Sec. 5.1 – Estimating with Finite Sums
Suppose f(x) is a non-negative, continuous function defined on some interval
[a,b]. Then by the area under the curve y = f(x) between x = a and x = b we
mean the area of the region bounded above by the graph of f(x), below by
the x axis, on the left by the vertical line x = a, and on the right by the
vertical line x = b.
The rectangular area approximations we will describe in this section depend
on subdividing the interval [a,b] into subintervals of equal length. For
example, dividing the interval [0,4] into four uniform pieces produces the
subintervals [0,1], [1,2], [2,3], and [3,4]. The next step is to approximate the
area above each subinterval with a rectangle, with the height of the
rectangle being chosen according to some rule. In particular, we will consider
the following rules:
left endpoint rule called LRAM (Left Rectangular Approximation Method)
The height of the rectangle is the value of the function f(x) at the lefthand endpoint of the subinterval. The width of each rectangle is
ba
x 
.
n
area of first rectangle = f (a)x
area of second rectangle = f ( x1 )x
area of third rectangle = f ( x2 )x
.
.
.
area of nth and last rectangle = f ( xn1 )x
The total area is the sum of the areas of the rectangles.
Example 1: Suppose f ( x)  1  x 2 . Use LRAM to approximate the area
bounded by the x-axis, and the lines x = 0 and x = 1 by dividing
the region into 4 subintervals.
2.0
y
0.2
x
0.2
1.0
1
1
3
, x = , x = , and x = 1,
4
2
4
which will divide the interval 0,1 into 4 regions. Then, we
We will draw lines at x = 0, x =
will inscribe rectangles in each of these regions, so that we
1
have 4 rectangles each having width x  . The inscribed
4
rectangles have areas:
1 1
f (0)  x  1  
4 4
17 1 17
1
f    x   
16 4 64
4
5 1 5
1
f    x   
4 4 16
2
25 1 25
3
f    x 
 
16 4 64
4
The approximate area is the sum of the areas of these 4 rectangles:
1 17 5 25 78
 

 1.21875
A= 
4 64 16 64 64
right endpoint rule called RRAM (Right Rectangular Approximation Method)
The height of the rectangle is the value of the function f(x) at the
right-hand endpoint of the subinterval. The width of each rectangle is
ba
x 
.
n
area of first rectangle = f ( x1 )x
area of second rectangle = f ( x2 )x
area of third rectangle = f ( x3 )x
.
.
.
area of nth and last rectangle = f ( xn )x
The total area is the sum of the areas of the rectangles.
Example 2: Approximate the area in Example 1 by using RRAM.
17 1 17
1
f    x   
16 4 64
4
5 1 5
1
f    x   
4 4 16
2
25 1 25
3
f    x 
 
16 4 64
4
1 1
f (1)  x  2  
4 2
The approximate area is the sum of the areas of these 4 rectangles:
17 5 25 1 47
 
 
 1.46875
A=
64 16 64 2 32
midpoint rule called MRAM (Midpoint Rectangular Approximation Method)
The height of the rectangle is the value of the function f(x) at the
ba
midpoint of the subinterval. The width of the rectangle is x 
.
n
a  x1
area of first rectangle = f ( x1 )x where x1 
(the midpt of  a, x1  )
2
x  x2
area of second rectangle = f ( x2 )x where x2  1
(the midpt of  x1 , x2  )
2
x  x3
area of third rectangle = f ( x3 )x where x3  2
(the midpt of  x2 , x3  )
2
.
.
.
x  xn
area of nth and last rectangle = f ( xn )x where xn  n1
(the midpt of
2
 xn1 , xn 
The total area is the sum of the areas of the rectangles.
Example 3: : Approximate the area in Example 1 by using MRAM.
65 1 65
1
f    x 
 
64 4 256
8
73 1 73
3
f    x 
 
64 4 256
8
89 1 89
5
f    x 
 
64 4 256
8
113 1 113
7
f    x 
 
64 4 256
8
The approximate area is the sum of the areas of these 4 rectangles:
65
73
89 113 85




 1.328125
A=
256 256 256 256 64
MRAM is more accurate that LRAM or RRAM.