Ch05 - Calculus (One and Several Variables) by Salas, Hille, Etgen Solution Manual PDF

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Calculus (One and Several Variables) by Salas, Hille, Etgen Solution Manual Chapter 5...

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SECTION 5.2 CHAPTER 5 SECTION 5.2 1. Lf (P ) = 0( 14 ) + 21 ( 41 ) + 1( 12 ) = 85 ,

Uf (P ) = 21 ( 41 ) + 1( 14 ) + 2( 21 ) =

11 8

5 97 , 2. Lf (P ) = 23 ( 31 ) + 41 ( 12 ) + 0( 41 ) + (−1)(1) = − 144 5 Uf (P ) = 1( 13 ) + 32 ( 12 ) + 14 ( 41 ) + 0(1) =

3. Lf (P ) = 14 ( 21 ) + 4. Lf (P ) =

97 144

1 1 16 ( 4 ) +

0( 41 ) =

9 , 64

Uf (P ) = 1( 12 ) + 41 ( 41 ) +

15 ( 1 ) + 3 ( 1 ) + 16 4 4 4

0( 21 ) =

27 , 64

Uf (P ) = 1( 14 ) +

5. Lf (P ) = 1( 12 ) + 89 ( 21 ) =

4 9 3 7 2 5 5 (25 ) + 5 ( 25 ) + 5 ( 25 )

=

9 5 7 1 3 ) + 1( 25 )= ) + 45 ( 25 ) + 53 ( 25 ) + 52 ( 25 Uf (P ) = 51 ( 25

7. Lf (P ) =

1 3 16 ( 4 ) +

8. Lf (P ) =

9 1 1 1 16 ( 4 ) + 16 ( 2 ) +

Uf (P ) = 1( 14 ) +

0( 21 ) +

1 1 1 1 16 ( 4 ) + 4 ( 2 )

0( 21 ) +

=

37 64

15 1 3 1 16 ( 4 ) + 4 ( 2 )

=

55 64

25 Uf (P ) = 89 ( 21 ) + 2( 12 ) = 16

17 16 ,

1 ) + 1( 3 ) + 6. Lf (P ) = 0( 25 5 25

1 1 16 ( 4 )

1 1 1 1 16 ( 4 ) + 4 ( 2 )

1 1 1 1 9 1 16 ( 2 ) + 16 ( 2 ) + 4 ( 4 ) +

=

1(21) =

      9. Lf (P ) = 0 π6 + 21 π3 + 0 π2 = 6π ,

19 25

Uf (P ) = 1( 34 ) +

3 , 16

=

14 , 25

1( 12 ) =

5 , 16

9 8

Uf (P ) =

10. Lf (P ) = 12 ( 3π ) + 0( π6 ) + (−1)( π2 ) = − π3 ,

1 1 1 1 16 ( 2 ) + 4 ( 4 ) +

 

1 π 2 6

    + 1 π3 + 1 π2 =

Uf (P ) = 1( π3 ) + 12 ( π6 ) + 0( π2 ) =

11π 12 5π 12

11. (a) Lf (P ) ≤ Uf (P ) but 3 ≤ 2.  1 (b) Lf (P ) ≤ f (x) dx ≤ Uf (P ) but 3 ≤ 2 ≤ 6. (c) Lf (P ) ≤



−1 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 1 xi−1 + 3 ≤ (xi−1 + xi ) + 3 ≤ xi + 3 2 Multiplying by Δxi = xi − xi−1 gives

2  1 (xi−1 + 3)Δxi ≤ x2i − xi−1 + 3(xi − xi−1 ) ≤ (xi + 3)Δxi . 2 that Summing from i = 1 to i = n, we find   2 2 1  Lf (P ) ≤ x12 − x 0 + 3(x1 − x0 ) · · · + 1 xn2 − xn−1 + 3(xn − xn−1 ≤ Uf (

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No

SECTION 5.2 The middle sum collapses to  1 1 2 2 xn − x0 + 3(xn − x0 ) = (b2 − a2 ) + 3(b − a) 2 2 Thus  b 1 (x + 3)dx = (b2 − a2 ) + 3(b − a) 2 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 ) (b) For each index i   −3xi ≤ − 32 xi + xi−1 ≤ −3xi−1 .

Multiplying by Δxi = xi − xi−1 gives

  2 ≤ −3xi−1 Δxi . −3xi Δxi ≤ − 32 xi 2 − xi−1

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

    Lf (P ) ≤ − 32 x1 2 − x0 2 − · · · − 23 xn 2 − x2n−1 ≤ Uf (P ).

The middle sum collapses to

  −23 xn 2 − x0 2 = − 23 (b2 − a2 ).

Thus

3 Lf (P ) ≤ − (b2 − a2 ) ≤ Uf (P ) so that 2



b a

3 −3x dx = − (b2 − a2 ). 2

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 ) (b) For each index i 1 + 2xi−1 ≤ 1 + (xi−1 + xi ) ≤ 1 + 2xi Multiplying by Δxi = xi − xi−1 gives   (1 + 2xi−1 ) Δxi ≤ (xi − xi−1 ) + xi 2 − x2i−1 ≤ (1 + 2xi ) Δxi .

Proceeding as before, we get

a 15.

−1 2

(x2 + 2x − 3) dx

b

(1 + 2x) dx = (b − a) + (b2 − a2 )

16.

0

(x3 − 3x) dx 3

√

17.



2π 0

2

t sin(2t + 1) dt

18.



1

4

t dt t2 + 1

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SECTION 5.2 19.

20.

21. Δx1 = Δx2 = 81 , m1 = 0, m2 =

1 4

Δx3 = Δx4 = Δx5 =

1 , 4

m3 = 21,

m4 = 1, m5 =

3 2

f (x1∗ ) = 18 , f (x∗2 ) = 83 , f (x∗3 ) = 43 , f (x∗4 ) = 54 , f (x5∗ ) = M1 = 41 , M 2 = 21, (a) Lf (P ) = 22.



25 32

M 3 = 1, M4 = (b) S ∗ (P ) =

15 16

3 , 2

3 2

M5 = 2

(c) Uf (P ) =

39 32

1

2x dx = 1.

0

23.

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

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

For each index i

  3 2 + xi−1 ≤ xi 3 x3i−1 ≤ 41 xi 3 + xi 2 xi−1 + xi xi−1

and thus by the hint

  x3i−1 (xi − xi−1 ) ≤ 41 xi 4 − x4i−1 ≤ xi 3 (xi − xi−1 ).

Adding up these inequalities, we find that

  Lf (P ) ≤ 14 xn 4 − x0 4 ≤ Uf (P ).  1 1 1 x3 dx = . Since xn = 1 and x0 = 0, the middle term is : 4 4 0 24. (a) Lf (P ) = x0 4 (x1 − x0 ) + x1 4 (x2 − x1 ) + · · · + xn−1 4 (xn − xn−1 ), Uf (P ) = x1 4 (x1 − x0 ) + x2 4 (x2 − x1 ) + · · · + xn 4 (xn − xn−1 )

(b) For each index i xi−1 4

3

2

5

2

3

xi + xi xi−1 + xi xi−1 + xi xi−1 + xi−1 ≤ Multiplying by Δxi = xi − xi−1 gives 4

4

≤ xi 4

 5 xi−1 4 Δxi ≤ 1  ≤ xi 4 Δxi . 5 xi − xi−1 5

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No

SECTION 5.2 Summing and collapsing the middle sum gives  1 Lf (P ) ≤ xn 5 − x0 5 ≤ Uf (P ), 5 Thus  1 1 1 x4 dx = (15 − 05 ) = . 5 5 0

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. b  1 Then a f (x) dx = 4 and ab g(x) dx = 3, but Lg (P ) = 3 and Lf (P ) = 2. 2 b b 27. Necessarily holds: Lg (P ) ≤ a g(x) dx < 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. b b 1 Then a f (x) dx = 6 and a g(x) dx = 4 , but Ug (P ) = 7 and Uf (P ) = 6. 2 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 , x 2 , . . . , 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. x > 0 for x ∈ [0, 2]. Thus, f is increasing on [0, 2]. 33. (a) f ′ (x) = √ 1 + x2 (b) Let P = {x0 , x1 , . . . , x n } be a regular partition of [0, 2] and let Δx = 2/n By Exercise 30,

0 2

f (x) dx − Lf (P ) ≤ |f (2) − f (0)|

It now follows that = 2.96 0 2 ∼ f (x) dx

0

2

f (x) dx − Lf (P ) < 0.1

if

√ 2( 5 − 1) ∼ 2.47 2 = = n n n n > 25.

(c) −2x 34. (a) f ′ (x) = (1 + x2 )2 < 0 on (0, 1)

⇒

f is decreasing.

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SECTION 5.2 (b) Uf (P ) − so need

1 0

f (x) dx ≤ |f (1) − f (0)|Δx = | 21 − 1| n1 =

1 . 2n

1 = 0.05, or n = 10. 2n

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



1

0

1 ∼ 0.78 dx = 1 + x2

35. Let S be the set of positive integers for which the statement is true. Since that k ∈ S. Then

1(2) = 1, 1 ∈ 2

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

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

Thus, 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 , . . . , x n } be a regular partition of [0, b]. Then Δx ib xi = , i = 0, 1, 2, . . . , n. n (a) Since f is increasing on [0, b],        b b 2b (n − 1)b Lf (P ) = f (0) + f +f + ···+ f n n n n   2b b (n − 1)b b = 0+ + + ···+ n n n n b2 [1 + 2 + · · · + (n − 1)] n2         b 2b (n − 1)b b Uf (P ) = f +f + ···+ f + f (b) n n n n   b b 2b (n − 1)b +b + = + ···+ n n n n =

(b)

=

b2 [1 + 2 + · · · + (n − 1) + n] n2

(c) By Exercise 35,  1 2 = b 1− 2    1 2 b2 n(n + 1) 1 2 n2 + n = b 1+ b = Lf (P ) = 2 · 2 n 2 2 n2

b2 (n − 1)n 1 2 = b Lf (P ) = 2 · 2 n 2



n2 − n n2



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

1 n 1 n





=

1 2 b (1 − ||P ||) 2

=

1 2 b (1 + ||P ||) 2

lim L (P ) = lim f

||P ||→0

lim ( )

1 b2 by the pinching theorem

1 2 Uf (P ) = 2 b , ||P ||→0

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No

SECTION 5.2

38. Let f (x) = x2 and let P = {x0 , x1 , x2 , . . . , x n } be a regular partition of [0, b]. Then Δx = b/n an ib xi = , i = 0, 1, 2, . . . , n. n (a) Since f is increasing on [0, b],        b b 2b (n − 1)b Lf (P ) = f (0) + f +f + ···+ f n n n n   b2 (n − 1)2 b2 b 4b2 = 0 + 2 + 2 + ···+ n n n n =

b3 [1 + 22 + · · · + (n − 1)2 ] n3

        2b (n − 1)b b b +f + ···+ f + f (b) Uf (P ) = f n n n n   2 b 4b2 n2 b2 b = + 2 + ···+ 2 2 n n n n

(b)

=

b3 [1 + 22 + · · · + n2 ] n3

(c) By Exercise 36, Lf (P ) =

b3 (n − 1)n(2n − 1) = b3 · 6 n3

b3 n(n + 1)(2n − 1) Uf (P ) = 3 · = b3 n 6





2n3 − 3n2 + n 6n3 2n3 + 3n2 + n 6n3





=

1 3 b = (2 − 3||P || + ||P ||2 ) 6

=

1 3 b = (2 + 3||P || + ||P ||2 ) 6

(d) For any partition P, Lf (P ) ≤ § ∗ (P ) ≤ Uf (P ). Since 1 lim Lf (P ) = lim Uf (P ) = b3 , ||P ||→0 3

||P ||→0

1 lim S ∗ (P ) = b3 by the pinching theorem. 3

||P ||→0

39. Choose each x∗i so that f (x∗i ) = mi . Then Si∗(P ) = Lf (P ). Similarly, choosing each x∗i so that f (x∗i ) = Mi gives S ∗i (P ) = Uf (P ). 1 Also, choosing each x∗i so that f (x∗i ) = (mi + Mi ) (they exist by the intermediate value theorem 2 gives 1 1 Si∗ (P ) = (m1 + M1 )Δx1 + · · · + (mn + Mn )Δxn 2 2 1 = [m1 Δx1 + · · · + mn Δxn + M1 Δx1 + · · · + Mn Δxn ] 2 1 = [Lf (P ) + Uf (P )]. 2

181 = 7.24, U 8.84 221 = ∼ ∼ 40. (a) Lf (P ) = f (P ) = 25 25 1 = 7.98 (c) S ∗ (P ) ∼ ∼ 8.04 (b) [L (P ) + U (P )] 402 =

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SECTION 5.3 ∼ 0.7105 U f (P ) =

∼ 0.6105, 41. (a) Lf (P ) = (b)

1 ∼ 0.6605 [Lf (P ) + Uf (P )] = 2

∼ 1.1824 U f (P ) =

∼ 1.0224, 42. (a) Lf (P ) = (b)

1 ∼ 1.1024 [Lf (P ) + Uf (P )] = 2

∼ 0.53138, 43. (a) Lf (P ) = (b)

(c) S ∗ (P ) ∼ = 0.6684

(c) S ∗ (P ) ∼ = 1.1074

Uf (P ) ∼ = 0.73138

1 ∼ 0.63138 [Lf (P ) + Uf (P )] = 2

= 0.63926 (c) S ∗ (P ) ∼

SECTION 5.3 1. (a)

(b)

(c)

  

5

f (x) dx = 0

(f)

2. (a)

(b)

(c)

(d)

(e)

     

2



2



5

f (x) dx = 1

f (x) dx = 1

0

(e)



2

f (x) dx = 4



f (x) dx = −

3

f (x) dx = 1

f (x) dx = 3

f (x) dx = −4

0

8





4

f (x) dx = 11 − 5 = 6

1

4

f (x) dx = −7

3

1



2

f (x) dx = 1





f (x) dx = −

f (x) dx = 5 − 6 = −1

0

5

f (x) dx −

4

4

f (x) dx = 4 − 6 = −2

0



8

1

8

f (x) dx = 4 + 1 = 5

1



f (x) dx = 0 (f)



1

3

 44

1

f (x) dx −



f (x) dx = −

f (x) dx = −

8

8



1

f (x) dx −

0

1

4

5

2

0

5

5



f (x) dx +

0

2

(d) 0





4

f (x) dx −



3

f (x) dx −

4

f (x) dx = 5 − 7 = −2

3

1

f (x) dx = 11 − (−2) = 13

8

f (x) dx = −6

3. With P =

1,

3 2

,2



1 , we have and f (x) = x 7 05≤

dx 

1

2

U (P )

5 < 1. 6

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No

SECTION 5.3

4. Using P = {0, 21 , 1}, we have 0.6 < 0.65 = Lf (P ) ≤

(d) F (2) =

6. (a) F (π) =



(c) F ′ ( π2 ) =

(e) −F (x) = 7. F ′ (x) =

1

1 dx ≤ Uf (P ) = 0.9 < 1. 1 + x2

0

√ (b) F ′ (x) = x x + 1

5. (a) F (0) = 0 



√ (c) F ′ (2) = 2 3

√ t t + 1 dt

2

(e) −F (x) =

0 π

sin π2 =



0

√ t t + 1 dt

x

(b) By Theorem 5.3.5, F ′ (x) = x sin x.  2π t sin t dt (d) F (2π) =

t sin t dt = 0 π

π 2



π 2

π

π

t sin t dt.

x

1 ; x2 + 9

(a)

1 10

√ 8. F ′ (x) = − x2 + 1

√ (a) − 2

√ 9. F ′ (x) = −x x2 + 1;

(a)

(b)

√ 2

1 9

(c)

4 37

(b) −1

√ (c) − 12 5

(b) 0

√ (c) − 14 5

(d)

−2x + 9) 2

(x2

−x (d) √ x2 + 1  √ x2 x2 + 1 + √ (d) − x2 + 1

10. F ′ (x) = sin πx

(a) 0

(b) 0

(c) 1

(d) π cos πx

11. F ′ (x) = cos πx;

(a) −1

(b) 1

(c) 0

(d) −π sin πx

12. F ′ (x) = (x + 1) 3

(a) 0

(b) 1

(c)

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.

27 8

(d) 3(x + 1) 2

(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 become  b f (x) dt = f (x) dt, trivially true. If c < a, we get a



a

b

f (t) dt +  ac



c

b

f (t) dt = −



c

a

f (t) dt +



b

c

The other possible cases are proved in a similar manner.

x−1

f (t) dt =



b

f (t) dt, as desired a

17.

F ′ (x) =

= 0 =⇒ x = 1 is a critical number. 1 + x2 2 (1 + x ) − 2x(x − 1) 1 > 0 means x = 1 is a local minimum. F ′′ (x) = so F ′′ (1) = 2

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SECTION 5.3 18.

F ′ (x) =

x−4 =0 1 + x2

F ′′ (x) =

1 (1 + x2 ) − 2x(x − 4) , so F ′′ (4) = > 0 means x = 4 (1 + x2 )2 17

=⇒

x = 4 is a critical number.

is a local minimum. 1 > 0 for x > 0. x Thus, F is increasing on (0, ∞);

1 < 0 for x > 0. x2 The graph of F is concave down on

19. (a) F ′ (x) =

(b) F ′′ (x) = −

there are no critical numbers.

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)

y 6 4 2 −1

1

2

3

x

21. (a) F is differentiable, therefore continuous (c) F ′ (1) = f (1) = 0

(b) F ′ (x) = f (x) f is differentiable; F ′′ (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 (c) G′ (1) = g(1) = 0

(b) G′ (x) = g (x) and g is differentiable; G′ (d) G′′ (x) = g ′ (x) < 0 for x < 1

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

G′′ (x) = g ′ (x) > 0 for x > 1

P1: PBU/OVY JWDD027-05

P2: PBU/OVY

QC: PBU/OVY

T1: PBU

JWDD027-Salas-v1

238

No

SECTION 5.3

23. (a)

(b)

y

y

4 6 2

4

−1

1

2

3

2

x −1

F (x) =

⎧ ⎨ 2x − 1 x2 + 5 2 2 ⎩

2x +