Problem :
Compute U_{3}(f, 0, 3) and L_{3}(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
M_{1} = f (0) = 4,
M_{2} = f (1) = 1,
M_{3} = f (3) = 1, and
m_{1} = f (1) = 1,
m_{2} = f (2) = 0,
m_{3} = f (2) = 0. Therefore,
U_{3}(f, 0, 3)   = M_{i} = (4 + 1 + 1) = 2 

L_{3}(f, 0, 3)   = m_{i} = (1 + 0 + 0) = 

We may conclude that
≤(x  2)^{2}dx≤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
xaxis. The area therefore counts as
negative, so the definite integral equals
 (1)(b  a) = a  b.