Aniq integralni



f (x)dx 

^x2k 2

• 
  x_2k
  

3. 
   f (x_2k ) 4 f (x_{2k 1} ) 
   
4. 
   f (x_{2k 2} )
   

3. 
   ^x2 k
   
4. 
   _b a
   
5. 
   6n
   
6. 
   6
   
7. 
   f (x_2k ) 4 f (x_{2k 1} ) 
   
8. 
   
   
9. 
   f (x_{2k 2} )
   
10. 
    
    
11. 
    (k 0,1,..., n 1)
    

3. 
   taqribiy hisoblanadi. Natijada
   
4. 
   
   

3. 
   b ^x₂ ^x₄
   
4. 
   ^x2 n
   

3. 
   _f (x)dx 
   
4. 
   _f (x)dx _f (x)dx ... 
   
5. 
   _f (x)dx 
   

3. 
   a x₀ x₂
   
4. 
   ^x2 n2
   

3. 
   ^{b a} [( f (x
   
4. 
   _6n 0
   
5. 
   ) 4 f (x₁ ) 
   
6. 
   f (x₂
   
7. 
   )) ( f (x₂
   
8. 
   ) 4 f (x₃ ) 
   

3. 
   f (x₄ )) ... ( f (x_2n2 ) 4 f (x_2n1 ) 
   
4. 
   b a
   
5. 
   f (x_2n ))] 
   

3. 
    [( f (x₀ ) 
   
4. 
   6n
   
5. 
   f (x_2n )) 4( f (x₁ ) 
   
6. 
   f (x₃ ) ...
   

3. 
   ... 
   
4. 
   hosil bo`ladi. Demak,
   
5. 
   f (x_2n1 )) 2( f (x₂ ) 
   
6. 
   f (x₄ ) ... 
   
7. 
   f (x_2n2 ))].
   

3. 
   
   

3. 
   b
   
4. 
   _f (x)dx 
   
5. 
   a
   
6. 
   b a
   
7. 
   
   
8. 
   
   
9. 
   6n
   
10. 
    [ f (x₀ ) 
    
11. 
    f (x_2n ) 4( f (x₁ ) 
    
12. 
    f (x₃ ) ...
    

3. 
   
   

3. 
   ... 
   
4. 
   f (x_2n1 )) 2( f (x₂ ) 
   
5. 
   f (x₄ ) ... 
   
6. 
   f (x_2n2 ))].
   
7. 
   (4 )
   

3. 
   
   

3. 
   formula Simpson formulasi deyiladi. Bu taqribiy formulaning hatoligi
   
4. 
   
   
5. 
   R_n^,
   
6. 
   
   
7. 
   
   
8. 
   f (x)
   
9. 
   
   
10. 
    funksiya
    
11. 
    
    
12. 
    
    
13. 
    [a, b]
    
14. 
    
    
15. 
    da uzluksiz
    

3. 
   _f (iv) _(x)
   
4. 
   hosilaga ega bo`lishi shartida,
   

3. 
   
   

3. 
   
   5
   _{R}(b a)
   
4. 
   _f (iv) _()
   
5. 
   ((a,b))
   

3. 
   
   
4. 
   bo`ladi. Demak,
   
5. 
   b
   
6. 
   n
   
7. 
   
   
8. 
   
   
9. 
   
   
10. 
    
    
11. 
    b a
    
12. 
    
    
13. 
    2880 n⁴
    

3. 
   _f (x)dx 
   
4. 
   a
   
5. 
   [ f (x₀ ) 
   
6. 
   6n
   
7. 
   f (x_2n ) 4( f (x₁ ) 
   
8. 
   
   
9. 
   5
   
10. 
    f (x₃ ) ... 
    
11. 
    f (x_2n1 )) 
    

3. 
   2( f (x₂ ) 
   
4. 
   f (x₄
   
5. 
   ) ... 
   
6. 
   f (x
   
7. 
   
   
8. 
   2n2
   
9. 
   _{))] }(b a)
   
10. 
    2880 n⁴
    
11. 
    f ^(iv) ().
    

3. 
   Misol. Ushbu
   
4. 
   
   
5. 
   
   2
   1
   
6. 
   _е^x dx
   
7. 
   0
   

3. 
   integral to`g`ri to`rtburchaklar, trapetsiyalar va Simpson formulalari yordamida taqribiy hisoblansin.
   
4. 
   ◄ [0,1] segmentni 5 ta teng bo`lakka bo`lamiz. Bunda bo`linish nuqtalari
   
5. 
   
   

3. 
   
   2
   x₀ 0, x₁ 0,2, x₂
   
4. 
   0,4, x₃
   
5. 
   0,6, x₄
   
6. 
   0,8, x₅
   
7. 
   1,0
   

3. 
   
   

3. 
   bo`lib, bu nuqtalarda
   
4. 
   f (x) e^x
   
5. 
   funksiyaning qiymatlari quyidagicha bo`ladi:
   

3. 
   
   
4. 
   f (x₀ ) 1,00000 ,
   
5. 
   f (x₁ ) 0,96079 ,
   
6. 
   f (x₂ ) 0,85214 ,
   
7. 
   f (x₃ ) 0,69768 ,
   
8. 
   f (x₄ ) 0,52729 ,
   
9. 
   f (x₅ ) 0,36788 .
   
10. 
    Har bir bo`lakning o`rtasini ifodalovchi nuqtalar
    

3. 
   x₁ 0,1 ,
   
4. 
   
   
5. 
   
   
6. 
   2
   
7. 
   x₃ 0,3 ,
   
8. 
   
   
9. 
   
   
10. 
    2
    
11. 
    x₅ 0,5 ,
    
12. 
    
    
13. 
    
    
14. 
    2
    
15. 
    x₇ 0,7 ,
    
16. 
    
    
17. 
    
    
18. 
    2
    
19. 
    x₉ 0,9
    
20. 
    
    
21. 
    
    
22. 
    2
    

3. 
   
   
4. 
   bo`lib, bu nuqtalardagi funksiyaning qiymatlari quyidagicha bo`ladi:
   
5. 
   
   
6. 
   f (x₁ ) 0,99005 ,
   
7. 
   
   
8. 
   
   
9. 
   2
   
10. 
    f (x ₃ ) 0,91393 ,
    
11. 
    
    
12. 
    
    
13. 
    2
    
14. 
    f (x₅ ) 0,77680 ,
    
15. 
    
    
16. 
    
    
17. 
    2
    
18. 
    f (x₇ ) 0,61263 ,
    
19. 
    
    
20. 
    
    
21. 
    5
    
22. 
    f (x₉ ) 0,44486 .
    
23. 
    
    
24. 
    
    
25. 
    5
    

1.