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Problem : Consider the function f (x) = x2 + 1 on the interval [0, 2]. Using four subdivisions, find the left-hand approximation, L4, of the area under the curve of f on the interval indicated.


Δx = = =  
L4 = f (0) + f () + f (1) + f ()  
  = 1 + +2 +  
  = =  

Problem : For the same function, using four subdivisions, find the right-hand sum, R4.


R4 = f () + f (1) + f () + f (2)  
  = +2 + + 5  
  = =  

Problem : For the same function, using four subdivisions, find the midpoint sum, M4.


M4 = f () + f () + f () + f ()  
  = + + +  
  = =  

Notice that on the interval in question, f is a strictly increasing function. If f is increasing on an interval, then Ln < Mn < Rn. If f is decreasing on an interval, then Rn < Mn < Ln.

Problem : Find

f (xk)Δx for f (x) = 2x on [0, 2]    

To solve this problem, notice that the graph of f (x) is a line, and the area in question is in the shape of a right triangle with base 2 and height f (2) = 4. So, the limit of the right hand sum, which is the area under the curve, is

(2)(4) = 4    

Problem : Find

f (xk)Δx for f (x) = on [0, 3]    

To solve this problem, notice that the graph of f (x) is a semicircle of radius 3 centered at the origin. The interval [0, 3] contains a section of the curve that is equivalent to a quarter-circle. Thus, the area under the curve is equal to one-fourth the area of a circle with radius 3, or Π(32) = Π.
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