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Chapter 2Combinatorial Analysis
主講人 :虞台文
Content Basic Procedure for Probability Calculation Counting
– Ordered Samples with Replacement– Ordered Samples without Replacement Permutations– Unordered Samples without Replacement Combinations
Binomial Coefficients Some Useful Mathematic Expansions Unordered Samples with Replacement Derangement Calculus
Chapter 2Combinatorial Analysis
Basic Procedure for
Probability Calculation
Basic Procedure for Probability Calculation
1. Identify the sample space .2. Assign probabilities to certain events
in A, e.g., sample point event P().3. Identify the events of interest.4. Compute the desired probabilities.
Chapter 2Combinatorial Analysis
Counting
Goal of Counting
Counting the number of elements in a particular set, e.g., a sample space, an event, etc.
? E
?E
Cases
Ordered Samples w/ Replacement
Ordered Samples w/o Replacement
– Permutations
Unordered Samples w/o Replacement
– Combinations
Unordered Samples w/ Replacement
Chapter 2Combinatorial Analysis
Ordered Samples
with Replacement
Ordered Samples
eat
ate
tea
elements in samples appearing in different orders are considered different.
Ordered Samples w/ Replacement
meet
teem
mete
1. Elements in samples appearing in different orders are considered different.
2. In each sample, elements are allowed repeatedly selected.
Ordered Samples w/ Replacement
Drawing k objects, their order is noted, among n distinct objects with replacement.
The number of possible outcomes is
ndistinctobjects
k
kn
Example 1
How many possible 16-bit binary words we may have?
2distinctobjects
16 1 16 {0,1}, 1, ,16ia a a i
162
Example 2
Randomly Choosing k digits from decimal number, Find the probability that the number is a valid octal number.
Randomly Choosing k digits from decimal number, Find the probability that the number is a valid octal number.
1 {0, ,9}, 1, ,k ia a a i k 10k
For any , P()=1/10k.
1 {0, ,7}, 1, ,k iE b b b i k 8kE
1( ) 8
10k
kP E 0.8k
Chapter 2Combinatorial Analysis
Ordered Samples
without Replacement
Permutations
Permutations
清心
也可
以
可以清心也以清心也可清心也可以心也可以清也可以清心
可以清心也以清心也可清心也可以心也可以清也可以清心
Ordered Samples w/o Replacement Permutations
Drawing k objects, their order is noted, among n distinct objects without replacement.
The number of possible outcomes is
ndistinctobjects
k
( 1)( 2) ( 1)n n n n k nkP
!
( )!
n
n k
Example 3
Five letters are to be selected without replacement from the alphabet (size 26) to form a word (possibly nonsense). Find the probabilities of the following events?1. Begin with an s.2. Contains no vowel.3. Begins and ends with a consonant.4. Contains only vowels.
Five letters are to be selected without replacement from the alphabet (size 26) to form a word (possibly nonsense). Find the probabilities of the following events?1. Begin with an s.2. Contains no vowel.3. Begins and ends with a consonant.4. Contains only vowels.
Example 3
Five letters are to be selected without replacement from the alphabet (size 26) to form a word (possibly nonsense). Find the probabilities of the following events?1. Begin with an s.2. Contains no vowel.3. Begins and ends with a consonant.4. Contains only vowels.
Five letters are to be selected without replacement from the alphabet (size 26) to form a word (possibly nonsense). Find the probabilities of the following events?1. Begin with an s.2. Contains no vowel.3. Begins and ends with a consonant.4. Contains only vowels.
Define
E1: word begins with an s.E2: word contains no vowel.E3: word begins and ends with a consonant.E4: word contains only vowels.
P(E1)=? P(E2)=? P(E3)=? P(E4)=?
Example 3Define
E1: word begins with an s.E2: word contains no vowel.E3: word begins and ends with a consonant.E4: word contains only vowels.
P(E1)=? P(E2)=? P(E3)=? P(E4)=?
1 2 3 4 5
{ , , }
,i
i j
a a za a a a a
a a i j
265P
For any , P()=1/||.
26 25 24 23 22
1 1 25 24 23 22E
2 21 20 19 18 17E
3 21 24 23 22 20E
4 5 4 3 2 1E
1 1
1( )
| |P E E
2 2
1( )
| |P E E
3 3
1( )
| |P E E
4 4
1( )
| |P E E
0.0385
0.3093
0.6462
21.52 10
Chapter 2Combinatorial Analysis
Unordered Samples
without Replacement
Combinations
Combinations
n distinctobjects
Choose k objectsChoose k objects
How many choices?How many choices?
Combinations
Drawing k objects, their order is unnoted, among n distinct objects w/o replacement, the number of possible outcomes is
nkC
n
k
( 1) ( 1)
!
n n n k
k
!
( )! !
n
n k k
This notation is preferred
n distinctobjects
Choose k objectsChoose k objects
How many choices?
How many choices?
More onn
k
( 1) ( 1)0
!0 0
n n n kn k
kk
k
Examples
( 1) ( 1)0
!0 0
n n n kn k
kk
k
2.5
3
2.5 1.5 0.5
3!
1
3
1 2 3
3!
1
0.3125
Example 4
The mathematics department consists of 25 full professors, and 15 associate professors, and 35 assistant professors. A committee of 6 is selected at random from the faculty of the department. Find the probability that all the members of the committee are assistant professors.
The mathematics department consists of 25 full professors, and 15 associate professors, and 35 assistant professors. A committee of 6 is selected at random from the faculty of the department. Find the probability that all the members of the committee are assistant professors.
x
Denoting the all-assistant event as E,
75
6
35
6E
35 751( )
6 6P E E
Example 5
A poker hand has five cards drawn from an ordinary deck of 52 cards. Find the probability that the poker hand has exactly 2 kings.
A poker hand has five cards drawn from an ordinary deck of 52 cards. Find the probability that the poker hand has exactly 2 kings.
x
Denoting the 2-king event as E,
52
5
4 48
2 3E
4 48
2 31( )
52
5
P E E
Example 6
Two boxes both have r balls numbered 1, 2, … , r. Two random samples of size m and n are drawn without replacement from the 1st and 2nd boxes, respectively. Find the probability that these two samples have exactly k balls with common numbers.
Two boxes both have r balls numbered 1, 2, … , r. Two random samples of size m and n are drawn without replacement from the 1st and 2nd boxes, respectively. Find the probability that these two samples have exactly k balls with common numbers.
11 22 33 rr
11 22 33 rr
m
n P(“k matches”) = ?
E
||=?
|E|=?
1( ) | |
| |E EP
Example 6
11 22 33 rr
11 22 33 rr
m
n| |
m n
r r
1( ) | |
| |E EP
| |n
Em
k
m
m
r r
k
# possible outcomes from the 1st box.
# possible k-matches.
# possible outcomes from the 2nd box for each k-match.
Example 6
11 22 33 rr
11 22 33 rr
m
n| |
m n
r r
1( ) | |
| |E EP
| |n
Em
k
m
m
r r
k
( )n
m
k kE
r
n
m
Pr
Example 6
11 22 33 rr
11 22 33 rr
m
n| |
m n
r r
1( ) | |
| |E EP
| |n
Em
k
m
m
r r
k
( )n
m
k kE
r
n
m
Pr
樂透和本例有何關係 ?
樂透和本例有何關係 ?
Example 6
11 22 33 rr
11 22 33 rr
m
n
1( ) | |
| |E EP
( )n
m
k kE
r
n
m
Pr
本式觀念上係由第一口箱子
出發所推得本式觀念上係
由第一口箱子出發所推得
Example 6
11 22 33 rr
11 22 33 rr
m
n
1( ) | |
| |E EP
( )nk k
m
n
r
Pr
m
E
觀念上，若改由第二口箱子
出發結果將如何 ?
觀念上，若改由第二口箱子
出發結果將如何 ?
Example 6
11 22 33 rr
11 22 33 rr
m
n
1( ) | |
| |E EP
( )nk k
m
n
r
Pr
m
E
k km
r
m
r
n n
Exercise11 22 33 rr
11 22 33 rr
m
n
1( ) | |
| |E EP
( )nk k
m
n
r
Pr
m
E
k km
r
m
r
n n
r r
r
k k
m m n n
n m
m
k k
n
r
證明
r r
r
k k
m m n n
n m
m
k k
n
r
證明
Chapter 2Combinatorial Analysis
Binomial Coefficients
Binomial Coefficients
0
1
2 2 2
3 3 2 2 3
4 4 3 2 2 3 4
1
2
3 3
4 6 4
x y
x y x y
x y x xy y
x y x x y xy y
x y x x y x y xy y
Binomial Coefficients
0
1
2 2 2
3 3 2 2 3
4 4 3 2 2 3 4
1
2
3 3
4 6 4
x y
x y x y
x y x xy y
x y x x y xy y
x y x x y x y xy y
?n
x y
Binomial Coefficients ?
nx y
nx y x y x y x y x y
1 2 2? ? ?n kn nkn nx x y x y x y y
n terms
Binomial Coefficients ?
nx y
nx y x y x y x y x y
1 2 2? ? ?n kn nkn nx x y x y x y y
n boxes
1
n 2
n
n
k
0
n
n
n
Binomial Coefficients
nx y x y x y x y x y
0
nn k k
k
yn
kx
0
nk n k
k
n
kx y
Facts: 0 for 1. n
k nk
0 for 02.n
kk
0
n k k
k kx y
n
0
k n k
k
n
kx y
n k k
k
xk
yn
k n k
k
n
kx y
Properties of Binomial Coefficients
Properties of Binomial Coefficients
0 0
1 1n n
k n k
k k
n n
k k
0
nn k n k
k
x y x yn
k
1 1n
2n
Properties of Binomial Coefficients
0 0
1 1 1n n
k k n k
k k
n n
k k
0
nn k n k
k
x y x yn
k
1 1n
0
Properties of Binomial Coefficients
0
nn k n k
k
x y x yn
k
Exercise
Properties of Binomial Coefficients
n 不 同 物 件 任 取 k 個
第一類取法 :
第二類取法 :
1n
k
1
1
n
k
Properties of Binomial Coefficients
1 1
1
n n n
k k k
1
0
1
1
2
0
2
1
2
2
0
0
3
0
3
1
3
2
3
3
4
0
4
1
4
2
4
3
4
4
Pascal Triangular
Properties of Binomial Coefficients
1
0
1
1
2
0
2
1
2
2
0
0
3
0
3
1
3
2
3
3
4
0
4
1
4
2
4
3
4
4
Pascal Triangular
1
1
1
1
1
1
1
1
1
2
3 3
4 6 4
1 1
1
n n n
k k k
Properties of Binomial Coefficients
吸星大法
1
1
n nk n
k k
Example 7-1
0
nk
k
nx
k
0
1n
k n k
k
nx
k
1
nx
Example 7-2
0
n
k
nk
k
0
k n
k
nk
k
1
k n
k
nk
k
1
1
1
k n
k
nn
k
1
1
1
k n
k
nn
k
kx+11
1 1
1
1 1
x n
x
nn
x
1
0
1x n
x
nn
x
1
0
1k n
k
nn
k
1
0
1n
k
nn
k
12nn
Fact:0
2n
n
k
n
k
?
Example 7-2
0
n
k
nk
k
1
k n
k
nk
k
1
1
1
k n
k
nn
k
kk+1
1
0
1n
k
nn
k
12nn
簡化版
Example 7-3
0
( 1)n
k
nk k
k
2
( 1)k n
k
nk k
k
kk+2
2( 1)2nn n
簡化版
2
1( 1)
1
k n
k
nn k
k
2
2( 1)
2
k n
k
nn n
k
2
0
2( 1)
n
k
nn n
k
?
Negative Binomial Coefficients
11
k
k k
n kn
Negative Binomial Coefficients
11
k
k k
n kn
How to memorize?
1 1
kn k
k k
n
k
k k (n) 1
Negative Binomial Coefficients
11
k
k k
n kn
這公式真的對嗎 ?
1 1
k
k
n
k
k k (n+k1) 1 1
k 1k
1
k
n
Negative Binomial Coefficients
( 1) ( 1)
!
n n n k
k
k
n
( 1) ( 1)1
!k kn n n
k
( 1) ( 1)1
!k n
k
nk n
11
k kn
k
Chapter 2Combinatorial Analysis
Some Useful Mathematic Expansions
Some Useful Mathematic Expansions
Some Useful Mathematic Expansions
1 1 z1
1 zz
z
2z z2z2 3z z
2z
3z
Some Useful Mathematic Expansions
1 1
1 1 ( )z z
2 431 ( ) )(( )) (z z zz
2 3 41 z zz z
Some Useful Mathematic Expansions
21
1 z
211
1z z
z
2 3 41 ? ? ? ?z z z z
2 21 1z z z z
Some Useful Mathematic Expansions
21
1 z
211
1z z
z
2 3 41 2 3 4 5z z z z
2 21 1z z z z
31
1 z
2 3 41 ? ? ? ?z z z z
2 2 21 1 1z z z z z z
1
1
n
z
2 3 41 ? ? ? ?z z z z
Some Useful Mathematic Expansions
211
1z z
z
1
1
n
z
1 ( )n
z
0
( ) 1k n k
k
nz
k
( ) 1n
z
0
( 1)k k
k
nz
k
0
1( 1) ( 1)k k k
k
n kz
k
0
1 k
k
n kz
k
Some Useful Mathematic Expansions
Some Useful Mathematic Expansions
2 2 1 2 31 1k k k kz z z z z z z z
2 1 21 1kz z z z z 11
1 1
kz
z z
11
1
kz
z
z值沒有任何限制
Some Useful Mathematic Expansions
Chapter 2Combinatorial Analysis
Unordered Samples with Replacement
Discussion
投返 非投返
有序
無序
knn
kP
n
k
?
Unordered Samples with Replacement
n不同物件任取 k個可重複選取
n不同物件，每一中物件均無窮多個從其中任取 k個
Unordered Samples with Replacement
0 1 2 3z z z z 0 1 2 3z z z z 0 1 2 3z z z z
此多項式乘開後zk之係數有何意義 ?
Unordered Samples with Replacement
0 1 2 3z z z z 0 1 2 3z z z z 0 1 2 3z z z z
2 3 2 3 2 31 1 1z z z z z z z z z
2 31n
z z z
1
1
n
z
0
1 k
k
n k
kz
Unordered Samples with Replacement
投返 非投返
有序
無序
knn
kP
n
k
1n k
k
Example 8
Suppose there are 3 boxes which can supply infinite red balls, green balls, and blue balls, respectively. How many possible outcomes if ten balls are chosen from them?
n = 3k = 10
3 10 1 12
10 10
Example 9
There are 3 boxes, the 1st box contains 5 red balls, the 2nd box contains 3 green balls, and the 3rd box contains infinite many blue balls. How many possible outcomes if k balls are chosen from them.
k=1有幾種取法k=2有幾種取法k=3有幾種取法k=4有幾種取法
觀察 :
Example 9
2 3 4 51 z z z z z 2 31 z z z 2 31 z z z
此多項式乘開後zk之係數卽為解
Example 9
2 3 4 51 z z z z z 2 31 z z z 2 31 z z z
6 41 1 1
1 1 1
z z
z z z
36 4 1
1 11
z zz
3
4 6 10 11
1z z z
z
4 6 10
0
3 11 j
j
jz z z z
j
4 6 10
0
21 j
j
jz z z z
j
此多項式乘開後zk之係數卽為解
Example 9
此多項式乘開後zk之係數卽為解
4 6 10
0
21 j
j
jz z z z
j
0 0 0 0
4 6 102 2 2 2j j j j
j j j j
z zj j j j
z z zj j j j
z z
0 0 0
4 10
0
62 2 2 2j j j j
j j j j
j j j jz z z z
j j j j
Coef(zk)=?
Example 9
4 6 1
0
0
0 0 0
2 2 2 2j j j j
j j j j
j j j jz z z z
j j j j
Coef(zk)=?
從 z0 開 始 從 z4 開 始 從 z6 開 始 從 z10 開 始
0 4 6 10
2 2 4 8
4 6 10k k k k
k k k k
k k k kz z z z
k k k k
jk4 jk6 jk10jk
Example 9
4 6 1
0
0
0 0 0
2 2 2 2j j j j
j j j j
j j j jz z z z
j j j j
Coef(zk)=?
從 z0 開 始 從 z4 開 始 從 z6 開 始 從 z10 開 始
0 4 6 10
2 2 4 8
4 6 10k k k k
k k k k
k k k kz z z z
k k k k
jk4 jk6 jk10jk
Example 9
0 4 6 10
2 2 4 8
4 6 10k k k k
k k k k
k k k kz z z z
k k k k
20 4
2 24 6
4( )
2 2 46 10
4 6
2 2 4 810
4 6 10
k
kk
k
k kk
k kCoef z
k k kk
k k k
k k k kk
k k k k
Example 9
20 4
2 24 6
4( )
2 2 46 10
4 6
2 2 4 810
4 6 10
k
kk
k
k kk
k kCoef z
k k kk
k k k
k k k kk
k k k k
2 2 4 8( )
4 6 10k k k k k
Coef zk k k k
Chapter 2Combinatorial Analysis
Derangement
Derangement
最後 ! ! !每一個人都拿到別人的帽子
錯排
Example 10
nkE 0
nE
n人中正好 k人拿對自己的帽子n人中正好 k人拿對自己的帽子 n人中無人拿
對自己的帽子n人中無人拿對自己的帽子
). ?3 ( nkEP
0 ). ?1 ( nEP
0 ) ?2. lim (n
nP E
Example 10
nkE 0
nE
). ?3 ( nkEP
0 ). ?1 ( nEP
0 ) ?2. lim (n
nP E
? ?nkE 0 ?nE
Example 10nkE
0nE
). ?3 ( nkEP
0 ). ?1 ( nEP
0 ) ?2. lim (n
nP E
n人中正好 k人拿對自己的帽子 n人中無人拿對自己的帽子
!n
20( ) ?EP 3
0( ) ?EP
1 2
2 1
1 2
2 3
3
1
3 1 2
1/2! 2/3!
Example 10nkE
0nE
). ?3 ( nkEP
0 ). ?1 ( nEP
0 ) ?2. lim (n
nP E
n人中正好 k人拿對自己的帽子 n人中無人拿對自己的帽子
!n
令 Ai表第 i個人拿了自己帽子
0 0( ) 1 ( )n nP E P E 1 21 ( )nP A A A
Example 10 ). ?3 ( nkEP
0 ). ?1 ( nEP
0 ) ?2. lim (n
nP E
Ai表第 i個人拿了自己帽子
0 1 2( ) 1 ( )nnP E P A A A
Example 10 ). ?3 ( nkEP
0 ). ?1 ( nEP
0 ) ?2. lim (n
nP E
Ai表第 i個人拿了自己帽子
0 1 2( ) 1 ( )nnP E P A A A
1 2 n
1
( 1)!( )
!i
nP A
n
. . .
. . .
Example 10 ). ?3 ( nkEP
0 ). ?1 ( nEP
0 ) ?2. lim (n
nP E
Ai表第 i個人拿了自己帽子
0 1 2( ) 1 ( )nnP E P A A A
1 2 n
1 2
( 1)!( )
!i
nP A
n
. . .
. . .
( 2)!( ) ,
!i j
nP A A i j
n
Example 10 ). ?3 ( nkEP
0 ). ?1 ( nEP
0 ) ?2. lim (n
nP E
Ai表第 i個人拿了自己帽子
0 1 2( ) 1 ( )nnP E P A A A
( 1)!( )
!i
nP A
n
( 2)!( ) ,
!i j
nP A A i j
n
( 3)!( ) ,
!i j k
nP A A A i j k
n
1
n
計 項1
n
計 項2
n
計 項2
n
計 項
3
n
計 項3
n
計 項
( )!
!k
n n kS
k n
! ( )!
!( )! !
n n k
k n k n
1
!k
Example 10 ). ?3 ( nkEP
0 ). ?1 ( nEP
0 ) ?2. lim (n
nP E
Ai表第 i個人拿了自己帽子
0 1 2( ) 1 ( )nnP E P A A A
( )!
!k
n n kS
k n
! ( )!
!( )! !
n n k
k n k n
1
!k 1
0
1 1 1( ) 1 1 1
2! 3! !nnP E
n
1
0
1 1 1( ) 1 1 1
2! 3! !nnP E
n
11 1 11 1
2! 3! !n
n
Example 10 ). ?3 ( nkEP
0 ). ?1 ( nEP
0 ) ?2. lim (n
nP E
Ai表第 i個人拿了自己帽子
0 1 2( ) 1 ( )nnP E P A A A
1
0
1 1 1( ) 1 1 1
2! 3! !nnP E
n
1
0
1 1 1( ) 1 1 1
2! 3! !nnP E
n
20
1 1( ) 1 1
2! 2P E
30
1 1 1( ) 1 1
2! 3! 3P E
40
1 1 1 3( ) 1 1
2! 3! 4! 8P E
Example 10 ). ?3 ( nkEP
0 ). ?1 ( nEP
0 ) ?2. lim (n
nP E
Ai表第 i個人拿了自己帽子
0 1 2( ) 1 ( )nnP E P A A A
1
0
1 1 1( ) 1 1 1
2! 3! !nnP E
n
11 1 1 11 1
1! 2! 3! !n
n
0
1 1 1 1( ) 1 1
1! 2! 3! !nnP E
n
Example 10 ). ?3 ( nkEP
0 ). ?1 ( nEP
0 ) ?2. lim (n
nP E
Ai表第個人拿了自己帽子
0 1 2( ) 1 ( )nnP E P A A A
1
0
1 1 1( ) 1 1 1
2! 3! !nnP E
n
11 1 1 11 1
1! 2! 3! !n
n
10im ( )l
n
nEP e
10im ( )l
n
nEP e
0
1 1 1 1( ) 1 1
1! 2! 3! !nnP E
n
Example 10 ). ?3 ( nkEP
0 ). ?1 ( nEP
0 ) ?2. lim (n
nP E
0
0( )!
n
nE
P En
( )!
nkn
k
EP E
n
0 0! ( )n nE n P E
?nkE
0
1 1 1 1( ) 1 1
1! 2! 3! !nnP E
n
Example 10 ). ?3 ( nkEP
0 ). ?1 ( nEP
0 ) ?2. lim (n
nP E
0
0( )!
n
nE
P En
( )!
nkn
k
EP E
n
0 0! ( )n nE n P E
?nkE
. . .. . .
. . .. . .
k matches n k mismatches
n
k
0n kE 0
!
n kEn
k n
0
1( ) ( )
!n n kkP E P E
k
0( )!
! ( )!
n kEn n k
k n n k
Example 10 ). ?3 ( nkEP
0 ). ?1 ( nEP
0 ) ?2. lim (n
nP E
52( ) ?P E 3
0
1
2!P E
0
1( ) ( )
!n n kkP E P E
k
1 1
2! 3
Remark
0
( ?)n
nk
k
P E
1
Chapter 2Combinatorial Analysis
Calculus
Some Important Derivatives
Derivatives for multiplications —
Derivatives for divisions —
Chain rule —
duvu v uv
dx
2
d v uv u v
dx u u
dy dy du
dx du dx ( )
( )
y f u
u g x
L’Hopital rule
Suppose as we hav( )
, .(
0e or
0)
f xx c
g x
( ) ( )lim lim
( ) ( )x c x c
f x f x
g x g x
Examples
Integration by Part
duv udv vdu
duv udv vdu uv udv vdu
Integration by Part
uv udv vdu udv uv vdu
b bb
aa audv uv vdu
The Gamma Function
1
0( ) , 0xx e dx
Example 12
1
0( ) , 0xx e dx
Example 12
1
0( ) , 0xx e dx
0(1) xe dx
0
xe
0 ( 1)
1
Example 12
1
0( ) , 0xx e dx
0
Example 12
1
0( ) , 0xx e dx
1
0( ) xx e dx
1
0
xx de
1 1
0 0
x xx e e dx 1 2
0 0( 1)x xx e x e dx
( 1)
( 1) ( 1)
Example 12
1
0( ) , 0xx e dx
( ) ( 1) ( 1)
( 1)( 2) ( 2)
( 1)( 2)( 3) (1)
( 1)!
Example 12
1
0( ) , 0xx e dx
2 / 2 ?xe dx
2 / 2 ?xe dx
Example 12
1
0( ) , 0xx e dx
2 / 2 ?xe dx
2 / 2 ?xe dx
2 2 2/ 2 / 2 / 2x x ye dx e dx e dy
2I
2 2
2 2
x y
I e dxdy
22 / 2
0 0
re rdrd
22
2 / 2
0 0 2r r
e d d
2
0d
2
2
Example 12
1
0( ) , 0xx e dx
2 / 2 2xe dx
Example 12
1/ 2
0
1
2xx e dx
2 / 2 2xe dx
1
0( ) , 0xx e dx
2et 2L /x y1/ 2
0
x x
xx e dx
222
2
2
1/ 22 2/ 2
0 2 2
y
y
yy ye d
2 / 2
0
2y y
ye ydy
y
2 / 2
02 ye dy
2 / 22
2ye dy
Example 12
1
0( ) , 0xx e dx
Example 12
1
0( ) , 0xx e dx
?3
2
?5
2
?7
2
1 1
2 2 2
3 3
2 2
3
4
5 5
2 2
15
8
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