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Problems

Problems

Problem : Compute U3(f, 0, 3) and L3(f, 0, 3) for f (x) = (x - 2)2.

We subdivide the interval [0, 3] into the three intervals [0, 1], [1, 2], [2, 3], so that M1 = f (0) = 4, M2 = f (1) = 1, M3 = f (3) = 1, and m1 = f (1) = 1, m2 = f (2) = 0, m3 = f (2) = 0. Therefore,


U3(f, 0, 3) = Mi = (4 + 1 + 1) = 2  
L3(f, 0, 3) = mi = (1 + 0 + 0) =  

We may conclude that (x - 2)2dx≤2.

Problem : Compute - 1dx.

This definite integral is equal to the area of a rectangle with height 1 unit and length (b - a) units lying below the x-axis. The area therefore counts as negative, so the definite integral equals - (1)(b - a) = a - b.

Problem : Compute xdx.

Since R(x, 0, 2) is a triangle with base of length 2 and a height of 2, we know that the area should be (2)(2) = 2. We check that this agrees with the Riemann sum definition:


Un(x, 0, 2)=  
 =i  
 =  
 =2 -  
 =2  

and similarly for Ln(x, 0, 2).