Aniq integralni



f (x)dx 

f (x_k ) 

2

f (x_{k 1} )

(x_{k 1} x_k )

(k 0,1,2,..., n 1)

x_k

taqribiy hisoblanadi. Natijada ushbu

b

_f (x)dx 

a

x<sub>1

</sub>f (x)dx 


1
x₀

x<sub>2

</sub>f (x)dx ... 

x₁

x<sub>n

</sub>f (x)dx 


2
^xn1

_f (x₀ ) 

2

f (x₁ ) _(x

x₀ ) 

f (x₁ ) 

2

f (x₂ ) _(x

• 
  x₁
  

) ...

... 

f (x_n1 ) 

2

f (x_n ) _(x

• 
  ^xn1
  

_{) }b a _


n
2

f (x₀ ) 

2

f (x_n ) _

f (x₁ ) 

f (x₂ ) ... 

f (x_n1 ) 

formulaga kelamiz. Demak,

^b b a f (x ) 

_f (x)dx  [ ⁰

f (x_n ) _

a

f (x₁ ) 

n

f (x₂ ) ... 

2

f (x_n1 )] .

(3)

3. 
   formula trapetsiyalar formulasi deyiladi.
   

3. 
   Bu taqribiy formulaning hatoligi
   
4. 
   R_n^, f (x) funksiya [a,b]
   
5. 
   da uzluksiz
   
6. 
   f ^(x)
   

3. 
   
   
4. 
   hosilaga ega bo`lishi shartida ,
   
5. 
   
   

3. 
   
   
4. 
   
   
5. 
   
   
6. 
   bo`ladi.
   
7. 
   Demak,
   
8. 
   R_n^
   
9. 
   _(b a)
   
10. 
    
    3
    12n²
    
11. 
    f ^
    
12. 
    ((a, b))
    

3. 
   
   

3. 
   b
   
4. 
   _f (x)dx 
   
5. 
   a
   
6. 
   b a _[
   
7. 
   n
   
8. 
   f (x₀ ) 
   
9. 
   2
   
10. 
    f (x_n ) _
    
11. 
    3
    
12. 
    f (x₁ ) 
    

• 
  f (x₂
  
3. 
   ) ... 
   
4. 
   f (x
   
5. 
   
   
6. 
   n1
   
7. 
   _{)] }(b a)
   
8. 
   12n²
   
9. 
   f ^().
   

3. 
   3⁰. Simpson formulasi. Bu holda
   
4. 
   f (x)
   
5. 
   funksiyaning
   

3. 
   
   
4. 
   b
   
5. 
   _f (x)dx
   
6. 
   a
   
7. 
   
   

3. 
   integralini taqribiy hisoblash uchun
   
4. 
   [a,b]
   
5. 
   segmentni
   
6. 
   a x₀ ,
   
7. 
   x₁,..., x_2k ,
   
8. 
   ^x2k 1^,
   

3. 
   ^x2k 2 ^{,..., x}2n2 ^{, x}2n1^{, x}2n ^b
   
4. 
   nuqtalar yordamida ²ⁿ
   
5. 
   ta teng bo`lakka bo`lib, har bir
   

3. 
   
   

3. 
   ^[x2k ^{, x}2k 2 ^]
   
4. 
   (k 0,1,2,..., n 1)
   
5. 
   bo`yicha integralni quyidagicha
   

3. 
   
   

3. 
   ^x2 k 2
   
4.