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

Lemma: Let ℝ. If

1. 
   
   
2. 
   
   

Proof of 1: By contradiction. Assume Then, Therefore, let By hypothesis, This is a contradiction. 
Proof of 2: By hypothesis, by a).


Properties of the Riemann Integral

Let be continuous on an interval and Then,

1. 
   Max-Min Rule.
   

Proof: Since given

1. 
   
   

Also,

2. 
   
   

Combining a) and b), we obtain In summary, we have Since was arbitrary, we have, by the above lemma applied to and to , that 

2. 
   Boundedness Rule.