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Thomas calculus, 12th edition, solution manual, chapter 5
Calculus Of Several Variables (Math 216)
جامعة المنصورة
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CHAPTER 5 INTEGRATION
5 AREA AND ESTIMATING WITH FINITE SUMS
1. f xa b œ x# Since f is increasing on Ò!ß "Ó, we use left endpoints to obtain
lower sums and right endpoints to obtain upper sums.
(a) ̃x œ " !# œ "# and x œ ̃i x œ # Ê a lower sum is # † "# œ "# ! "# œ")
œ!
"
i i i #
i
!ˆ ‰ Š ˆ ‰‹
(b) ̃x œ " !% œ "% and x œ ̃i x œ % Ê a lower sum is † " œ " ! " "# $ œ "% † () œ$#(
œ!
$
i i i # # #
i
!ˆ ‰ 4 4 4 Š ˆ ‰ 4 ˆ ‰ ˆ ‰ 4 ‹
(c) ̃x œ " !# œ "# and x œ ̃i x œ # Ê an upper sum is # † "# œ "# "# +1 œ&)
œ
# #
i
i i
i 1
2
!ˆ ‰ Šˆ ‰ ‹
(d) ̃x œ " !% œ "% and x œ ̃i x œ % Ê an upper sum is † " œ " " "# $ +1 œ "% † $!"' œ"&$#
œ"
% # # # #
i i i
i
!ˆ ‰ 4 4 4 Šˆ ‰ 4 ˆ ‰ ˆ ‰ 4 ‹ ˆ ‰
2. f xa b œ x$ Since f is increasing on Ò!ß "Ó, we use left endpoints to obtain
lower sums and right endpoints to obtain upper sums.
(a) ̃x œ " !# œ "# and x œ ̃i x œ # Ê a lower sum is # † "# œ "# ! "# œ"'"
œ!
" $ $ $
i i i
i
!ˆ ‰ Š ˆ ‰‹
(b) ̃x œ " !% œ "% and x œ ̃i x œ % Ê a lower sum is † " œ " ! " "# $ œ #&'$' œ'%*
œ!
$ $
i i i $ $ $ $
i
!ˆ ‰ 4 4 4 Š ˆ ‰ 4 ˆ ‰ ˆ ‰ 4 ‹
(c) ̃x œ " !# œ "# and x œ ̃i x œ # Ê an upper sum is # † "# œ "# "# +1 œ "# † ) œ"'
œ
$ $ $
i i i
i 1
2
!ˆ ‰ Šˆ ‰ ‹
(d) ̃x œ " !% œ "% and x œ ̃i x œ % Ê an upper sum is † " œ " " "# $ +1 œ œ "!!#&' œ#&'%
œ"
% $ $ $ $
i i i $
i 4 4 4 4 4
!ˆ ‰ Šˆ ‰ ˆ ‰ ˆ ‰ ‹
258 Chapter 5 Integration
3. f xa b œ " x Since f is decreasing on 1Ò ß 5 , we use left endpoints to obtainÓ
upper sums and right endpoints to obtain lower sums.
(a) ̃x œ & "# œ # and x œ " ̃i x œ " # Êi a lower sum is " † # œ # "$ "& œ"'"&
####### œ"
####### #
####### i
####### i
! x i ˆ ‰
(b) ̃x œ & "% œ 1 and x œ " ̃i x œ " i Ê a lower sum is " † " œ " "# "$ "% "& œ(('!
####### œ"
####### %
####### i
####### i
! x i ˆ ‰
(c) ̃x œ & "# œ # and x œ " ̃i x œ " # Êi an upper sum is " † # œ # " "$ œ)$
####### œ!
####### "
####### i
####### i x
! ˆ ‰
i
(d) ̃x œ & "% œ 1 and x œ " ̃i x œ " i Ê an upper sum is " † " œ " " "# "$ "% œ#&"#
####### œ!
####### $
####### i
####### i
! x i ˆ ‰
4. f xa b œ % x# Since f is increasing on Ò#ß !Ó and decreasing on Ò!ß #Ó, we use
left endpoints on Ò#ß !Ó and right endpoints on Ò!ß #Óto obtain
lower sums and use right endpoints on Ò#ß !Óand left endpoints
on Ò!ß #Óto obtain upper sums.
(a) ̃x œ # ##a b œ # and x i œ # ̃i x œ # # Êi a lower sum is# † ˆ% a# b# ‰ # † a% # #bœ!
(b) ̃x œ # #% œ " and x œ # ̃i x œ # i Ê a lower sum is % x # † " % x # † "
####### œ! œ$
####### a b " %
####### i i i
####### i i
! ˆ a b ‰ !ˆ a b‰
œ " ˆ ˆ% a# b #‰ ˆ% a" b #‰ a% " #b a% # #b‰œ '
(c) ̃x œ # ##a b œ # and x i œ # ̃i x œ # # Êi a upper sum is# † ˆ% a b! # ‰ # † a% ! #bœ "'
(d) ̃x œ # #% œ " and x œ # ̃i x œ # i Ê a upper sum is % x # † " % x # † "
####### œ" œ#
####### a b # $
####### i i i
####### i i
! ˆ a b ‰ !ˆ a b‰
œ " ˆ ˆ% a" b #‰ a% ! # b a% ! #b a% " #b‰œ "%
5. f xa b œ x# Using 2 rectangles Ê ̃x œ " !# œ "# Ê "# ˆf ˆ ‰"% fˆ ‰$%‰
œ "# "% $% œ "!$# œ"'&
####### # #
Š ˆ ‰ ˆ ‰‹
Using 4 rectangles Ê ̃x œ " !% œ"%
Ê "% ˆ fˆ ‰ ") fˆ ‰ $) fˆ ‰ &) fˆ ‰()‰
œ "% ") $) &) () œ#"'%
####### # # # #
Š ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰‹
260 Chapter 5 Integration
13. (a) Because the acceleration is decreasing, an upper estimate is obtained using left end-points in summing
acceleration † ?t. Thus, ?t œ 1 and speed ̧ 32 19 11 7 4 œ74 ft/sec
(b) Using right end-points we obtain a lower estimate: speed ̧ 19 11 7 4 2
œ 45 ft/sec
(c) Upper estimates for the speed at each second are:
t 0 1 2 3 4 5
v 0 32 51 63 70 74.
Thus, the distance fallen when t œ 3 seconds is s ̧ 32 51 63 œ146 ft.
14. (a) The speed is a decreasing function of time Êright end-points give an lower estimate for the height (distance)
attained. Also
t 0 1 2 3 4 5
v 400 368 336 304 272 240
gives the time-velocity table by subtracting the constant g œ32 from the speed at each time increment
? t œ 1 sec. Thus, the speed ̧240 ft/sec after 5 seconds.
(b) A lower estimate for height attained is h ̧ 368 336 304 272 240 œ1520 ft.
15. Partition [ !ß #] into the four subintervals [0 0], [0 1], [1 1], and [1 2]. The midpoints of these ß ß ß ß
subintervals are m " œ 0, m# œ 0, m $ œ 1, and m %œ1. The heights of the four approximating
rectangles are f(m )" œ (0) $ œ 641 , f(m )# œ (0) $ œ 2764 , f(m )$ œ (1) $ œ 12564 , and f(m )% œ (1)$œ 34364
Notice that the average value is approximated by "# " "# "# "# "# $""'
####### $ $ $ $
’ ˆ ‰ ˆ ‰ 4 ˆ ‰ ˆ ‰ 43 ˆ ‰ ˆ ‰ 45 ˆ ‰ ˆ ‰ 47 “œ
. We use this observation in solving the next several exercises.
approximate area under
curve f(x) x
œ
œ
####### "
length of [!ß# ] † ” $ •
16. Partition [1 9] into the four subintervals [ß "ß $], [3 ß &], [ &ß (], and [ (ß *]. The midpoints of these subintervals are
m " œ 2, m # œ 4, m $ œ 6, and m% œ 8. The heights of the four approximating rectangles are f(m )" œ"#,
f(m )# œ " 4 , f(m )$ œ " 6 , and f(m )% œ " 8. The width of each rectangle is ?x œ2. Thus,
Area ̧ 2 ˆ ‰"# 2 ˆ ‰" 4 2 ˆ ‰" 6 2 ˆ ‰" 8 œ 125 # Ê average value ̧ length of [area "ß* ] œ 8 œ 9625.
####### ˆ 2512 ‰
17. Partition [0 2] into the four subintervals [0 0], [0 1], [1 1], and [1 2]. The midpoints of the subintervalsß ß ß ß ß
are m " œ 0, m # œ 0, m $ œ 1, and m %œ1. The heights of the four approximating rectangles are
f(m )" "# sin # "# "# 1, f(m )# " sin # "# "# 1, f(m )$ " sin# "# "
####### #
œ 14 œ œ œ 2 341 œ œ œ 2 541 œ Š È 2 ‹
œ "# "# œ 1, and f(m )% œ " sin # œ "# " œ 1. The width of each rectangle is x œ"#. Thus,
####### #
####### 2 4
####### 7
####### 2
1 Š ‹
È?
Area ̧ (1 1 1 1) ˆ ‰"# œ 2 Ê average value ̧ length of [0 2] area ß œ # 2 œ1.
18. Partition [0 4] into the four subintervals [0 1], [1 2 ], [2 3], and [3 4]. The midpoints of the subintervalsß ß ß ß ß ß
are m " œ "# , m # œ # 3 , m $ œ # 5 , and m %œ# 7. The heights of the four approximating rectangles are
f(m )" 1 cos 1 cos 0 (to 5 decimal places),
####### % %
œ Š Š ‹‹ œ ˆ ˆ ‰‰ œ
####### 1 ˆ "#‰ 1
####### 4 8
f(m )# 1 cos 1 cos 0, f(m ) 1 cos 1 cos
####### % % % %
œ Š Š 1 ‹‹ œ ˆ ˆ 1 ‰‰ œ œ Š Š 1 ‹‹ œ ˆ ˆ 1 ‰‰
####### ˆ 3 # ‰ ˆ 5 #‰
####### 4 8 4 8
####### 3 5
####### 3
œ 0, and f(m )% œ 1 cos œ 1 cos œ0. The width of each rectangle is
####### % %
Š Š ‹‹ ˆ ˆ ‰‰
####### 1 ˆ 7 #‰ 1
####### 4 8
####### 7
?x œ ". Thus, Area ̧ (0)(1) (0)(1) (0)(1) (0)(1) œ 2 Êaverage
value ̧ length of [0 4]area ß œ 2 4 œ 58.
Section 5 Area and Estimating with Finite Sums 261
19. Since the leakage is increasing, an upper estimate uses right endpoints and a lower estimate uses left
endpoints:
(a) upper estimate œ (70)(1) (97)(1) (136)(1) (190)(1) (265)(1) œ758 gal,
lower estimate œ (50)(1) (70)(1) (97)(1) (136)(1) (190)(1) œ543 gal.
(b) upper estimate œ (70 97 136 190 265 369 516 720) œ2363 gal,
lower estimate œ (50 70 97 136 190 265 369 516) œ1693 gal.
(c) worst case: 2363 720t œ 25,000 Ê t ̧31 hrs;
best case: 1693 720t œ 25,000 Ê t ̧32 hrs
20. Since the pollutant release increases over time, an upper estimate uses right endpoints and a lower estimate
uses left endpoints:
(a) upper estimate œ (0)(30) (0)(30) (0)(30) (0)(30) (0)(30) (0)(30) œ60 tons
lower estimate œ (0)(30) (0)(30) (0)(30) (0)(30) (0)(30) (0)(30) œ46 tons
(b) Using the lower (best case) estimate: 46 (0)(30) (0)(30) (0)(30) (0)(30) œ126 tons,
so near the end of September 125 tons of pollutants will have been released.
21. (a) The diagonal of the square has length 2, so the side length is È #. Area œ ŠÈ #‹ œ
####### #
(b) Think of the octagon as a collection of 16 right triangles with a hypotenuse of length 1 and an acute angle measuring
#"' 1 œ 1 ).
Area œ "' ˆ ‰ˆ"# sin ) 1 ‰ˆcos ) 1 ‰ œ % sin% 1 œ # È# ̧ #Þ)#)
(c) Think of the 16-gon as a collection of 32 right triangles with a hypotenuse of length 1 and an acute angle measuring
#$# 1 œ"' 1.
Area œ $# ˆ ‰ˆ"# sin "' 1 ‰ˆcos "' 1 ‰ œ ) sin) 1 œ # È# ̧ $Þ!'"
(d) Each area is less than the area of the circle, 1. As n increases, the area approaches 1.
22. (a) Each of the isosceles triangles is made up of two right triangles having hypotenuse 1 and an acute angle measuring
## 1 n œ 1 n . The area of each isosceles triangle is A Tœ # ˆ ‰ˆ"# sin 1 n ‰ˆcos 1 n ‰œ"# sin #n 1.
(b) The area of the polygon is A P nA T n sin n , so nsinn
####### n n
####### sin
œ œ # # # # œ † œ
####### Ä_ Ä_
1 lim 1 lim 1 # 1 # n 11
####### ˆ n‰
(c) Multiply each area by r .#
A T œ "# r sin# #n 1
A P œ n#r sin# #n 1
lim A r
####### n Ä_ P
œ 1
23-26. Example CAS commands:
Maple:
with( Student[Calculus1] );
f := x -> sin(x);
a := 0;
b := Pi;
plot( f(x), x=a., title="#23(a) (Section 5)" );
N := [ 100, 200, 1000 ]; # (b)
for n in N do
Xlist := [ a+1.(b-a)/ni $ i=0. ];
Ylist := map( f, Xlist );
end do:
for n in N do # (c)
Section 5 Sigma Notation and Limits of Finite Sums 263
7. (a)! 2 2 2 2 2 2 2 1 2 4 8 16 32
6 k 1 k 1 œ
œ "" #" $" %" &" '"œ
(b)! 2 2 2 2 2 2 2 1 2 4 8 16 32
5 k 0 k œ
œ! " # $ % &œ
(c)! 2 2 2 2 2 2 2 1 2 4 8 16 32
4 k k 1 œ"
œ "" !" "" #" $" %"œ
All of them represent 1 2 4 8 16 32
8. (a) !( 2) ( 2) ( 2) ( 2) ( 2) ( 2) ( 2) 1 2 4 8 16 32
6 k 1 k 1 œ
œ "" #" $" %" &" '"œ
(b) !( 1) 2 ( 1) 2 ) 2 ( 1) 2 ( 1) 2 ( 1) 2 ( 1) 2 1 2 4 8 16 32
5 k 0 k k œ
œ ! ! Ð" " " # # $ $ % % & &œ
(c)! ( 1) 2 ) 2 ( ) 2 ( ) 2 ( 1) 2 ( ) 2
3 k 2 k 1 k 2 œ
œ Ð" #" ## " "" "# " !" !# "" "# "#" ##
( 1) $" 2 $#œ 1 2 4 8 16 32;
(a) and (b) represent 1 2 4 8 16 32; (c) is not equivalent to the other two
9. (a)! 1
4 k œ 2
####### ( ) ( 1) ( ) ( )
####### k 1 2 1 3 1 4 1 3
####### " " "
####### #
k" œ #" $" %"œ " "
(b)! 1
2 k œ 0
####### ( ) ( 1) ( ) ( )
####### k 1 0 1 1 1 2 1 3
####### " " "
####### #
k œ! " #œ " "
(c)! 1
1 kœ"
####### ( ) ( 1) ( ) ( )
####### k 2 1 2 0 2 1 2 3
####### " " "
####### #
k œ " ! "œ " "
(a) and (c) are equivalent; (b) is not equivalent to the other two.
10. (a) !(k 1) (1 1) (2 1) (3 1) (4 1) 0 1 4 9
4 k œ 1
# œ # # # #œ
(b) !(k 1) ( 1 1) (0 1) (1 1) (2 1) (3 1) 0 1 4 9 16
3 k œ 1
# œ # # # # #œ
(c) !k ( 3) ( 2) ( 1) 9 4 1
" kœ 3
# œ # # #œ
(a) and (c) are equivalent to each other; (b) is not equivalent to the other two.
11.! k 12.! k 13.!
6 4 4 k œ 1 k œ 1 k œ 1
####### # "
####### #k
14.! 2k 15.! ( 1) 16. !( 1)
5 5 5 k 1 k 1 k 1 k 1 k œ œ œ
"
####### k 5
####### k
17. (a)! 3a 3 !a 3( 5) 15
n n k 1 k 1 k k œ œ
œ œ œ
(b)! !b (6) 1
n n k 1 k 1 k œ œ
####### b
####### 6 6 6
k œ " œ " œ
(c)! (a b )! a !b 5 6 1
n n n k 1 k 1 k 1 k k k k œ œ œ
œ œ œ
(d)! (a b )! a !b 5 6 11
n n n k 1 k 1 k 1 k k k k œ œ œ
œ œ œ
(e)! (b 2a )! b 2 !a 6 2( 5) 16
n n n k 1 k 1 k 1 k k k k œ œ œ
œ œ œ
264 Chapter 5 Integration
18. (a)! 8a 8! a 8(0) 0 (b)! 250b 250 !b 250(1) 250
n n n n k 1 k 1 k 1 k 1 k k k k œ œ œ œ
œ œ œ œ œ œ
(c)! (a 1)! a! 1 0 n n (d)! (b 1)! b! 1 n
n n n n n n k 1 k 1 k 1 k 1 k 1 k 1 k k k k œ œ œ œ œ œ
œ œ œ œ œ "
19. (a)! k 55 (b) !k 385
10 10 k œ 1 k œ 1
œ 10(10# 1) œ #œ 10(10 1)(2(10) 6 1)œ
(c)! k ’ “ 55 3025
10 k œ 1
####### $ #
####### #
####### #
œ 10(10 1) œ œ
20. (a)! k 91 (b) !k 819
13 13 k œ 1 k œ 1
œ 13(13# 1) œ #œ 13(13 1)(2(13) 6 1)œ
(c)! k ’ “ 91 8281
13 k œ 1
####### $ #
####### #
####### #
œ 13(13 1) œ œ
21.! 2k 2! k 2 Š ‹ 56 22.! !k Š ‹
7 7 5 5 k œ 1 k œ 1 k œ 1 k œ 1
œ œ 7(7 # " ) œ 151 k œ 151 œ 151 5(5 #1) œ 1
23.! a 3 k b! 3 !k 3(6) 73
6 6 6 k œ 1 k œ 1 k œ 1
# œ # œ 6(6 " )(2(6) 6 1)œ
24.! ak 5 b! k! 5 5(6) 61
6 6 6 k œ 1 k œ 1 k œ 1
# œ # œ 6(6 " )(2(6) 1) œ
####### 6
25.! k(3k 5)! a3k 5k b 3! k 5 !k 3 Š ‹ 5 Š ‹ 240
5 5 5 5 k œ 1 k œ 1 k œ 1 k œ 1
œ # œ # œ 5(5 1)(2(5) 6 1) 5(5 #1) œ
26.! k(2k 1)! a2k k b 2! k !k 2 Š ‹ 308
7 7 7 7 k œ 1 k œ 1 k œ 1 k œ 1
œ # œ # œ 7(7 1)(2(7) 6 1) 7(7 #1)œ
27.! Œ! k ! k Œ! k Š ‹ Š ‹ 3376
5 5 5 5 k œ 1 k œ 1 k œ 1 k œ 1
####### k
####### 225 2 5 25
$ œ œ 5(5 1) 5(5 1) œ
####### $ $
####### " "
####### # # # #
####### $ # $
28. Œ! k ! Œ! k ! k Š ‹ Š ‹ 588
7 7 7 7 k œ 1 k œ 1 k œ 1 k œ 1
####### # #
####### " $ "
####### # #
####### # #
k 4 œ 4 œ 7(7 1) 4 7(7 1) œ
$
29. (a)! 3 3 a b 21 (b)! 7 7 a b 3500
7 500 k œ 1 k œ 1
œ 7 œ œ 500 œ
(c) Let j œ k 2 Ê k œ j 2; if k œ 3 Ê j œ 1 and if k œ 264 Ê j œ 262 Ê! 10 œ! 10 œ 10 a bœ 2620
264 262 k œ 3 j œ 1
262
30. (a) Let j œ k 8 Ê k œ j 8; if k œ 9 Ê j œ 1 and if k œ 36 Ê j œ 28 Ê! k œ! a j 8 bœ! j ! 8
36 28 28 28 k œ 9 j œ 1 j œ 1 j œ 1
œ 28 28 a 2 1 b ) a 28 bœ 630
(b) Let j œ k 2 Ê k œ j 2; if k œ 3 Ê j œ 1 and if k œ 17 Ê j œ 15 Ê! k œ! a j 2 b
17 15 k œ 3 j œ 1
####### 2 2
œ! a j 4j 4 b œ! j ! 4j ! 4 œ 4 † 4 15a b
15 15 15 15 j œ 1 j œ 1 j œ 1 j œ 1
####### 2 2 15 15 1 2 15 1 15 15 1
####### 6 2
####### a ba a b b a b
œ 1240 480 60 œ 1780
266 Chapter 5 Integration
36. (a) (b) (c)
37. kx " x ! k œ k1 0 k œ 1, xk # x "k œ k1 1 k œ 0, xk $ x # k œ k 2 1 k œ 0, xk % x $k œ k2 2 kœ0,
and xk & x %k œ k 3 2 k œ 0; the largest is l lP œ1.
38. kx " x ! k œ k 1 ( 2) k œ 0, xk # x "k œ k 0 ( 1) k œ 1, xk $ x #k œ k 0 ( 0) kœ0,
k x % x $ k œ k0 0 k œ 0, and xk & x %k œ k 1 0 k œ 0; the largest is l lP œ1.
39. f xa b œ " x# Let ̃x œ " ! n œ "n and c i œ ̃i x œni. The right-hand sum is
! a c b! Š ˆ ‰‹ !an ib
####### i 1 i 1 i 1
####### n n n
####### i n n n n
####### i
####### œ œ œ
" # " œ " " # œ "$ # #
œ nn n œ " n œ " n nn n
####### i 1
####### $ n n n n $ #
$ $ $ $
####### " # $
####### œ
####### # " # "
####### ' '
!i a ba b
œ " "
####### #
####### ' Ä_
####### œ
####### # "
$n "#
n . Thus, lim c
####### n i 1
####### n
!a i bn
œ lim " œ " œ
####### nÄ_
####### #
####### ' $ $
Π"
$n "# n
40. f xa b œ #x Let ̃x œ $ ! n œ $n and c i œ ̃i x œ$ni. The right-hand sum is
! ˆ ‰!!
####### i i i
####### n n n
####### i n ni n n n n n n
####### n n
####### œ" œ" œ"
####### $ ' $ ") ") " * *
# c œ † œ # iœ # † # œ
####### a b #
Thus, lim lim lim.
####### n i n n
####### n
####### i n n
####### Ä_ n n Ä_ n Ä_ n
####### œ"
! ' † $ œ * # *# œ ˆ * *‰œ *
41. f xa b œ x# " Let ̃x œ $ ! n œ $n and c i œ ̃i x œ$ni. The right-hand sum is
! ac b! Š ˆ ‰ ‹ !Š ‹
####### i i i
####### n n n
####### i n n n n n
####### i i
####### œ" œ" œ"
# " $ œ $ # " $ œ $ *# # "
œ #( $ † œ #( $
####### œ"
####### # " # "
####### n i n n '
####### n n n n
! i n $Š a ba b‹
œ * # $# $ œ # $Þ
####### a n n nb ")
####### n
$ # $ #n( *#
n Thus,
lim c lim.
####### n i n
####### n
####### Ä_ i n Ä_
####### œ"
####### # $ ")
!a "b œ Œ # $ œ * $ œ "#
#n( *# n
Section 5 Sigma Notation and Limits of Finite Sums 267
42. f xa b œ $x # Let ̃x œ " ! n œ "n and c i œ ̃i x œni. The right-hand sum is
! ˆ ‰! ˆ ‰ ˆ ‰! Š ‹
####### i i i
####### n n n
####### i n n n n n
####### i n n n
####### œ" œ" œ"
####### # " # " $ # $ " # "
$ c œ $ œ $ i œ $ '
####### a ba b
œ # $# œ # $ "
####### #
####### Ä_ œ"
####### n n n #
####### n n n
####### i
####### n
####### i
$ # $ $ "#
n n . Thus, lim! cˆ ‰
œ lim œ œ "
####### nÄ_
####### #
####### # #
Œ
$ "#
n n .
43. f xa b œ x x # œ x a " xb Let ̃x œ " ! n œ "n and c i œ ̃i x œni. The right-hand sum is
! a b! Š ˆ ‰‹!!
####### i i i i
####### n n n n
####### i i n ni ni n n n
####### œ" œ" œ" œ"
c c# " œ # " œ "# i "$ i#
œ n" # n n# " n"$ n n " ' # n " œ n # n # n # n $'n n$ n
$Š a b ‹ Š a ba b‹
œ " # '
####### #
####### Ä_ œ"
####### " # "
$ "#
n n n. Thus, lim c c
####### n i
####### n
!a i i bn
œ lim œ œ
####### nÄ_
####### "
####### # ' # ' '
####### # " # &
” Š ‹ Œ •
"n $n n"#
.
44. f xa b œ $x #x # Let ̃x œ " ! n œ "n and c i œ ̃i x œni. The right-hand sum is
! a b! Š ˆ ‰‹!!
####### i i i i
####### n n n n
####### i i n n n n n
####### i i
####### œ" œ" œ" n œ"
$ c # c# " œ $ # # " œ $# i #$ i#
œ n$ # n n# " n#$ n n " ' # n " œ $ n # $ n # n # n $$n# n "
nullŠ a b ‹ Š a ba b‹
œ $
####### $
####### # $
####### #
####### Ä_ œ"
$n $n n"#. Thus, lim c c# "
####### n i
####### n
!a i i bn
œ lim œ œ
####### nÄ_
####### $
####### # $ # $ '
####### # $ # "$
” Š ‹ Œ •
$n $n n"#
.
45. f xa b œ 2x 3 Let ̃x œ " ! n œ "n and c i œ ̃i x œni. The right-hand sum is
! a b! Š ˆ ‰‹! Š ‹
####### i i i
####### n n n
####### i
####### 3 3
####### n n n
####### i 3
####### n n
####### n n 2
####### œ" œ" œ"
####### " " # # "
2c œ 2 œ 4 i œ 4
####### a b
œ # " œ # "œ
####### n n 2n "
####### 4n n
####### # # n # 2n
null####### a b # "#
4
n n .
Thus, lim 2c lim.
####### n i n
####### n
####### i
####### 3
####### Ä_ n Ä_
####### œ"
####### " " "
!a b œ ” # • œ#
#n "# n
Section 5 The Definite Integral 269
12. (a) ' ! g(t) dt ' $ g(t) dt 2 (b) ' $ g(u) du '$g(t) dt 2
####### $!!!
œ œ È œ œÈ
(c) ' $ [ g(x)] dx ' $ g(x) dx 2 (d) ' $ dr '$g(t) dt 2 1
####### !!!!
œ œ È g(r) œ " œ Š "‹ ŠÈ ‹œ
####### È 2 È 2 È 2
13. (a) ' $ f(z) dz ' ! f(z) dz '!f(z) dz 7 3 4
####### % % $
œ œ œ
(b) ' % f(t) dt '$f(t) dt 4
####### $ %
œ œ
14. (a) ' h(r) dr ' h(r) dr ' h(r) dr 6 0 6
####### " " "
####### $ $ "
œ œ œ
(b) ' $ h(u) du œ ' " h(u) du œ '"h(u) du œ 6
####### " $ $
Œ
15. The area of the trapezoid is A œ "#(B b)h
œ "# (5 2)(6) œ 21 Ê # # 3 dx
####### %
' ˆ x ‰
œ21 square units
16. The area of the trapezoid is A œ "#(B b)h
œ "# (3 1)(1) œ 2 Ê "Î#( 2x 4) dx
####### $Î#
'
œ 2 square units
17. The area of the semicircle is A œ "# 1 r # œ"# 1 (3)#
œ 9 # Ê $ 9 x dx œ 9 # square units
####### $
1 ' È # 1
270 Chapter 5 Integration
18. The graph of the quarter circle is A œ " 4 1 r # œ" 41 (4)#
œ 4 1 Ê '% 16 x dx œ 4 1 square units
####### !
È
19. The area of the triangle on the left is A œ "# bh œ"#(2)(2)
œ 2. The area of the triangle on the right is A œ "#bh
œ "# (1)(1) œ"#. Then, the total area is 2.
Ê '# x dx œ2 square units
####### "
k k
20. The area of the triangle is A œ "# bh œ "#(2)(1) œ 1
Ê '" 1 x dx œ1 square unit
####### "
a k kb
21. The area of the triangular peak is A œ "# bh œ "#(2)(1) œ1.
The area of the rectangular base is S œ jw œ (2)(1) œ2.
Then the total area is 3 Ê '" 2 x dx œ3 square units
####### "
a k kb
22. y œ 1 È 1 x # Ê y 1 œ È 1 x#
Ê (y 1) # œ 1 x # Ê x # (y 1) #œ1, a circle with
center ( !ß ") and radius of 1 Ê y œ 1 È 1 x #is the
upper semicircle. The area of this semicircle is
A œ "# 1 r # œ "# 1 (1) #œ# 1. The area of the rectangular base
is A œ jw œ (2)(1) œ 2. Then the total area is 2 1
Ê '" 1 1 x dx œ 2 square units
####### "
####### #
Š ‹
È 1
272 Chapter 5 Integration
41. ' $ 7 dx 7(1 3) 14 42. ' ! 5x dx 5 '!x dx 5 10
####### "
œ œ œ œ # # œ
####### 2 2
’ 2 # 0 #“
43. ' ! (2t 3) dt 2 '" t dt '!3 dt 2 3(2 0) 4 6 2
####### "
####### # #
####### 2 2
œ œ ’ 2 # 0 #“ œ œ
44. ' ! t 2 dt ' ! t dt '! 2 dt # # 2 2 0 1 2 1
####### È 2 È 2 È 2 Š È 2 ‹
Š È ‹ È 0 È ’ È “
œ œ – — œ œ
null#45. ' 1 dz ' 1 dz ' dz ' 1 dz ' z dz 1[1 2]
####### # # # # "
####### " " " " #
####### # # # # # # # #
ˆ z ‰ œ z œ " œ " ’ 2 1 “œ " "ˆ ‰ 3 œ 7
####### 4
null46. ' $ (2z 3) dz ' $ 2z dz ' $ 3 dz 2 ' ! z dz '$ 3 dz 2 3[0 3] 9 9 0
####### !!! $!
œ œ œ ’ # #“ œ œ
####### 3 # 0 #
47. ' " 3u du 3 ' " u du 3 ' ! u du '! u du 3 3 3 7
####### # # # "
# œ # œ ” # # • œ Š ’ 2 0 “ ’ " 0 “‹ œ ’ 2 1 “œ ˆ ‰ 7 œ
####### 3 3 3 3 3 3 3
$ $ $ $ $ $
48. ' "Î# 24u du 24 ' "Î# u du 24 ' ! u du '! u du 24 24 7
####### " " " "Î#
# œ # œ # # œ œ œ
- — ” • ’ “
####### 1
####### 3 3 3
####### $ ˆ "#‰$ ˆ 78 ‰
49. ' 3x x 5 dx 3 ' x dx ' x dx ' 5 dx 3 5[2 0] (8 2) 10 0
####### !!!!
####### # # # #
####### # #
a b œ œ ’ “ ’ # #“ œ œ
####### 2 0 2 0
####### 3 3
$ $ # #
50. ' " 3x x 5 dx ' ! 3x x 5 dx 3 ' ! x dx ' ! x dx '! 5 dx
####### ! " " " "
a # b a # b
œ œ ” •
œ ’ 3 Š 13 03 ‹ Š 1 0 ‹ 5(1 0) “œ ˆ 3 5 ‰œ 7
$ $ # #
####### # # # #
51. Let ?x œ b n 0 œ bnand let x ! œ 0, x "œ?x,
x # œ 2 ?x ß á ß x n "œ (n 1) ?x, x nœ n ?x œb.
Let the c 's be the right end-points of the subintervalsk
Ê c " œ x , c" # œx , and so on. The rectangles#
defined have areas:
f(c )" ?x œ f( ?x) ?x œ 3( ?x) # ?x œ3( ?x)$
f(c )# ?x œ f(2 ?x) ?x œ 3(2 ?x) # ?x œ3(2) (# ?x)$
f(c )$ ?x œ f(3 ?x) ?x œ 3(3 ?x) # ?x œ3(3) (# ?x)$
ã
f(c )n ?x œ f(n ?x) ?x œ 3(n ?x) # ?x œ3(n) (# ?x)$
Then S n f(c )k x 3k ( x)
n n k 1 k 1
œ! œ!
œ œ
? # ? $
œ 3( ?x) $ ! k # œ 3 Š ‹ Š ‹
n k œ 1
####### b
####### n 6
####### $ n(n 1)(2n 1)
$
œ b 2 3 n n Ê 3x dx œ lim b 2 3 n n œb.
####### $ b $
####### # # # #
####### " "
####### !
ˆ ‰ ' # ˆ ‰ $
n Ä _
Section 5 The Definite Integral 273
52. Let ?x œ b n 0 œ bnand let x ! œ 0, x "œ?x,
x # œ 2 ?x ß á ß x n "œ (n 1) ?x, x nœ n ?x œb.
Let the c 's be the right end-points of the subintervalsk
Ê c " œ x , c" # œx , and so on. The rectangles#
defined have areas:
f(c )" ?x œ f( ?x) ?x œ 1 ?( x) # ?x œ1 ?( x)$
f(c )# ?x œ f(2 ?x) ?x œ 1 (2 ?x) # ?x œ 1 (2) ( # ?x)$
f(c )$ ?x œ f(3 ?x) ?x œ 1 (3 ?x) # ?x œ 1 (3) ( # ?x)$
ã
f(c )n ?x œ f(n ?x) ?x œ 1 (n ?x) # ?x œ 1 (n) ( # ?x)$
Then S n f(c )k x k ( x)
n n k 1 k 1
œ! œ!
œ œ
? 1 # ? $
œ 1? ( x) $ ! k# œ 1 Š ‹ Š ‹
n k œ 1
####### b
####### n 6
####### $ n(n 1)(2n 1)
$
œ 1 6 b 2 3 n n Ê x dx œ lim 1 b 6 2 3 n n œ 1 b 3.
####### $ b $ $
ˆ # ‰ ˆ #‰
####### " "
####### !
' 1
n Ä _
53. Let ?x œ b n 0 œ bnand let x ! œ 0, x "œ?x,
x # œ 2 ?x ß á ß x n "œ (n 1) ?x, x nœ n ?x œb.
Let the c 's be the right end-points of the subintervalsk
Ê c " œ x , c" # œx , and so on. The rectangles#
defined have areas:
f(c )" ?x œ f( ?x) ?x œ 2( ?x)( ?x) œ2( ?x)#
f(c )# ?x œ f(2 ?x) ?x œ 2(2 ?x)( ?x) œ2(2)( ?x)#
f(c )$ ?x œ f(3 ?x) ?x œ 2(3 ?x)( ?x) œ2(3)( ?x)#
ã
f(c )n ?x œ f(n ?x) ?x œ 2(n ?x)( ?x) œ 2(n)( ?x)#
Then S n f(c )k x 2k( x)
n n k 1 k 1
œ! œ!
œ œ
??
œ 2( ?x) # ! k œ 2 Š ‹ Š ‹
n k œ 1
####### b
####### n 2
####### # n(n 1)
nullœ b # ˆ 1 "n ‰ Ê !2x dx œ lim b # ˆ 1 "n‰œb .#
####### b
'
n Ä _
54. Let ?x œ b n 0 œ bnand let x ! œ 0, x "œ?x,
x # œ 2 ?x ß á ß x n "œ (n 1) ?x, x nœ n ?x œb.
Let the c 's be the right end-points of the subintervalsk
Ê c " œ x , c" # œx , and so on. The rectangles#
defined have areas:
f(c )" ?x œ f( ?x) ?x œ ˆ ?# x 1 ‰( ?x) œ #" ( ?x) #?x
f(c )# ?x œ f(2 ?x) ?x œ ˆ 2 ?# x 1 ‰( ?x) œ #" (2)( ?x) #?x
f(c )$ ?x œ f(3 ?x) ?x œ ˆ 3 ?# x 1 ‰( ?x) œ #" (3)( ?x) #?x
ã
f(c )n ?x œ f(n ?x) ?x œ ˆ n ? # x 1 ‰( ?x) œ #" (n)( ?x) #?x
Then S n f(c )k x k( x) x ( x) k x 1 (n)
n n n n k 1 k 1 k 1 k 1
œ! œ! ˆ ‰ œ! ! œ Š ‹ Š ‹ˆ ‰
œ œ œ œ
? "# ? # ? "# ? # ? "# bn n(n 2 1) bn
nullnullœ " 4 b # 1 n 1 b Ê ! #x 1 dx œ lim " 4 b # 1 "n b œ " 4 b #b.
####### b
ˆ ‰ ' ˆ ‰ ˆ ˆ ‰ ‰
n Ä _
Section 5 The Definite Integral 275
60. av(f) œ Š 1 1 ( 2) ‹ #at t bdt
####### "
'
œ " # t dt " #t dt
####### " "
####### #
####### 3 3
' '
œ " ! t dt " ! t dt "
####### " #
####### # #
####### # #
#######
####### 3 3 3
' ' Š 1 # ( 2)#‹
œ " 3 Š 13 ‹ " 3 Š ( 3 2)‹ "# œ# 3.
$ $
61. (a) av(g) œ Š " ‹ "a k kx 1 bdx
####### "
####### 1 ( 1)
'
œ "# ( x 1) dx "# (x 1) dx
####### "!
####### ! "
' '
œ "# " x dx "# " 1 dx "# ! x dx "# !1 dx
####### !! " "
' ' ' '
œ "# Š 0 # ( #1) ‹ "# (0 ( 1)) "# Š 1 # 0 # ‹ "#(1 0)
# #œ "#.
(b) av(g) œ ˆ " ‰ " a k kx 1 bdx œ "# " (x 1) dx
####### $ $
####### 3 1
' '
œ "# " x dx "# "1 dx œ "# # # "#(3 1)
####### $ $
' ' Š 3 # 1 #‹
œ 1.
(c) av(g) œ Š " ‹ "a k kx 1 bdx
####### $
####### 3 ( 1)
'
œ " " x 1 dx " " x 1 dx
####### " $
####### 4 4
' a k k b ' a k k b
œ " 4 ( 1 2) œ" 4 (see parts (a) and (b) above).
62. (a) av(h) œ Š " ‹ k kx dx œ ( x) dx
####### 0 ( 1) " "
####### 0 0
' '
œ '" x dx œ # # œ #".
####### 0
####### 0 # ( 1)#
276 Chapter 5 Integration
(b) av(h) œ ˆ " ‰ k kx dx œ x dx
####### " "
####### 1 0 0
' '
œ Š "# # ‹œ "#.
####### # 0 #
(c) av(h) œ Š " ‹ k kx dx
####### "
####### "
####### 1 ( 1)
'
œ "# " x dx x dx
####### "
Π' k k ' k k
####### 0
####### 0
œ "# ˆ "# ˆ "# ‰‰œ "#(see parts (a) and (b)
above).
63. Consider the partition P that subdivides the interval a, b into n subintervals of widthÒ Ó ̃x œ b n aand let c kbe the right
endpoint of each subinterval. So the partition is P œ Öa, a b n a, a # ab n a b ,... , a n ba n a b × and c kœ a k ba n ab.
We get the Riemann sum! f ca b x! c! n c ba a. As nb and P
####### k k k
####### n n n
####### k b n a n n
####### c b a c b a
####### œ" œ" œ"
̃ œ † œ a b " œ a b† œ Ä _ m m Ä!
this expression remains c ba a. Thus,b 'ac dx œ c ba a .b
####### b
64. Consider the partition P that subdivides the interval 0, 2 into n subintervals of widthÒ Ó ̃x œ 2 n 0 œ 2 n and let c kbe the
right endpoint of each subinterval. So the partition is P œ Ö0, 2 n , 2 † 2 n ,... , n † 2 n œ 2 × and c kœ k † 2 n œ2kn. We get the
Riemann sum! f ca b x! ’ 2 ˆ ‰ 1 “! ˆ 1 ‰! k! 1 n 2.
####### k k k k k
####### n n n n n
####### k 2kn 2 n 2 n 4kn n 8 2 n n 8 2 n n
####### n n 4 n
####### œ" œ" œ" œ" œ"
####### " "
̃ œ † œ œ 2 œ 2 † # † œ
####### a b a b
As n Ä _ and m m Ä !P the expression 4 nn 2 has the value 4 2 œ 6. Thus, 2x 1 dx œ6.
####### 0
####### 2
a "b ' a b
65. Consider the partition P that subdivides the interval a, b into n subintervals of widthÒ Ó ̃x œ b n aand let c kbe the right
endpoint of each subinterval. So the partition is P œ Öa, a b n a, a # ab n a b ,.. ., a n ba n a b × and c kœ a k ba n ab.
We get the Riemann sum! f ca b x! c ˆ ‰ ! Ša ‹ !Ša ‹
####### k k k k
####### n n n n
####### k k
####### b a b a b a
####### n n n n n n
####### k b a ak b a k b a
####### œ" œ" œ" œ"
####### # # #
####### #
̃ œ œ a b œ a b a b
null#œ b n a a n k n k œ b n a† na n † n †
####### k k k
####### n a b a n b a n a b a n n b a n n n
####### œ" œ" œ"
####### # # # # # " " # "
Π# '
! a b ! a b ! a b a b a b a ba b
# $# $œ a b a ab # a ba a b # † " ' † " # " œ ab a ab # a ba ab# † " " ' † "
####### n #
####### n n
####### ab a b $ an ba n b ab ab$
null"n $n n"#As n Ä _ and m m Ä !P this expression has value ab a ab # a ba ab# † " a b 'ab †
$
œ ba # a $ ab # #a b # a $ "$ ab $ $b a # $ba # a $ b œ b$ a$ . Thus, ax dx# œ b$ a$.
####### $ $ b $ $
'
66. Consider the partition P that subdivides the interval Ò1, 0 into n subintervals of width Ó ̃x œ 0 na 1 bœn 1 and let c kbe
the right endpoint of each subinterval. So the partition is P œ Ö1, 1 1 n , 1 2 † 1 n ,.. ., 1 n † 1 nœ 0 ×and
Thomas calculus, 12th edition, solution manual, chapter 5
Course: Calculus Of Several Variables (Math 216)
University: جامعة المنصورة
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