Definition (Definite Integral): Let be continuous on the closed interval

Proof: by the Non-Negative Rule. By the Linear Rule, we then obtain 

10. 
    Increasing Rule.
    

Proof: by the Positive Rule. By the Linear Rule, we then obtain 

11. 
    Zero Width Interval Rule.
    

12. 
    Order Integration Rule.
    

13. 
    Additive Rule.
    

Proof: Since is continuous on , it is continuous on and on . Therefore,

1. 
   of ,
   
2. 
   of ,
   
3. 
   of ,
   

Let and be a partitions of and , respectively, such that and Let . Then and . So,

Thus, Since was arbitrary, 

14. 
    If is continuous on the smallest interval that contains , then , no matter what the order of are Interval Additive Rule.
    

Proof: Consider the case The other cases are handled in the same way. Now, Thus, The other cases are done in the same way. 


Exercises:

1. 
   Let be a polynomial and Show that
   
2. 
   Let be continuous on the closed interval If show that