We can apply the Midpoint Rule to the approximate any definite integral - not just those representing area. The basic steps are summerized as follows.
Using the Midpoint Rule
to Approximate a Definite Integral
To approximate the definite integral
∫ a b f ( x)     dx
by the Midpoint Rule, use the following steps.

   1.  Divide the interval [a,b] into n subintervals, each of width
△ x = b - a n

   2.  Find the midpoint of each subinterval
M i d p o i n t s = { x 1 , x 2 , x 3 , &dots; , x n }
   3.  Evaluate f at each of these midpoints and form the following sum
∫ a b f ( x ) d x ≈ b - a n [ f ( x 1 ) + f ( x 2 ) + f ( x 3 ) + &dots; + f ( x n ) ]


EXAMPLE 1
Use the Midpoint Rule with n = 4 to approximate the definite integral

∫ 1 2 1 x     dx

SOLUTION
Using four subintervals, each of width

△ x = 2 - 1 4 = 1 4

we divide the interval [1,2] into the following subintervals.

[ 1 , 5 4 ] [ 5 4 , 3 2 ] [ 3 2 , 7 4 ] [ 7 4 , 2 ]

Since the midpoints of these subintervals are given by

Midpoints = {   9 8 ⏞ x 1 ,   11 8 ⏞ x 2 ,   13 8 ⏞ x 3 ,   15 8 ⏞ x 4   }

we approximate the definite integral as follows.

∫ 1 2 1 x     dx ≈ b - a n [ f ( x 1 ) + f ( x 2 ) + f ( x 3 ) + f ( x 4 ) ] =
= 1 4 [ 8 9 + 8 11 + 8 13 + 8 15 ] ≈ 0.691

The area represented by this integral is shown in the next Figure

For the region in the previous Figure we can find the exact area with a definite integral. That is,

∫ 1 2 1 x     dx = [ ln x ] 1 2 = ln 2 - ln 1 =
= ln 2 - 0 = ln 2 ≈ 0.693


EXAMPLE 2
Use the Midpoint Rule with n = 5 to approximate the definite integral

∫ 0 1 1 x 2 +1     dx

SOLUTION
Using four subintervals, each of width

△ x = 1 - 0 5 = 1 5

we divide the interval [0, 1] into the following subintervals.

[ 0 , 1 5 ] [ 1 5 , 2 5 ] [ 2 5 , 3 5 ] [ 3 5 , 4 5 ] [ 4 5 , 1 ]

Since the midpoints of these subintervals are given by

Midpoints = {   1 10 ⏞ x 1 ,     3 10 ⏞ x 2 ,     5 10 ⏞ x 3 ,     7 10 ⏞ x 4 ,     9 10 ⏞ x 5 }

we approximate the definite integral as follows.

∫ 0 1 1 x 2 +1     dx ≈ b - a n [ f ( x 1 ) + f ( x 2 ) + f ( x 3 ) + f ( x 4 ) + f ( x 5 ) ] =
= 1 5 [ 1 1.01 + 1 1.09 + 1 1.25 + 1 1.49 + 1 1.81 ] ≈ 0.786

The area represented by this integral is shown in the next Figure.


For the region in the previous Figure we can find the exact area with a definite integral. That is,

∫ 0 1 1 x 2 +1     dx = [ arctan x ] 0 1 = arctan 1 - arctan 0 = π 4 ≈ 0.785

You can look for the more examples and test your ability for solving these problems.