

∫ π/2

0

(1 + cosx− 1) dx

= [2 sinx]π/20 = 2



SECTION 5.5 253

37.

A ∼=∫ 1.32

0

[3x + 1 − (x3 + 2x)] dx +∫ 2

1.32

[x3 + 2x− (3x + 1)] dx

=∫ 1.32

0

(x + 1 − x3) dx +∫ 2

1.32

(x3 − x− 1) dx

=[12 x

2 + x− 14 x

4]1.320

+[14 x

4 − 12 x

2 − x]21.32

= 2.86

38.A =

∫ √ 1+√17

2

−√

1+√17

2

(4 − x2 − x4 + 2x2) dx

∼= 11.34

39. h ∼= 9.44892

40. h ∼= 0.0355

PROJECT 5.5

1. (a) g(x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

−1, x = 0

0, 0 < x ≤ 1

1, 1 < x ≤ 2...

4, 4 < x ≤ 5.

∫ 5

0

g(x) dx =∫ 1

0

g(x) dx +∫ 2

1

g(x) dx + · · · +∫ 5

4

g(x) dx = 0 + 1 + 2 + 3 + 4 = 10.



254 SECTION 5.5

(b) g(x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

0, 0 ≤ x < 1

1, 1 ≤ x < 2...

4, 4 ≤< x < 5

5 x = 5.∫ 5

0

g(x) dx =∫ 1

0

g(x) dx +∫ 2

1

g(x) dx + · · · +∫ 5

4

g(x) dx = 0 + 1 + 2 + 3 + 4 = 10.

(c) g(x) =

⎧⎨⎩1 x = 1, 2, · · · 5

0 otherwise;

∫ 5

0

g(x) dx = 0.

2. (a)∫ 2

0

g(x) dx =∫ 1

0

(2 − x) dx +∫ 2

1

(2 + x) dx =[2x− 1

2x2]10

+[2x + 1

2x2]21

= 5.

1 2x

2

4

y

(b)∫ 5

0

g(x) dx =∫ 2

0

x2 dx +∫ 5

2

x dx =[

13x

3]20

+[

12x

2]52

= 796 .

2 5x

4

5

y

(c)∫ 2π

0

g(x) dx =∫ π/2

0

cosx dx +∫ π

π/2

sinx dx +∫ 2π

π

12 dx =

[sinx

]π/20

−[cosx

]ππ/2

+[

12x]2ππ

= 2 + 12π

2π 2 ππ

x

1

y



SECTION 5.6 255

3. (a) For 0 ≤ x < 1, G(x) =∫ x

0

(2 − t) dt =[2t− 1

2 t2]x0

= 2x− 12x

2

For 1 ≤ x ≤ 2, G(x) =∫ 1

0

(2 − t) dt +∫ x

1

(2 + t) dt = 32 +[2t + 1

2 t2]x1

= 2x + 12x

2 − 1

Thus, G(x) =

⎧⎨⎩2x− 1

2x2, 0 ≤ x < 1

2x + 12x

2 − 1, 1 ≤ x ≤ 2.

limh→0−

G(1 + h) −G(1)h

= limh→0−

2(1 + h) − 12 (1 + h)2 − 3

2

h= lim

h→0−

h− 12h

2

h= 1

limh→0+

G(1 + h) −G(1)h

= limh→0+

2(1 + h) + 12 (1 + h)2 − 1 − 3

2

h= lim

h→0+

3h + 12h

2

h= 3

Thus, G is not differentiable at x = 1.

(b) G(x) =

⎧⎨⎩

13x

3, 0 ≤ x < 2

12x

2 + 23 , 2 ≤ x ≤ 5

.

limh→0−

G(2 + h) −G(2)h

= limh→0−

13 (2 + h)3 − 8

3

h= lim

h→0−

4h + 2h2 + 13h

3

h= 4

limh→0+

G(2 + h) −G(2)h

= limh→0+

12 (2 + h)2 + 2

3 − 83

h= lim

h→0+

2h + 12h

2

h= 2

Thus, G is not differentiable at x = 2.

(c) G(x) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩

sinx, 0 ≤ x < π/2

1 − cosx, π/2 ≤ x < π

2 + 12x− π/2, π ≤ x ≤ 2π

.

G is not differentiable at x = π/2, π;

see the graph of G:

π2

π 2 πx

1

y

SECTION 5.6

1.∫

dx

x4=∫

x−4 dx = −13x−3 + C

2.∫

(x− 1)2 dx =∫

(x2 − 2x + 1) dx =13x3 − x2 + x + C



256 SECTION 5.6

3.∫

(ax + b) dx =12ax2 + bx + C

4.∫

(ax2 + b) dx =13ax3 + bx + C

5.∫

dx√1 + x

=∫

(1 + x)−1/2 dx = 2(1 + x)1/2 + C

6.∫

x3 + 1x5

dx =∫

x−2 + x−5 dx = −x−1 − 14x−4 + C

7.∫ (

x3 − 1x2

)dx =

∫(x− x−2) dx =

12x2 + x−1 + C

8.∫ (√

x− 1√x

)dx =

∫(x1/2 − x−1/2) dx =

23x3/2 − 2x1/2 + C

9.∫

(t− a)(t− b) dt =∫

[t2 − (a + b)t + ab] dt =13t3 − a + b

2t2 + abt + C

10.∫

(t2 − a)(t2 − b) dt =∫ (

t4 − (a + b)t2 + ab)dt =

15t5 − 1

3(a + b)t3 + abt + C

11.∫

(t2 − a)(t2 − b)√t

dt =∫

[t7/2 − (a + b)t3/2 + abt−1/2] dt

= 29 t

9/2 − 25 (a + b)t5/2 + 2abt1/2 + C

12.∫ (

2 −√x)(2 +

√x)dx =

∫(4 − x) dx = 4x− 1

2x2 + C

13.∫

g(x)g′(x) dx =12[ g(x)]2 + C

14.∫

sinx cosx dx =12

sin2 x + C

15.∫

tanx sec2 x dx =∫

secxd

dx[secx] dx =

12

sec2 x + C∫tanx sec2 x dx =

∫tanx

d

dx[tanx] dx =

12

tan2 x + C

16.∫

g′(x)[g(x)]2

dx = − 1g(x)

+ C

17.∫

4(4x + 1)2

dx =∫

4(4x + 1)−2 dx = −(4x + 1)−1 + C

18.∫

3x2

(x3 + 1)2dx = − 1

x3 + 1+ C, (use Exercise 16)



SECTION 5.6 257

19. f(x) =∫

f ′(x) dx =∫

(2x− 1) dx = x2 − x + C.

Since f(3) = 4, we get 4 = 9 − 3 + C so that C = −2 and

f(x) = x2 − x− 2.

20. f(x) =∫

(3 − 4x) dx = 3x− 2x2 + C, f(1) = 6 =⇒ f(x) = −2x2 + 3x + 5

21. f(x) =∫

f ′(x) dx =∫

(ax + b) dx =12ax2 + bx + C.

Since f(2) = 0, we get 0 = 2a + 2b + C so that C = −2a− 2b and

f(x) = 12ax

2 + bx− 2a− 2b.

22. f(x) =∫

(ax2 + bx + c) dx =13ax3 +

12bx2 + cx + K,

f(0) = 0 =⇒ f(x) =a

3x3 +

b

2x2 + cx

23. f(x) =∫

f ′(x) dx =∫

sinx dx = − cosx + C.

Since f(0) = 2, we get 2 = −1 + C so that C = 3 and

f(x) = 3 − cosx.

24. f(x) =∫

cosx dx = sinx + C, f(π) = 3 =⇒ f(x) = 3 + sinx

25. First,

f ′(x) =∫

f ′′(x) dx =∫

(6x− 2) dx = 3x2 − 2x + C.

Since f ′(0) = 1, we get 1 = 0 + C so that C = 1 and

f ′(x) = 3x2 − 2x + 1.

Next,

f(x) =∫

f ′(x) dx =∫

(3x2 − 2x + 1) dx = x3 − x2 + x + K.

Since f(0) = 2, we get 2 = 0 + K so that K = 2 and

f(x) = x3 − x2 + x + 2.

26. f ′(x) =∫

−12x2 dx = −4x3 + C, f ′(0) = 1 =⇒ f ′(x) = −4x3 + 1

f(x) =∫

(−4x3 + 1) dx = −x4 + x + K, f(0) = 2 =⇒ f(x) = −x4 + x + 2



258 SECTION 5.6

27. First,

f ′(x) =∫

f ′′(x) dx =∫

(x2 − x) dx =13x3 − 1

2x2 + C.

Since f ′(1) = 0, we get 0 = 13 − 1

2 + C so that C = 16 and

f ′(x) = 13x

3 − 12x

2 + 16 .

Next,

f(x) =∫

f ′(x) dx =∫ (

13x3 − 1

2x2 +

16

)dx =

112

x4 − 16x3 +

16x + K.

Since f(1) = 2, we get 2 = 112 − 1

6 + 16 + K so that K = 23

12 and

f(x) =x4

12− x3

6+

x

6+

2312

=112

(x4 − 2x3 + 2x + 23).

28. f ′(x) =∫

(1 − x) dx = x− x2

2+ C, f ′(2) = 1 =⇒ f ′(x) = x− x2

2+ 1

f(x) =∫

(x− x2

2+ 1) dx =

x2

2− x3

6+ x + K, f(2) = 0 =⇒ f(x) = −x3

6+

x2

2+ x− 8

3

29. First,

f ′(x) =∫

f ′′(x) dx =∫

cosx dx = sinx + C.

Since f ′(0) = 1, we get 1 = 0 + C so that C = 1 and

f ′(x) = sinx + 1.

Next,

f(x) =∫

f ′(x) dx =∫

(sinx + 1) dx = − cosx + x + K.

Since f(0) = 2, we get 2 = −1 + 0 + K so that K = 3 and

f(x) = − cosx + x + 3.

30. f ′(x) =∫

sinx dx = − cosx + C, f ′(0) = −2 =⇒ f ′(x) = − cosx− 1

f(x) =∫

(− cosx− 1) dx = − sinx− x + K, f(0) = 1 =⇒ f(x) = 1 − sinx− x

31. First,

f ′(x) =∫

f ′′(x) dx =∫

(2x− 3) dx = x2 − 3x + C.

Then,

f(x) =∫

f ′(x) dx =∫

(x2 − 3x + C) dx =13x3 − 3

2x2 + Cx + K.



SECTION 5.6 259

Since f(2) = −1, we get

(1) −1 = 83 − 6 + 2C + K;

and, from f(0) = 3, we conclude that

(2) 3 = 0 + K.

Solving (1) and (2) simultaneously, we get K = 3 and C = − 13 so that

f(x) = 13x

3 − 32x

2 − 13x + 3.

32. f ′(x) =∫

(5 − 4x) dx = 5x− 2x2 + C,

f(x) =∫

(5x− 2x2 + C) dx =52x2 − 2

3x3 + Cx + K

f(1) =52− 2

3+ C + K = 1, f(0) = K = −2 =⇒ f(x) = −2

3x3 +

52x2 +

76x− 2

33.d

dx

[∫f(x) dx

]= f(x);

∫d

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

34.∫

[f(x)g′′(x) − g(x)f ′′(x)] dx =∫

[f(x)g′′(x) + f ′(x)g′(x) − f ′(x)g′(x) − g(x)f ′′(x)] dx

=∫ (

d

dx[f(x)g′(x)] − d

dx[f ′(x)g(x)]

)dx = f(x)g′(x) − g(x)f ′(x) + C

35. (a) x(t) =∫

v(t) dt =∫

(6t2 − 6) dt = 2t3 − 6t + C.

Since x(0) = −2, we get −2 = 0 + C so that C = −2 and

x(t) = 2t3 − 6t− 2. Therefore x(3) = 34.

Three seconds later the object is 34 units to the right of the origin.

(b) s =∫ 3

0

| v(t)| dt =∫ 3

0

| 6t2 − 6 | dt =∫ 1

0

(6 − 6t2) dt +∫ 3

1

(6t2 − 6) dt

= [6t− 2t3]10 + [2t3 − 6t]31 = 4 + [36 − (−4)] = 44.

The object traveled 44 units.

36. (a) v(t) =∫

a(t) dt =∫

(t + 2)3 dt =14(t + 2)4 + C,

v(0) = 3 =⇒ v(t) =14(t + 2)4 − 1

(b) x(t) =∫ [

(t + 2)4

4− 1]dt =

(t + 2)5

20− t + K, x(0) = 0 =⇒ x(t) =

(t + 2)5

20− t− 8

5

37. (a) v(t) =∫

a(t) dt =∫

(t + 1)−1/2 dt = 2(t + 1)1/2 + C.

Since v(0) = 1, we get 1 = 2 + C so that C = −1 and

v(t) = 2(t + 1)1/2 − 1.



260 SECTION 5.6

(b) We know v(t) by part (a). Therefore,

x(t) =∫

v(t) dt =∫

[2(t + 1)1/2 − 1] dt =43(t + 1)3/2 − t + C.

Since x(0) = 0, we get 0 = 43 − 0 + C so that

C = − 43 and x(t) = 4

3 (t + 1)3/2 − t− 43 .

38. (a) x(t) =∫

t(1 − t) dt =t2

2− t3

3+ C, x(0) = −2 =⇒ x(t) =

t2

2− t3

3− 2

x(10) = −8563

: 28513

units to the left of the origin.

(b) s =∫ 10

0

|v(t)| dt =∫ 1

0

t(1 − t) dt +∫ 10

1

−t(1 − t) dt =[t2

2− t3

3

]10

−[t2

2− t3

3

]101

=8513

= 28323

units.

39. (a) v0 = 60 mph = 88 feet per second. In general, v(t) = at + v0. Here, in feet and

seconds, v(t) = −20t + 88. Thus v(t) = 0 at t = 4.4 seconds.

(b) In general, x(t) = 12at

2 + v0t + x0. Here we take x0 = 0. In feet and seconds

x(t) = −10(4.4)2 + 88(4.4) = 10(4.4)2 = 193.6 ft.

40. Let acceleration = a. Then v(t) =∫

a dt = at + v0.

x(t) =∫

v(t) dt =∫

(at + v0) dt =12at2 + v0t + x0 =

12

[v(t) + 2v0] t + x0

41. [v(t)]2 = (at + v0)2 = a2t2 + 2av0t + v02

= v02 + a(at2 + 2v0t)

= v02 + 2a( 1

2at2 + v0t)

(set x(t) = 12at

2 + v0t + x0)

= v02 + 2a [x(t) − x0]

42. (a) v(t) = at + v0, and by Exercise 40 x(t) =12

[v(t) + 2v0] t, so

a =v(t) − v0

t=

v(t) − v0

2x(t)(v(t) + 2v0) =

58.72 + (58.7)(88) − 2(88)2

2(264)= −13.02 ft/sec2

[Note 60 mph = 88 ft/sec, 40 mph = 5823 ft/sec.]

(b) t =2x(t)

v(t) + 2v0=

2(264)58 2

3 + 176= 2.24 sec

(c) We don’t know x(t), so we will use t =v(t) − v0

a=

0 − 88−13.02

= 6.8 sec

(d) x(t) =12

[v(t) + 2v0] t = (88)(6.8) = 598.4 ft



SECTION 5.6 261

43. The car can accelerate to 60 mph (88 ft/sec) in 20 seconds thereby covering a distance of 880 ft. It candecelerate from 88 ft/sec to 0 ft/sec in 4 seconds thereby covering a distance of 176 ft. At full speed,88 ft/sec, it must cover a distance of

52802

− 880 − 176 = 1584 ft.

This takes158488

= 18 seconds. The run takes at least 20 + 18 + 4 = 42 seconds.

44. v(t) =∫

sin t dt = − cos t + C, v(0) = v0 =⇒ v(t) = − cos t + v0 + 1

x(t) =∫

(− cos t + v0 + 1) dt = − sin t + (v0 + 1)t + K, x(0) = x0 =⇒ x(t) = x0 + (v0 + 1)t− sin t

45. v(t) =∫

a(t) dt =∫

(2A + 6Bt) dt = 2At + 3Bt2 + C.

Since v(0) = v0, we have v0 = 0 + C so that v(t) = 2At + 3Bt2 + v0.

x(t) =∫

v(t) dt =∫

(2At + 3Bt2 + v0) dt = At2 + Bt3 + v0t + K.

Since x(0) = x0, we have x0 = 0 + K so that K = x0 and

x(t) = x0 + v0t + At2 + Bt3.

46. v(t) =∫

cos t dt = sin t + C, v(0) = v0 =⇒ v(t) = sin t + v0

x(t) =∫

(sin t + v0) dt = − cos t + v0t + K, x(0) = x0 =⇒ x(t) = x0 + 1 + v0t− cos t

47. x′(t) = t2 − 5, y′(t) = 3t,

x(t) = 13 t

3 − 5t + C. y(t) = 32 t

2 + K.

When t = 2, the particle is at (4, 2). Thus, x(2) = 4 and y(2) = 2.

4 = 83 − 10 + C =⇒ C = 34

3 . 2 = 6 + K =⇒ K = −4.

x(t) = 13 t

3 − 5t + 343 , y(t) = 3

2 t2 − 4.

Four seconds later the particle is at (x(6), y(6)) = (1603 , 50).

48. x(t) =∫

(t− 2) dt =t2

2− 2t + C, x(4) = 3 =⇒ x(t) =

t2

2− 2t + 3

y(t) =∫ √

t dt =23t3/2 + K, y(4) = 1 =⇒ y(t) =

23t3/2 − 13

3

5 seconds later, t = 9, so position is (x(9), y(9)) =(

512 , 41

3

).

49. Since v(0) = 2, we have 2 = A · 0 + B so that B = 2. Therefore

x(t) =∫

v(t) dt =∫

(At + 2) dt =12At2 + 2t + C.



262 SECTION 5.6

Since x(2) = x(0) − 1, we have

2A + 4 + C = C − 1 so that A = − 52 .

50. x(t) =∫

(At2 + 1) dt =13At3 + t + C

x(1) − x(0) = (13A + 1 + C) − C =

A

3+ 1 = 0 =⇒ A = −3

Distance traveled =∫ 1/

√3

0

(1 − 3t2) dt +∫ 1

1/√

3

(3t2 − 1) dt =[t− t3

]1/√3

0+[t3 − t

]11/

√3

=4√

39

51. x(t) =∫

v(t) dt =∫

sin t dt = − cos t + C

Since x(π/6) = 0, we have 0 = −√

32

+ C so that C =√

32

and x(t) =√

32 − cos t.

(a) At t = 11π/6 sec. (b) We want to find the smallest t0 > π/6 for

which x(t0) = 0 and v(t0) > 0. We get

t0 = 13π/6 seconds.

52. x(t) =∫

cos t dt = sin t + C, x(π

6

)= sin

π

6+ C = 0 =⇒ x(t) = sin t− 1

2

(a) x(t) = 0 at t = 56π sec.

(b) x(t) = 0 and v(t) > 0 =⇒ t =13π6

sec.

53. The mean-value theorem. With obvious notation

x(1/12) − x(0)1/12

=4

1/12= 48.

By the mean-value theorem there exists some time t0 at which

x′(t0) =x(1/12) − x(0)

1/12.

54. (Taking the direction of motion as positive, speed and velocity are the same.) Let v be the speed of

the motorcycle at time 0, the time when the brakes are applied. The distance between the motorcycle

and the hay wagon t time units later is given by

d(t) = −12at2 + (v1 − v)t + s

[v1t + s gives the position of the hay wagon, 12at

2 + vt gives the position of the motorcycle]. Collision

can be avoided only if the quadratic

d(t) = −12at2 + (v1 − v)t + s



SECTION 5.7 263

remains positive. This can be true only if the discriminant of the quadratic,

B2 − 4AC = (v1 − v)2 + 2as = (v − v1)2 + 2as

remains negative. Observe that

(v − v1)2 + 2as < 0 iff v < v1 +√

2|a|s

55.v′(t)

[v(t)]2= 2 =⇒ −[v(t)]−1 = 2t− v0

−1.

=⇒ [v(t)]−1 = v0−1 − 2t =⇒ v(t) =

1v0

−1 − 2t=

v0

1 − 2tv0.

56. dsd

dx

(∫x2 − x3 + x4

√x

dx

)=

x2 − x3 + x4

√x

; ds

∫d

dx

(x2 − x3 + x4

√x

)dx =

x2 − x3 + x4

√x

+ C

57.∫

(cosx− 2 sinx) dx = sinx + 2 cosx + C and so

d

dx

(∫(cosx− 2 sinx) dx

)=

d

dx[sinx + 2 cosx] = cosx− 2 sinx;

d

dx[cosx− 2 sinx] = − sinx− 2 cosx and so

∫d

dx[cosx− 2 sinx] dx =

∫(− sinx− 2 cosx) dx = cosx− 2 sinx + C

58. f(x) = x + 2√x− 6 59. f(x) = sinx + 2 cosx + 1

60. f(x) = 3x + 2 − 2 cosx− 3 sinx 61. 112 x

4 − 12 x

3 + 52 x

2 + 4x− 3

SECTION 5.7

1.

{u = 2 − 3x

du = −3 dx

};

∫dx

(2 − 3x)2=∫

(2 − 3x)−2 dx = −13

∫u−2 du =

13u−1 + C

= 13 (2 − 3x)−1 + C

2.

{u = 2x + 1

du = 2 dx

};

∫dx√

2x + 1=

12

∫du√u

=√u + C =

√2x + 1 + C

3.

{u = 2x + 1

du = 2 dx

};

∫ √2x + 1 dx =

∫(2x + 1)1/2 dx =

12

∫u1/2 du =

13u3/2 + C

=13(2x + 1)3/2 + C

4.

{u = ax + b

du = a dx

};

∫ √ax + b =

1a

∫ √u du =

23a

u3/2 + C =23a

(ax + b)3/2 + C



264 SECTION 5.7

5.

{u = ax + b

du = a dx

};

∫(ax + b)3/4 dx =

1a

∫u3/4 du =

47a

u7/4 + C

=47a

(ax + b)7/4 + C

6.

{u = ax2 + b

du = 2ax dx

};

∫2ax(ax2 + b)4 dx =

∫u4 du =

15u5 + C =

15(ax2 + b)5 + C

7.

{u = 4t2 + 9

du = 8t dt

};

∫t

(4t2 + 9)2dt =

18

∫du

u2= −1

8u−1 + C = −1

8(4t2 + 9

)−1 + C

8.

{u = t2 + 1

du = 2t dt

};

∫3t

(t2 − 1)2dt =

32

∫du

u2= − 3

2u+ C =

−32(t2 + 1)

+ C

9.

{u = 1 + x3

du = 3x2 dx

};

∫x2(1 + x3)1/4 dx =

13

∫u1/4 du =

415

u5/4 + C =415

(1 + x3)5/4 + C

10.

{u = a + bxn

du = nbxn−1 dx

};

∫xn−1

√a + bxn dx =

1bn

∫ √u du =

23bn

u3/2 + C =2

3bn(a + bxn)3/2 + C

11.

{u = 1 + s2

du = 2s ds

};

∫s

(1 + s2)3ds =

12

∫du

u3= −1

4u−2 + C = −1

4(1 + s2)−2 + C

12.

{u = 6 − 5s2

du = −10s ds

};

∫2s

3√

6 − 5s2ds = −1

5

∫u−1/3 du = − 3

10u2/3 + C =

−310(6 − 5s2

)2/3 + C

13.

{u = x2 + 1

du = 2x dx

};∫

x√x2 + 1

dx =∫ (

x2 + 1)−1/2

x dx =12

∫u−1/2 du = u1/2 + C =

√x2 + 1 + C

14.

{u = 1 − x3

du = −3x2 dx

};

∫x2

(1 − x3)2/3= −1

3

∫du

u2/3= −u1/3 + C = −(1 − x3)1/3 + C

15.

{u = x2 + 1

du = 2x

};

∫5x(x2 + 1

)−3dx =

52

∫u−3 du = −5

4u−2 + C = −5

4(x2 + 1)−2 + C

16.

{u = 1 − x4

du = −4x3 dx

};

∫2x3(1 − x4)−1/4 dx = −1

2

∫u−1/4 du = −2

3u3/4 + C = −2

3(1 − x4)3/4 + C



SECTION 5.7 265

17.

{u = x1/4 + 1

du = 14x

−3/4 dx

};

∫x−3/4

(x1/4 + 1

)−2

dx = 4∫

u−2 du = −4u−1 + C = −4(x1/4 + 1)−1 + C

18.

{u = x2 + 3x + 1

du = (2x + 3) dx

};

∫4x + 6√

x2 + 3x + 1dx = 2

∫du√u

= 4√u + C = 4

√x2 + 3x + 1 + C

19.

{u = 1 − a4x4

du = −4a4x3 dx

};

∫b3x3

√1 − a4x4

dx = − b3

4a4

∫u−1/2 du = − b3

2a4u1/2 + C = − b3

2a4

√1 − a4x4 + C

20.

{u = a + bxn

du = bnxn−1 dx

};

∫xn−1

√a + bxn

dx =1bn

∫du√u

=2bn

√u + C =

2bn

√a + bxn + C

21.

{u = x2 + 1

du = 2x dx

};

∫x(x2 + 1

)3dx =

12

∫ 2

1

u3 du =18u4 + C =

18(x2 + 1)4 + C

∫ 1

0

x(x2 + 1) dx =[18(x2 + 1)4

]10

=18[16 − 1] =

158

22.

{u = 4 + 2x3

du = 6x2 dx

x = −1 ⇒ u = 2

x = 0 ⇒ u = 4

};

∣∣∣∣∫ 0

−1

3x2(4 + 2x3

)2dx =

12

∫ 4

2

u2 du =[16u3

]42

=283

23. 0; the integrand is an odd function

24.

{u = r2 + 16

du = 2r dr

r = 0 ⇒ u = 16

r = 3 ⇒ u = 25

};

∣∣∣∣∫ 3

0

r√r2 + 16

dr =12

∫ 25

16

du√u

=[√

u]2516

= 1

25.

{u = a2 − y2

du = −2y dy

};

∫y√

a2 − y2 dy = −12

∫u1/2 du = − 1

3u3/2 + C = − 1

3 (a2 − y2)3/2 + C

∫ a

0

y√a2 − y2 dy = − 1

3

[(a2 − y2)3/2

]a0

= 13 (a2)3/2 = 1

3 |a|3



266 SECTION 5.7

26.

⎧⎪⎪⎨⎪⎪⎩

u = 1 − y3

a3

du = −3y2

a3dy

y = −a ⇒ u = 2

y = 0 ⇒ u = 1

⎫⎪⎪⎬⎪⎪⎭

∣∣∣∣∣∣∣∣∫ 0

−a

y2

(1 − y3

a3

)−2

dy = −a3

3

∫ 1

2

u−2 du =a3

3

∫ 2

1

u−2 du = −a3

3

[1u

]21

=a3

6

27.

{u = 2x2 + 1

du = 4x dx

};

∫x√

2x2 + 1 dx =14

∫u1/2 du 1

6u3/2 + C = 1

6 (2x2 + 1)3/2 + C

∫ 2

0

x√

2x2 + 1 dx =[16 (2x2 + 1)3/2

]20

= 133

28.

{u = 2x2 + 1

du = 4x dx

x = 0 ⇒ u = 1

x = 2 ⇒ u = 9

};

∣∣∣∣∣∫ 2

0

x

(2x2 + 1)2dx =

14

∫ 9

1

u−2 du =[− 1

4u

]91

=29

29.

{u = 1 + x−2

du = −2x−3 dx

};

∫x−3(1 + x−2)−3 dx = −1

2

∫u−3 du = 1

4u−2 + C = 1

4 (1 + x−2)−2 + C

∫ 2

0

x−3(1 + x−2)−3 dx =[14(1 + x−2)−2

]21

=39400

30.

{u = (x + 2)(x + 3)

du = (2x + 5)dx

x = 0 ⇒ u = 6

x = 1 ⇒ u = 12

};

∫ 1

0

2x + 5(x + 2)2(x + 3)2

dx =∫ 12

6

1u2

du =[− 1u

]126

=112

31.

{u = x + 1

du = dx

};

∫x√x + 1 dx =

∫(u− 1)

√u du =

∫(u3/2 − u1/2) du

= 25u

5/2 − 23u

3/2 + C = 25 (x + 1)5/2 − 2

3 (x + 1)3/2 + C

32.

{u = x− 1

du = dx

};

∫2x

√x− 1 dx =

∫2(u + 1)

√u du = 2

∫(u3/2 + u1/2) du

=45u5/2 +

43u3/2 + C =

45(x− 1)5/2 +

43(x− 1)3/2 + C

33.

{u = 2x− 1

du = dx

};

∫x√

2x− 1 dx =12

∫(u− 1)

2√u du =

14

∫(u3/2 + u1/2) du

=110

u5/2 +16u3/2 + C =

110

(2x− 1)5/2 +16(2x− 1)3/2 + C



SECTION 5.7 267

34.

{u = 2t + 3

du = 2 dt

};

∫t(2t + 3)8 dt =

12

∫12(u− 3)u8 du =

14

∫(u9 − 3u8) du

=140

u10 − 112

u9 + C =140

(2t + 3)10 − 112

(2t + 3)9 + C

35.∫

1√x√x + x

dx =∫

1√x√√

x + 1dx

{u =

√x + 1

du = dx/2√x

};

∫1

√x√√

x + 1dx = 2

∫u−1/2 du = 4

√u + C = 4

√√x + 1 + C

36.

{u = x2 + 1

du = 2x dx

x = −1 ⇒ u = 2

x = 0 ⇒ u = 1

};

∣∣∣∣∫ 0

−1

x3(x2 + 1

)6dx =

12

∫ 1

2

(u− 1)u6 du

=12

∫ 1

2

(u7 − u6) du =[

116

u8 − 114

u7

]12

= −25516

+12714

= −769112

37.

{u = x + 1

du = dx

x = 0 ⇒ u = 1

x = 1 ⇒ u = 2

};

∣∣∣∣∫ 1

0

x + 3√x + 1

dx =∫ 2

1

u + 2√u

du =∫ 2

1

(u1/2 + 2u−1/2) du =[23u3/2 + 4u1/2

]21

=163

√2 − 14

3

38. Set u = x− 1. Then du = dx, x = u + 1, x2 = u2 + 2u + 1; u(2) = 1, u(5) = 4.∫ 5

2

x2

√x− 1

dx =∫ 4

1

(u3/2 + 2u1/2 + u−1/2) du =[25u5/2 +

43u3/2 + 2u1/2

]41

=35615

39.

{u = x2 + 1

du = 2xdx

};

∫x√x2 + 1 dx =

12

∫ √u du =

13u3/2 + C =

13(x2 + 1)3/2 + C.

Also, 1 =13(02 + 1) + C =⇒ C =

23. Thus y =

13(x2 + 1)

32 +

23.

40.

⎧⎨⎩

u = 1 +√x

du =1

2√xdx

⎫⎬⎭ ; −

∫1

2√x(1 +

√x)2

dx = −∫

u−2 du =1u

+ C =1

1 +√x

+ C.

Also,13

=1

1 +√

4+ C =⇒ C = 0 Thus y =

11 +

√x

41.∫

cos (3x + 1) dx = −13

sin (3x + 1) + C 42.∫

sin 2πx dx = − 12π

cos 2πx + C

43.∫

csc2 πx dx = − 1π

cotπx + C 44.∫

sec 2x tan 2x dx =12

sec 2x + C



268 SECTION 5.7

45.

{u = 3 − 2x

du = −2 dx

};∫

sin (3 − 2x) dx =∫

− 12 sinu du = 1

2 cosu + C = 12 cos (3 − 2x) + C

46.

{u = sinx

du = cosx dx

};∫

sin2 x cosx dx =∫

u2 du =13u3 + C =

13

sin3 x + C

47.

{u = cosx

du = − sinx dx

};∫

cos4 x sinx dx =∫

−u4 du = −15u5 + C = −1

5cos5 x + C

48.

{u = x2

du = 2x dx

};∫

x sec2 x2 dx =12

∫sec2 u du =

12

tanu + C =12

tanx2 + C

49.

{u = x1/2

du = 12x

−1/2 dx

};

∫x−1/2 sinx1/2 dx =

∫2 sinu du = −2 cosu + C = −2 cosx1/2 + C

50.

{u = 1 − 2x

du = −2 dx

};∫

csc(1 − 2x) cot(1 − 2x) dx = −12

∫cscu cotu du =

12

cscu + C =12

csc(1 − 2x) +

C

51.

{u = 1 + sinx

du = cosx dx

};

∫ √1 + sinx cosx dx =

∫u1/2 du =

23u3/2 + C = 2

3 (1 + sinx)3/2 + C

52.

{u = 1 + cosx

du = − sinx dx

};

∫sinx√

1 + cosxdx = −

∫du√u

= −2√u + C = −2

√1 + cosx + C

53.

{u = sin πx

du = π cos πx dx

};∫

sin πx cos πx dx =1π

∫u du =

12π

u2 + C =12π

sin2 πx + C

54.

{u = sin πx

du = π cos πx dx

};∫

sin2 πx cos πx dx =1π

∫u2 du =

13π

u3 + C =13π

sin3 πx + C

55.

{u = cos πx

du = −π cos πx dx

};

∫cos2 πx sin πx dx = − 1

π

∫u2 du = − 1

3πu3 + C = − 1

3πcos3 πx + C

56.∫ (

1 + tan2 x)sec2 x dx =

∫sec2 x dx +

∫tan2 x sec2 x dx

= tanx +∫

u2 du = tanx +13u3 + C = tanx +

13

tan3 x + C

57.

{u = sin x2

du = 2x cos x2 dx

};∫

x sin3 x2 cos x2 dx =12

∫u3 du =

18u4 + C =

18

sin4 x2 + C



SECTION 5.7 269

58.

{u = sin (x2 − π)

du = 2x cos (x2 − π) dx

};

∫x sin4(x2 − π) cos (x2 − π) dx =

12

∫u4 du =

110

u5 + C =110

sin5(x2 − π) + C

59.

{u = 1 + tanx

du = sec2 x dx

};

∫sec2 x√1 + tanx

dx =∫

u−1/2 du = 2u1/2 + C = 2(1 + tanx)1/2 + C

60.

{u = 2 + cot 2x

du = −2 csc2 2x dx

};

∫csc2 2x√2 + cot 2x

dx = −12

∫u−1/2 du = −u1/2 + C = −

√2 + cot 2x + C

61.

{u = 1/x

du = −1/x2 dx

};∫

cos (1/x)x2

dx = −∫

cos u du = − sin u + C = − sin (1/x) + C.

62.

{u = 1/x

du = −1/x2 dx

};∫

sin (1/x)x2

dx = −∫

sin u du = cos u + C = cos (1/x) + C.

63.

{u = tan (x3 + π)

du = 3x2 sec2 (x3 + π) dx

};

∫x2 tan (x3 + π) sec2(x3 + π) dx =

13

∫u du =

16u2 + C =

16

tan2(x3 + π) + C

64.∫ (

x sin2 x− x2 sin x cos x)dx =

∫x sin x(sin x− x cos x) dx

{u = sin x− x cos x

du = x sin x dx

};

∫x sin x(sin x− x cos x) dx =

∫u du = 1

2 u2 + C = 1

2 (sin x− x cos x)2 + C

65.

{u = sin x

du = cos x dx

x = −π ⇒ u = 0

x = π ⇒ u = 0

};

∣∣∣∣∫ π

−π

sin4 x cos x dx =∫ 0

0

u4 du = 0.

66.∫ π/3

−π/3

secx tanx dx = [secx]π/3−π/3 = 0 67.∫ 1/3

1/4

sec2 πx dx =1π

[tanπx]1/31/4 =1π

(√

3 − 1)

68.∫ 1

0

cos2(π

2x)

sin(π

2x)dx =

−23π

[cos3

π

2x]10

=23π



270 SECTION 5.7

69.

{u = cos x

du = − sin x dx

x = 0 ⇒ u = 1

x = π/2 ⇒ u = 0

};

∣∣∣∣∫ π/2

0

sin x cos3 x dx = −∫ 0

1

u3 du =∫ 1

0

u3 du =[u4

4

]10

=14.

70.∫ π

0

x cosx2 dx =12[sinx2

]π0

=12

sinπ2

71.∫

sin2 x dx =∫

1 − cos 2x2

dx =12x− 1

4sin 2x + C

72.∫

cos2 dx =∫

1 + cos 2x2

dx =12x +

14

sin 2x + C

73.∫

cos2 5x dx =∫

1 + cos 10x2

dx =12x +

120

sin 10x + C

74.∫

sin2 3x dx =∫

1 − cos 6x2

dx =12x− 1

12sin 6x + C

75.∫ π/2

0

cos2 2x dx =∫ π/2

0

1 + cos 4x2

dx =[12x +

18

sin 4x]π/20

=π

4

76.∫ 2π

0

sin2 x dx =[12x− 1

4sin 2x

]2π0

= π

77. A =∫ π

2

0

[cosx− (− sinx)] dx = [sinx− cosx]π20 = 2

78. A =∫ 1/4

0

(cosπx− sinπx) dx =1π

[sinπx + cosπx]1/40 =1π

(√

2 − 1)

79. A =∫ 1/4

0

(cos2 πx− sin2 πx

)dx =

∫ 1/4

0

cos 2πx dx =12π

[sin 2πx]1/40 =12π

80.∫ 1/4

0

(cos2 πx + sin2 πx) dx =∫ 1/4

0

1 dx = [x]1/40 =14

81. A =∫ 1/4

1/6

(csc2 πx− sec2 πx) dx =[

1π

(− cotπx− tanπx)]1/41/6

=1π

(−2 + cot

π

6+ tan

π

6

)=

1π

(−2 +

√3 +

1√3

)=

13π

(4√

3 − 6)

82. (a)

{u = sinx

du = cosx dx

};∫

sin x cos x dx =∫

u du = 12u

2 + C = 12 sin2 x + C



SECTION 5.8 271

(b)

{u = cos x

du = − sin x dx

};∫

sin x cos x dx = −∫

u du = − 12u

2 + C ′ = − 12 cos2 x + C ′

(c) C ′ = C + 12

83. (a)

{u = secx

du = secx tanx dx

};∫

sec2 x tanx dx =∫

u du = 12u

2 + C = 12 sec2 x + C

(b)

{u = tanx

du = sec2 x dx

};∫

sec2 x tanx dx =∫

u du = 12u

2 + C ′ = 12 tan2 x + C ′

(c) C ′ = 12 + C

84. (a) Set u = x− c. Then dx = du; u(a + c) = a, u(b + c) = b.∫ b+c

a+c

f(x− c) dx =∫ b

a

f(u) du =∫ b

a

f(x) dx

(b) Set u = x/c. Then du = (1/c) dx; u(ac) = a, u(bc) = b.

1c

∫ bc

ac

f(x/c) dx =∫ b

a

f(u) du =∫ b

a

f(x) dx

85. A = 4∫ r

0

√r2 − x2 dx = 4

∫ π/2

0

√r2 − r2 sin2 u (r cosu) du (x = r sinu)

= 4∫ π/2

0

r2 cos2 u du = 4r2

[12u +

14

sin 2u]π/20

= πr2

86. A =4ba

∫ a

0

√a2 − x2 dx =

4ba

(area of circle of radius a

4

)=

4ba

(πa2

4

)= πab

SECTION 5.8

1. Yes;∫ b

a

[f(x) − g(x)] dx =∫ b

a

f(x) dx−∫ b

a

g(x) dx > 0.

2. No; take, for example, the function f(x) = x and g(x) = 0 on [− 12 , 1].

3. Yes; otherwise we would have f(x) ≤ g(x) for all x ∈ [a, b ] and it would follow that∫ b

a

f(x) dx ≤∫ b

a

g(x) dx.

4. No; take, for example, the function f(x) = 0 and g(x) = −1 on [0, 1].

5. No; take f(x) = 0 , g(x) = −1 on [ 0, 1 ].



272 SECTION 5.8

6. Yes;∫ b

a

|f(x)| dx ≥∫ b

a

f(x) dx and we are assuming that∫ b

a

f(x) dx >

∫ b

a

g(x) dx.

7. No; take, for example, any odd function on an interval of the form [−c, c].

8. Yes; if f(x) �= 0 for each x ∈ [a, b], then by continuity either f(x) > 0 for all x ∈ [a, b],

or f(x) < 0 for all x ∈ [a, b]. In either case∫ b

a

f(x) dx �= 0

9. No;∫ 1

−1

x dx = 0 but∫ 1

−1

|x| dx �= 0.

10. Yes;∣∣∣∣∫ b

a

f(x) dx∣∣∣∣ = |0| = 0 11. Yes; Uf (P ) ≥

∫ b

a

f(x) dx = 0.

12. No; if f(x) = 0 for all x ∈ [a, b], then

∫ b

a

f(x) dx = 0, and Uf (P ) = 0 for all P.

13. No; Lf (P ) ≤∫ b

a

f(x) dx = 0.

14. No; take f(x) = x on [−1.1]; ∫ 1

−1

x dx = 0 but∫ 1

−1

x2 dx =23

15. Yes;∫ b

a

[f(x) + 1] dx =∫ b

a

f(x) dx +∫ b

a

1 dx = 0 + b− a = b− a.

16.d

dx

[∫ b

u

f(t) dt]

=d

du

[∫ b

u

f(t) dt]du

dx=

d

du

[−∫ u

b

f(t) dt]du

dx= −f(u)

du

dx.

17.d

dx

[∫ 1+x2

0

dt√2t + 5

]=

1√2 (1 + x2) + 5

d

dx

(1 + x2

)=

2x√2x2 + 7

18.d

dx

[∫ x2

1

dt

t

]=

1x2

2x =2x.



SECTION 5.8 273

19.d

dx

[∫ a

x

f(t) dt]

=d

dx

[−∫ x

a

f(t) dt]

= −f(x)

20.d

dx

[∫ x3

0

dt√1 + t2

]=

3x2

√1 + x6

.

21.d

dx

[∫ 3

x2

sin t

tdt

]= − d

dx

[∫ x2

3

sin t

tdt

]= − sin(x2)

x2(2x) = − 2 sin(x2)

x

22.d

dx

[∫ 4

tanx

sin(t2) dt]

= − sin(tan2 x

)sec2 x.

23.d

dx

[∫ √x

1

t2

1 + t2dt

]=

x

1 + x· 12√x

=√x

2(1 + x)

24.d

dx

[∫ v

u

f(t) dt]

=d

dx

[∫ v

a

f(t) dt−∫ u

a

f(t) dt]

= f(v)dv

dx− f(u)

du

dx.

25.d

dx

[∫ x2

x

dt

t

]=

1x2

d

dx

(x2)− 1

x

d

dx(x) =

2xx2

− 1x

=1x

26.d

dx

[∫ x2+x

√x

dt

2 +√t

]=

12 +

√x2 + x

(2x + 1) − 12 + 4

√x· 12√x

27.d

dx

[∫ 2x

tanx

t√

1 + t2 dt

]= 2x

√1 + (2x)2 (2) − tanx

√1 + tan2 x (sec2 x)

= 4x√

1 + 4x2 − tanx sec2 x | secx|

28.d

dx

[∫ 1/x

3x

cos 2t dt

]= cos

(2x

) (−1x2

)− cos 6x (3) = −cos(2/x)

x2− 3 cos 6x

29. Set h(x) = g(x) − f(x) and apply (5.8.2) to h.

30. Suppose f(c) > 0 for some c ∈ (a, b). Then by Exercise 48, Section 2.4, there exists δ > 0

such that f(x) > 0 for all x ∈ (c− δ, c + δ). Also, we can choose δ such that (c− δ, c + δ) ⊂ (a, b).



274 SECTION 5.8

Then∫ b

a

|f(x)| dx ≥∫ c+δ

c−δ

|f(x)|dx > 0, a contradiction. The same holds if f(c) < 0 for some c.

Thus f(x) = 0 for all x ∈ (a, b). Then since f is continuous on [a, b], we must have

f(a) = f(b) = 0 , so f(x) = 0 for all x ∈ [a, b].

31. H(x) =∫ x3−4

2x

x dt

1 +√t

= x

∫ x3−4

2x

dt

1 +√t,

H ′(x) = x ·[

3x2

1 +√x3 − 4

− 21 +

√2x

]+ 1 ·

∫ x3−4

2x

dt

1 +√t,

H ′(2) = 2[

123

− 23

]+∫ 4

4

dt

1 +√t︸︷︷︸ =

203

= 0

32. H(x) =1x

∫ x

3

[2t− 3H ′(t)] dt,

H ′(x) =−1x2

∫ x

3

[2t− 3H ′(t)] dt +1x

[2x− 3H ′(t)] ,

H ′(3) =−132

∫ 3

3

[2t− 3H ′(t)] dt +13

[2 · 3 − 3H ′(3)]

=−132

· 0 + 2 −H ′(3) =⇒ H ′(3) = 1.

33. (a) Let u = −x. Then du = − dx; and u = 0 when x = 0, u = a when x = −a.∫ 0

−a

f(x) dx = −∫ 0

a

f(−u) du =∫ a

0

f(−u) du =∫ a

0

f(−x) dx

(b)∫ a

−a

f(x) dx =∫ 0

−a

f(x) dx +∫ a

0

f(x) dx = −∫ 0

a

f(u) du +∫ a

0

f(x) dx

∧u = −x, du = −dx

=∫ a

0

f(u) du +∫ a

0

f(x) dx =∫ a

0

[f(x) + f(−x)] dx

34. (a)∫ a

−a

f(x) dx =∫ 0

−a

f(x) dx +∫ a

0

f(x) dx

In first integral, use u = −x, du = −dx, u(−a) = a, u(0) = 0, x = −u, and note that

f(x) = f(−u) = −f(u) since f is odd. Then∫ 0

−a

f(x) dx = −∫ 0

a

f(−u) du =∫ a

0

f(−u) du = −∫ a

0

f(u) du

So∫ a

−a

f(x) dx = −∫ a

0

f(u) du +∫ a

0

f(x) dx = 0



SECTION 5.9 275

(b) As above, but now f(x) = f(−u) = f(u) since f is even, so∫ 0

−a

f(x) dx = −∫ 0

a

f(−u) du =∫ a

0

f(−u) du =∫ a

0

f(u) du,

hence∫ a

−a

f(x) dx = 2∫ a

0

f(x) dx

35.∫ π/4

−π/4

(x + sin 2x) dx = 0 since f(x) = x + sin 2x is an odd function.

36.t3

1 + t2is an odd function, so

∫ 3

−3

t3

1 + t2dt = 0

37.∫ π/3

−π/3

(1 + x2 − cosx) dx = 2∫ π/3

0

(1 + x2 − cosx) dx since f(x) = 1 + x2 − cosx is an even function.

2∫ π/3

0

(1 + x2 − cosx) dx = 2[x +

13x3 − sinx

]π/30

=23π +

281

π3 −√

3

38. 2x and sinx are odd, and x2 and cos 2x are even, so∫ π/4

−π/4

(x2 − 2x + sinx + cos 2x) dx = 2∫ π/4

0

(x2 + cos 2x) dx = 2[x3

3+

12

sin 2x]π/40

=π3

96+ 1

SECTION 5.9

1. A.V. =1c

∫ c

0

(mx + b) dx =1c

[m2x2 + bx

]c0

=mc

2+ b; at x = c/2

2. A.V. =12

∫ 1

−1

x2 dx =12

[x3

3

]1−1

=13; at x = ±

√3

3.

3. A.V. = 12

∫ 1

−1

x3 dx = 0 since the integrand is odd; at x = 0

4. A.V. =13

∫ 4

1

x−2 dx =13

[− 1x

]41

=13· 34

=14; at x = 2.

5. A.V. =14

∫ 2

−2

|x| dx =12

∫ 2

0

|x| dx =12

∫ 2

0

x dx =12

[x2

2

]20

= 1; at x = ±1

6. A.V. =116

∫ 8

−8

x1/3 dx = 0 (odd function); at; x = 0

7. A.V. =12

∫ 2

0

(2x− x2

)dx =

12

[x2 − x3

3

]20

=23; at x = 1 ± 1

3

√3



276 SECTION 5.9

8. A.V. =13

∫ 3

0

(3 − 2x) dx =13[3x− x2

]30

= 0; at x =32.

9. A.V. =19

∫ 9

0

√x dx =

19

[23x3/2

]90

= 2; at x = 4

10. A.V. =14

∫ 2

−2

(4 − x2) dx =14

[4x− x3

3

]2−2

=83; at x = ±2

√3

3.

11. A.V. =12π

∫ 2π

0

sinx dx =12π

[− cosx]2π0 = 0; at x = π

12. A.V. =1π

∫ π

0

cosx dx =1π

[sinx]π0 = 0; at x =π

2.

13. A.V. =1

b− a

∫ b

a

xn dx =1

b− a

[xn+1

n + 1

]ba

=bn+1 − an+1

(n + 1)(b− a).

14. (a) for constant f, f(b)(b− a) =∫ b

a

f(x) dx

(b) for increasing f, f(b)(b− a) >∫ b

a

f(x) dx

(c) for decreasing f, f(b)(b− a) <∫ b

a

f(x) dx

15. Average of f ′ on [a, b ] =1

b− a

∫ b

a

f ′(x) dx =1

b− a[f(x)]ba =

f(b) − f(a)b− a

.

16. (a) True, because∫ b

a

(f + g) dx =∫ b

a

f dx +∫ b

a

g dx.

(b) True, because∫ b

a

αf dx = α

∫ b

a

f dx.

(c) False; take f(x) = g(x) = x on [0, 1] : A.V. (f g) =13, (A.V.(f))(A.V.(g)) =

12· 12

=14.

(d) False; take f(x) = x2 and g(x) = x on [0,1]:

A.V.(f/g) = A.V.(x) =12,

A.V.(f)A.V.(g)

=1/31/2

=23.

17. Distance from (x, y) to the origin:√x2 + y2. Since y = x2, D(x) =

√x2 + x4.

On[0,√

3]

, A.V.=1√3

∫ √3

0

x√

1 + x2 dx =1√3

[13(1 + x2)3/2

]√3

0

=1

3√

37 =

79

√3.



SECTION 5.9 277

18. Distance from (x, y) to the origin:√

x2 + y2. Since y = mx, D(x) =√x2 + m2x2 = |x|

√1 + m2.

On [0, 1], A.V.=1

1 − 0

∫ 1

0

x√

1 + m2 dx =[

12

√1 + m2 x2

]10

=√

1 + m2

2

19. The distance the stone has fallen after t seconds is given by s(t) = 16t2.

(a) The terminal velocity after x seconds is s′(x) = 32x. The average velocity

iss(x) − s(0)

x− 0= 16x. Thus the terminal velocity is twice the average velocity.

(b) For the first 12x seconds, aver. vel. =

s(

12x)− s(0)

12x− 0

= 8x.

For the next 12x seconds, aver. vel. =

s(x) − s(

12x)

x− 12x

= 24x.

Thus, for the first 12x seconds the average velocity is one-third of the average velocity

during the next 12x seconds.

20. Obvious since∫ a

−a

f(x) dx = 0

21. Suppose f(x) �= 0 for all x in (a, b). Then, since f is continuous, either

f(x) > 0 on (a, b) or f(x) > 0 on (a, b).

In either case,∫ b

a

f(x) dx �= 0.

22.1

(a + 2n) − a

∫ a+2n

a

sinπx dx =12n

[− 1π

cosπx]a+2n

a

= − 12nπ

(cos(aπ + 2nπ) − cos aπ) = 0

Similarly for the average value of cosπx on [a, a + 2n].

23. (a) v(t) − v(0) =∫ t

0

a du; v(0) = 0. Thus v(t) = at.

x(t) − x(0) =∫ t

0

v(u) du; x(0) = x0. Thus x(t) =∫ t

0

au du + x0 =12at2 + x0.

(b) vavg =1

t2 − t1

∫ t2

t1

at dt =1

t2 − t1

[12at2]t2t1

=at22 − at212(t2 − t1)

=v(t1) + v(t2)

2

24. Let c be the point that divides the rod into two pieces of equal mass:∫ c

0

kx dx =∫ L

c

kx dx =⇒ 12kc

2 = 12kL

2 − 12kc

2 =⇒ kc2 = 12kL

2 =⇒ c =√

22

L



278 SECTION 5.9

25. (a) M =∫ 6

0

12√x + 1

dx = 12∫ 6

0

1√x + 1

dx = 24[√

x + 1]60

= 24(√

7 − 1)

xMM =∫ 6

0

12x√x + 1

dx = 12∫ 7

1

(u1/2 − u−1/2

)du

∧u = x + 1, du = dx, x = u− 1

= 12[23 u

3/2 − 2u1/2]71

= 16 + 32√

7;

xM =16 + 32

√7

24(√

7 − 1)=

4√

7 + 23√

7 − 3

(b) A.V.=16

∫ 6

0

12√x + 1

dx =16

[24(√

7 − 1)] = 4(√

7 − 1)

26.∫ b

a

(x− xM )λ(x) dx =∫ b

a

xλ(x) dx− xM

∫ b

a

λ(x) dx = xM M − xM M = 0

27. (a) M =∫ L

0

k√x dx = k

[23x3/2

]L0

=23kL3/2

xMM =∫ L

0

x(k√x)dx =

∫ L

0

kx3/2 dx =[25kx5/2

]L0

=25kL5/2

xM =(

25kL

5/2) / (

23kL

3/2)

= 35L

(b) M =∫ L

0

k (L− x)2 dx =[−1

3k (L− x)3

]L0

=13kL3

xMM =∫ L

0

x[k (L− x)2

]dx =

∫ L

0

k(L2x− 2Lx2 + x3

)dx

= k[12L

2x2 − 23Lx

3 + 14x

4]L0

= 112kL

4

xM =(

112kL

4)/(

13kL

3)

= 14L

28. xM M =∫ b

a

xλ(x) dx

=∫ x1

x0

xλ(x) dx +∫ x2

x1

xλ(x) dx + · · · +∫ xn

xn−1

xλ(x) dx

= xM1 M1 + xM2 M2 + · · · + xMnMn

29. 14LM = 1

8LM1 + xM2M2

xM2 =1M2

(14LM − 1

8LM1

)=

L

8M2(2M −M1)



SECTION 5.9 279

30. By Exercise 28, xM M = xM1 M1 + xM2 M2, so23LM =

14LM1 +

78LM2.

Also, M1 + M2 = M . Solving gives: M1 =13M, M2 =

23M .

31. Let M =∫ a+L

a kx dx, where a is the point of the first cut.

Thus M =[kx2

2

]a+L

a

=k

2(2aL + L2). Hence a =

2M − kL2

2kL, and a + L =

2M + kL2

2kL.

32. Yes. Suppose g(x) < 0 on [a, b]. Let m be the minimum value of f and M the maximum value of f on

[a, b]. Then

m ≤ f(x) ≤ M and mg(x) ≥ f(x)g(x) ≥ Mg(x) since g(x) < 0.

The proof now proceeds exactly as in the proof of Theorem 5.9.3, only the inequalities are reversed.

33. If f is continuous on [a, b ], then, by Theorem 5.2.5, F satisfies the conditions of the mean-value

theorem of differential calculus (Theorem 4.1.1). Therefore, by that theorem, there is at least

one number c in (a, b) for which

F ′(c) =F (b) − F (a)

b− a.

Then

∫ b

a

f(x) dx = F (b) − F (a) = F ′(c)(b− a) = f(c)(b− a).

34.

(min of f

on [c− h, c + h]

)≤

(average of f

on [c− h, c + h]

)≤

(max of f

on [c− h, c + h]

).

By continuity, as h → 0+

(min of f

on [c− h, c + h]

)→ f(c) and

(max of f

on [c− h, c + h]

)→ f(c).

By the pinching theorem the middle term must also tend to f(c).

35. If f and g take on the same average value on every interval [a, x], then

1x− a

∫ x

a

f(t) dt =1

x− a

∫ x

a

g(t) dt.

Multiplication by (x− a) gives ∫ x

a

f(t) dt =∫ x

a

g(t) dt.

Differentiation with respect to x gives f(x) = g(x). This shows that, if the averages are

everywhere the same, then the functions are everywhere the same.



280 SECTION 5.9

36. Partition [a, b] into n subintervals of equal lengthb− a

n, where P = {x0, . . . , xn}

and x∗i is a point from [xi−1, xi]. Then the average value of f on [a, b] is:

1b− a

∫ b

a

f(x) dx =1

b− alim

||P ||→0

[f(x∗

1)(b− a

n

)+ · · · + f(x∗

n)(b− a

n

)]

=(

1b− a

)limn→∞

b− a

n[f(x∗

1) + · · · + f(x∗n)]

= limn→∞

1n

[f(x∗1) + · · · + f(x∗

n)],

which is the limit of arithmetic averages of values of f on [a, b].

37. Let P = {a = x0, x1, x2, . . . , xn = b} be a partition of the interval [a, b]. Then∫ b

a

f(x) dx =∫ x1

x0

f(x) dx +∫ x2

x1

f(x) dx + · · · +∫ xn

xn−1

f(x) dx

By the mean-value theorem for integrals, there exists a number x∗i ∈ (xi−1, xi) such that∫ xi

xi−1

f(x) dx = f (x∗i ) (xi − xi−1) = f (x∗

i ) Δxi, i = 1, 2, . . . , n

Thus ∫ b

a

f(x) dx = f (x∗1) Δx1 + f (x∗

2) Δx2 + · · · + f (x∗n) Δxn

38. (a) A.V.=13

∫ 2

−1

(x3 − x + 1) dx =13

[x4

4− x2

2+ x

]2−1

=74

(b) x3 − x + 1 =74; at x ∼= 1.263

39. (a) A.V. =1π

∫ π

0

sinx dx =1π

[− cosx]π0 =2π

(b) sinx =2π

=⇒ x = 0.690 (c)

40. (a) A.V. =125π

∫ π/6

−π/4

2 cos 2x dx =125π

[sin 2x]π/6−π/4 =125π

[√3

2+ 1

]=

65π

(√

3 + 2) ∼= 1.426

(b) 2 cos 2x = 1.426 at x ∼= ±0.389



REVIEW EXERCISES 281

41. (a) f(x) = 0 at a ∼= −3.4743 and b ∼= 3.4743.

(b)

− 3 −2 − 1 1 2 3x

25

50y

(c)1

b− a

∫ b

a

(−x4 + 10x2 + 25) dx ∼= 36.0984;

solving f(c) = 36.0984 for c we get c ∼= ±2.9545, ±1.1274

42. (a) f(x) = 0 at a ∼= −1.8364 and b ∼= 1.8364.

(b)

− 1 1x

8

y

(c)1

b− a

∫ b

a

(−x4 + x2 + 8

)dx ∼= 6.8496;

solving f(c) = 6.8496 for c we get c ∼= ±1.2975

REVIEW EXERCISES

1.∫

x3 − 2x + 1√x

dx =∫

(x5/2 − 2√x + x−1/2) dx =

27x

72 − 4

3x

32 + 2x

12 + C

2.∫

(x3/5 − 3x5/3) dx =58x

85 − 9

8x

83 + C

3.

{u = 1 + t3

du = 3t2 dt

};

∫t2(1 + t3)10 dt =

13

∫u10 du = 1

33u11 + C = 1

33 (1 + t3)11 + C

4.∫

(1 + 2√x)2 dx =

∫(1 + 4

√x + 4x) dx = x + 8

3x3/2 + 2x2 + C

5.

⎧⎨⎩ u = t2/3 − 1

du =3

3t1/3dt

⎫⎬⎭ ;∫ (

t2/3 − 1)2

t1/3dt = 3

2

∫u2 du = 1

2u3 + C = 1

2 (t2/3 − 1)3 + C

6.

{u = x2 − 2

du = 2x dx

};

∫x√x2 − 2 dx =

12

∫u1/2 du =

13u3/2 + C =

13(x2 − 2)3/2 + C



282 REVIEW EXERCISES

7. Set u = 2 − x. Then du = −dx and x = 2 − u.∫x√

2 − x dx = −∫

(2 − u)u1/2 du =∫ (

u3/2 − 2u1/2)du

= 25 u

5/2 − 43 u

3/2 + C = 25 (2 − x)5/2 − 4

3 (2 − x)3/2 + C

8.

{u = 2 + 2x3

du = 6x2 dx

};

∫x2(2 + 2x3)4dx = 1

6

∫u4 du = 1

30u5 + C = 1

30 (2 + 2x3)5 + C

9.

⎧⎨⎩

u = 1 +√x

du =1√xdx

⎫⎬⎭ ;

∫(1 +

√x)5√

xdx =

∫2u5 du = 1

3u6 + C = 1

3 (1 +√x)6 + C

10.

{u = 1/x

du = −dx/x2

};

∫sin(1/x)

x2dx = −

∫sinu du = cos(1/x) + C

11.

{u = 1 + sinx

du = cosx dx

};

∫cosx√

1 + sinxdx =

∫1√udu = 2

√u + C = 2

√1 + sinx + C

12.∫

(sec θ − tan θ)2 dθ =∫ (

sec2 θ − 2 sec θ tan θ + tan2 θ)dθ

=∫ (

2 sec2 θ − 2 sec θ tan θ − 1)dθ = 2 tan θ − 2 sec θ − θ + C

13.∫

(tan 3θ − cot 3θ)2 dθ =∫

(tan2 3θ + cot2 3θ − 2) dθ =∫

(sec2 3θ + csc2 3θ − 4) dθ

=13

tan 3θ − 13

cot 3θ − 4θ + C

14.

{u = x2

du = 2x dx

};

∫x sin3 x2 cosx2dx = 1

2

∫sin3 u cosu du = 1

8 sin4 u + C = 18 sin4

(x2)2 + C

15.∫

11 + cos 2x

dx =∫

11 + 2 cos2 x− 1

dx = 12

∫1

cos2 xdx = 1

2

∫sec2 x dx =

12

tanx + C

16.∫

11 − sin 2x

dx =∫

11 − sin 2x

1 + sin 2x1 + sin 2x

dx =∫

(sec2 2x + sec 2x tan 2x) dx

= 12 tan 2x + 1

2 sec 2x + C

17.

{u = secπx

du = π secπx tanπx dx

};

∫sec3 πx tanπx dx =

∫sec2 πx(secπx tanπx) dx =

∫1πu2 du =

13π

u3 + C =13π

(secπx)3 + C

18.

{u = 1 + bx2

du = 2bx dx

};

∫ax√

1 + bx2 dx =a

2b

∫u1/2 du =

a

3bu3/2 + C =

a

3b(1 + bx2)3/2 + C




19. Set u = 1 + bx. Then du = b dx and x = 1b (u− 1).∫

ax√

1 + bx dx =a

b2

∫u1/2(u− 1) du =

a

b2

∫ (u3/2 − u1/2

)du

=a

b2

[25u5/2 − 2

3u3/2

]+ C =

2a5b2

(1 + bx)5/2 − 2a3b2

(1 + bx)3/2 + C

20. Set u = 1 + bx, du = b dx and x2 =1b2(u2 − 2u + 1

)∫

ax2√

1 + bx dx =a

b3

∫ (u2 − 2u + 1

)u1/2 du

=a

b3

(27u7/2 − 4

5u5/2 +

23u3/2

)+ C

=2a7b3

(1 + bx)7/2 − 4a5b3

(1 + bx)5/2 +2a3b3

(1 + bx)3/2 + C

21.

{u = 1 + g2(x)

du = 2g(x)g′(x)dx

};∫

g(x)g′(x)√1 + g2(x)

dx =12

∫1√udu =

√u + C =

√1 + g2(x) + C

22. Set u = g(x). Then du = g′(x) dx;∫

g′(x)g3(x)

dx =∫

u−3 du = −12g−2(x) + C =

12g2(x)

+ C

23.∫ 2

−1

(x2 − 2x + 3) dx =∫ 2

−1

x2 dx−∫ 2

−1

2x dx +∫ 2

−1

3 dx = 13

[x3]2−1

−[x2]2−1

+ 3 [x]2−1 = 9

24. Set u = 1 + x2. Then du = 2x dx, u(0) = 1, u(1) = 2.∫ 1

0

x

(x2 + 1)3dx =

12

∫ 2

1

u−3 du = −14

[u−2]21

=316

.

25. Set u = sin 2x. Then du = 2 cos 2x dx, u(0) = 0, u(π/4) = 1.

∫ π/4

0

sin3 2x cos 2xdx = 12

∫ 1

0

u3 du = 18

[u4]10

=18

26.∫ π/8

0

(tan2 2x + sec2 2x

)dx =

∫ π/8

0

(2 sec2 2x− 1

)dx =

[tan 2x− x

]π/80

= 1 − π

8

27. Set u = x3 + 3x− 6. Then du = 3(x2 + 1

)dx, u(0) = −6, u(2) = 8.

∫ 2

0

(x2 + 1)(x3 + 3x− 6)1/3 dx = 13

∫ 8

−6

u1/3 du = 14

[u4/3

]8−6

= 4 − 14(6)4/3

28. Set u = 1 + x1/3. Then du = 13x

−2/3 dx, u(1) = 2, u(8) = 3.∫ 8

1

(1 + x1/3

)2x2/3

dx = 3∫ 3

2

u2 du =[u3]32

= 19




29. (a)∫ 3

2

f(x) dx =∫ 3

0

f(x) dx−∫ 2

0

f(x) dx = −2

(b)∫ 5

3

f(x) dx =∫ 3

0

f(x) dx +∫ 5

3

f(x) dx−∫ 2

0

f(x) dx = 6;

(c) Mean-value theorem: there exists a c ∈ (3, 5) such that f(c) =12

∫ 5

3

f(x) dx = 4.

(d) If f(x) ≥ 0 on [2, 3], then∫ 3

2

f ≥ 0. But∫ 3

2

f = −2 < 0.

30.∫ 8

−2

g(x) dx =∫ 8

−2

[f(x) + 3] dx =∫ 8

−2

f(x) dx +∫ 8

−2

3 dx = 4 +[3x]8−2

= 4 + 30 = 34

31. A =∫ 1

−2

[(4 − x2) − (x + 2)

]dx =

∫ 1

−2

(−x2 − x + 2)dx =92

− 2 − 1 1 2x

1

2

3

4y

32. A =∫ 3

−2

[(4 − x2) − (−2 − x)

]dx =

∫ 3

−2

(6 + x− x2) dx =1256

− 2 − 1 1 2 3x

− 6

− 4

− 2

2

4

y

33. A =∫ 3

0

(3y − y2) dy =92

2 4 6 8 10x

1

2

3

4

y

34. A =∫ 4

0

√x dx +

∫ 6

4

(6 − x) dx;

A =∫ 2

0

(6 − y − y2) dy =223

1 2 3 4 5 6x

1

2

3

4y




35. A =∫ 1

0

[x3]dx +

∫ 2

1

(2 − x) dx =34

A =∫ 1

0

(2 − y − y1/3] dy =34

1 2x

1

y

36. A =∫ 2

−1

[14 (x2 − x4) − (−x− 1)

]dx

=∫ 2

−1

[1 + x + 1

4x2 − 1

4x4]dx =

185

− 1 1 2x

y

37.d

dx

(∫ x

0

dt

1 + t2

)=

11 + x2

38.d

dx

∫ x2

0

dt

1 + t2=

11 + (x2)2

(2x) =2x

1 + x4

39. Fix a number a. Then

d

dx

(∫ x2

x

dt

1 + t2

)=

d

dx

[∫ x2

a

dt

1 + t2−∫ x

a

dt

1 + t2

]=

2x1 + x4

− 11 + x2

40.d

dx

(∫ sinx

0

dt

1 − t2

)=

11 − sin2 x

cosx =1

cosx= secx

41.d

dx

(∫ cosx

0

dt

1 − t2

)=

11 − cos2 x

(− sinx) = − 1sinx

= − cscx

42. f ′(x) = x√

1 + x2 =⇒ f(x) =∫

x√

1 + x2 dx = 13 (x2 + 1)3/2 + C. Since γ passes through (0, 1), 1 =

13 + C. Therefore, C = 2

3 and y = 13 (x2 + 1)3/2 + 2

3 .




43. (a) Yes, at x = 0.

(b) F ′(x) =1

x2 + 2x + 2> 0 =⇒ F increases on (−∞,∞).

(c) F ′′(x) =−2(x + 1)x2 + 2x + 2

. The graph of F is concave up on (−∞,−1) and concave down on (−1,∞)

(d)

− 4 − 2 2 4x

− 2

− 1

1

y

44.∫ x

0

tf(t) dt = x sinx + cosx− 1 ⇒ xf(x) =d

dx

∫ x

0

tf(t) dt =d

dx(x sinx + cosx− 1) = x cosx

(a) πf(π) = −π; f(π) = −1 (b) For x �= 0, f(x) = cosx and f ′(x) = sinx.

45. favg =14

∫ 4

0

x√x2 + 9

dx =14

[√x2 + 9

]40

=12

46.∫ π

0

f(x) dx =∫ π

0

(x + 2 sinx) dx =[12x2 − 2 cosx

]π0

= 4 +π2

2;

favg =1

π − 0

∫ π

0

f(x) dx =1π

[4 +

π2

2

]=

4π

+π

2

47. favg =12π

∫ a+2π

a

cosx dx =[sinx

]a+2π

a= sin(a + 2π) − sin a = 0

48.∫ β

α

f(x)dx 49.∫ β

α

|f(x)|dx

50.12

[∫ β

α

|f(x)|dx +∫ β

α

f(x)dx]

51.12

[∫ β

α

|f(x)|dx−∫ β

α

f(x)dx]




52. λ(x) = k(2a− x), k > 0. for some positive number k.

M =∫ a

0

k(2a− x)dx = k[2ax− 1

2x2]a0

= 32ka

2

xMM =∫ a

0

kx(2a− x)dx = k[ax2 − 1

3x3]a0

= 23ka

3; xM =23ka

3

32ka

2=

4a9

.

53. λ(x) = k(

14a− x

)for 0 ≤ x ≤ 1

4a and λ(x) = k(x− 14a) for 1

4a ≤ x ≤ a; k > 0.

M =∫ a/4

0

k(

14a− x

)dx +

∫ a

a/4

k(x− 1

4a)dx = 5

16ka2

xMM =∫ a/4

0

kx(

14a− x

)dx +

∫ a

a/4

kx(x− 1

4a)dx =

41192

ka3; xM =41192ka

3

516ka

2=

41a60

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Calculus one and several variables 10E Salas solutions manual ch05