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CHAPTER

4

The Definite Integral
5.

4.1 Concepts Review
1. 2 ⋅
5(6) = 30; 2(5) = 10 2

m =1

∑ (−1)m 2m−2

8

= (−1)1 2−1 + (−1) 2 20 + (−1)3 21 +(−1) 4 22 + (−1)5 23 + (−1)6 24 +(−1)7 25 + (−1)8 26 1 = − + 1 − 2 + 4 − 8 + 16 − 32 + 64 2 85 = 2

2. 3(9) – 2(7) = 13; 9 + 4(10) = 49 3. inscribed; circumscribed 4. 0 + 1 + 2 + 3 = 6

Problem Set 4.1
1.

k =1

∑ (k − 1) = ∑k − ∑ 1
k =1 k =1

6

6

6

6.

(−1)k 2k k =3 ( k + 1)

∑

7

=

6(7) − 6(1) 2 = 15 =

(−1)3 23 (−1) 4 24 + 4 5 +

2.

∑ i2 =
i =1 7

6

6(7)(13) = 91 6

(−1)5 25 (−1)6 26 (−1)7 27 + + 6 7 8 1154 =− 105

1 1 1 1 3. ∑ = + + k =1 k + 1 1 + 1 2 + 1 3 + 1

7.

n =1

∑ n cos(nπ) = ∑ ( −1)
n =1

6

6

n

⋅n

1 1 1 1 + + + 4 +1 5 +1 6 +1 7 +1 1 1 1 1 11 1 = + + + + + + 2 3 4 5 6 7 8 1443 = 840 481 = 280 +
4.

= –1 + 2 – 3 + 4 – 5 + 6 =3
8.
⎛ kπ ⎞ k sin ⎜ ⎟ ⎝ 2 ⎠ k =−1

∑

6

l =3

∑ (l + 1)2 = 42 + 52 + 62 + 72 + 82 + 92 = 271

8

⎛ π⎞ ⎛π⎞ = − sin ⎜ − ⎟ + sin ⎜ ⎟ + 2sin(π) ⎝ 2⎠ ⎝2⎠ 3π ⎞ ⎛ ⎛ 5π ⎞ +3sin ⎜ ⎟ + 4sin(2π) +5sin ⎜ ⎟ + 6sin(3π) ⎝ 2 ⎠ ⎝ 2 ⎠ =1+1+0–3+0+5+0 =4

9. 1 + 2 + 3 +

+ 41 = ∑ i
i =1

41

10. 2 + 4 + 6 +8 +

+ 50 = ∑ 2i
i =1

25

11. 1 +

1 1 + + 2 3

+

1 100 1 =∑ 100 i =1 i

Instructor’s Resource Manual Section 4.1 249 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing fromthe publisher.

1 1 1 12. 1 − + − + 2 3 4

−

1 100 (−1)i +1 =∑ 100 i =1 i + a99 = ∑ a2i −1
i =1 50

20.

∑ [(i − 1)(4i + 3)]
= ∑ (4i 2 − i − 3) = 4∑ i 2 − ∑ i − ∑ 3
i =1 i =1 i =1 i =1 10 10 10 i =1 10

10

13. a1 + a3 + a5 + a7 + 14.

f ( w1 )Δx + f ( w2 )Δx +
= ∑ f ( wi )Δx
i =1 n

+ f ( wn )Δx

= 4(385) – 55 – 3(10) = 1455
21.

15.

∑ (ai + bi )
= ∑ ai + ∑ bii =1 i =1 i =1 10 10

10

k =1

∑ (k 3 − k 2 ) =

10

k =1

∑ k3 −∑ k2
k =1

10

10

= 3025 − 385 = 2640

= 40 + 50 = 90

22.

k =1

∑ 5k 2 (k + 4) = ∑ (5k 3 + 20k 2 )
k =1 10 10 k =1 k =1

10

10

16.

n =1

∑ (3an + 2bn )
10 10 n =1 n =1

10

= 5 ∑ k 3 + 20 ∑ k 2

= 3∑ an + 2 ∑ bn

= 5(3025) + 20(385) = 22,825
23.

= 3(40) + 2(50) = 220
17.

∑(2i 2 − 3i + 1) = 2∑ i 2 − 3∑ i + ∑1
i =1 i =1 i =1 i =1

n

n

n

n

=
p =0

∑ (a p +1 − b p +1 ) ∑ a p − ∑ bp
p =1 p =1 10 10

9

2n(n + 1)(2n + 1) 3n(n + 1) − +n 6 2
2n3 + 3n 2 + n 3n 2 + 3n − +n 3 2 4n3 − 3n 2 − n 6
n

= =
n

=

= 40 − 50 = –10

24.

∑ (2i − 3)2 = ∑ (4i 2 − 12i + 9)
i =1

18.

q =1

∑ (aq − bq − q)
q =1

10

= 4∑ i 2 − 12∑ i + ∑ 910

n

i =1 n

n

=

∑ aq − ∑ bq − ∑ q
q =1 q =1

10

10

i =1

i =1

i =1

= =

4n(n + 1)(2n + 1) 12n(n + 1) − + 9n 6 2 4n3 − 12n 2 + 11n 3

= 40 − 50 − = −65
100

10(11) 2

25.

19.

∑ (3i − 2)
i =1 100 i =1 100 i =1

S = 1 + 2 + 3 + + (n − 2) + (n − 1) + n + S = n + (n − 1) + (n − 2) + + 3 + 2 + 1 2S = (n + 1) + (n + 1) + (n + 1) + 2S = n(n + 1) n(n + 1)S= 2 + (n + 1) + (n + 1) + (n + 1)

= 3∑ i − ∑ 2 = 3(5050) − 2(100) = 14,950

250 Section 4.1 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.

26. S − rS = a + ar + ar 2 +
− (ar + ar 2 +

+ ar n
+ ar n + ar n +1 )
n +1

27. a.

= a − ar n +1
a − ar = S (1 − r ); S = 1− r

1− ⎛1⎞ ∑ ⎜ 2 ⎟ = 12 k =0 ⎝ ⎠ 2
10 k

(1)

11

⎛1⎞ = 2 − ⎜ ⎟ , so ⎝2⎠

10

⎛1⎞ ⎛1⎞ ∑ ⎜ 2 ⎟ = 1− ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠ k =1
b.

10

k

10

=

1023 . 1024

k =0 10 k =1

∑ 2k =

10

1 − 211 = 211 − 1, so −1

∑ 2k = 211 − 2 =...