 ```Chapter 13 – 4a The Definite Integral
Approximating Areas Using Left and Right Sums
y = 2x
12
Example 1. Find the approximate area between the graph of y = 2x and the xaxis from x = 2 to x = 4. The actual area is 12.
10
8
6
4
2
Using the sum of the areas of rectangles whose heights are the values of y for
the left x-coordinate:
1
1
1
0
0
1
2
3
4
5
1
𝐴𝑟𝑒𝑎 ≈ 4 ∙ 2 + 5 ∙ 2 + 6 ∙ 2 + 7 ∙ 2
y = 2x
12
1
𝐴𝑟𝑒𝑎 ≈ (4 + 5 + 6 + 7) ∙ 2 = 11
10
8
6
4
2
0
0
1
2
3
4
5
Using the sum of the areas of rectangles whose heights are the values of y for the right x-coordinate:
1
1
1
1
𝐴𝑟𝑒𝑎 ≈ 5 ∙ 2 + 6 ∙ 2 + 7 ∙ 2 + 8 ∙ 2
y = 2x
12
1
𝐴𝑟𝑒𝑎 ≈ (5 + 6 + 7 + 8) ∙ 2 = 13
10
8
6
4
2
0
0
1
2
3
4
5
To improve the accuracy of the approximation, increase the number of rectangles. Suppose we use 8 rectangles
y = 2x
y = 2x
12
12
10
10
8
8
6
6
4
4
2
2
0
0
0
1
2
3
4
5
0
1
2
3
4
5
1
4
Left Sum: 𝐴𝑟𝑒𝑎 ≈ (4 + 4.5 + 5 + 5.5 + 6 + 6.5 + 7 + 7.5) ∙ = 11.5
1
Right Sum: 𝐴𝑟𝑒𝑎 ≈ (4.5 + 5 + 5.5 + 6 + 6.5 + 7 + 7.5 + 8) ∙ 4 = 12.5
Example 2. Find the approximate area between the graph of 𝑦 =
√−𝑥 2 + 2𝑥 + 3 and the x-axis from x = 0 to x = 2. The region is a semi-circle
1
2
y = sqrt(4x - x2)
2
and the actual area is 𝜋𝑟 2 = 2𝜋 ≈ 6.283.
1
y = sqrt(4x - x2)
0
0
1
2
2
1
0
0
1
2
3
4
Using the left sum:
𝐴𝑟𝑒𝑎 ≈ 0 ∙ 1 + 1.732051 ∙ 1 + 2 ∙ 1 + 1.732051 ∙ 1 =(0 + 1.732051 + 2 + 1.732051) ∙ 1 = 5.464
Using the right sum:
y = sqrt(4x - x2)
2
1
0
0
1
2
3
4
𝐴𝑟𝑒𝑎 ≈ 1.732051 ∙ 1 + 2 ∙ 1 + 1.732051 ∙ 1 + 0 ∙= (1.732051 + 2 + 1.732051 + 0) ∙ 1 = 5.464
y = sqrt(4x - x2)
y = sqrt(4x - x2)
2
2
1
1
0
0
0
1
2
3
4
0
1
2
3
4
𝐿𝑒𝑓𝑡 𝑆𝑢𝑚 ≈ (0 + 1.32 + 1.73 + 1.94 + 2.00 + 1.94 + 1.73 + 1.32) ∙
1
= 5.991
2
𝑅𝑖𝑔ℎ𝑡 𝑆𝑢𝑚 ≈ (1.32 + 1.73 + 1.94 + 2.00 + 1.94 + 1.73 + 1.32 + 0) ∙
1
= 5.991
2
3
4
Riemann Sum
The left sum and the right sum are special instances of a Riemann Sum. Assume that a function f(x) is continuous
on the interval [a,b]. Divide the interval [a,b] into n sub-intervals of width∆𝑥 =
𝑏−𝑎
.
𝑛
These subintervals have
endpoints 𝑥0 , 𝑥1 , 𝑥2 , 𝑥3 , … , 𝑥𝑛 where 𝑥0 = 𝑎, 𝑥𝑛 = 𝑏, and 𝑥𝑖 − 𝑥𝑖−1 = ∆𝑥 for any value of i from 1 to n. (This
implies that 𝑥𝑖 = 𝑎 + 𝑖∆𝑥 for any i from 1 to n.) If 𝑥𝑖−1 ≤ 𝑐𝑖 ≤ 𝑥𝑖 (i.e., 𝑐𝑖 is in the ith subinterval) the Riemann
sum is given by:
𝑛
𝑆𝑛 = ∑ 𝑓(𝑐𝑖 ) ∆𝑥
𝑖=1
The left sum is a Riemann sum in which every 𝑐𝑖 = 𝑥𝑖−1 (the left end of the subinterval):
𝑛
𝑆𝑛 = ∑ 𝑓(𝑥𝑖−1 ) ∆𝑥
𝑖=1
The right sum is a Riemann sum in which every 𝑐𝑖 = 𝑥𝑖 (the right end of the subinterval):
𝑛
𝑆𝑛 = ∑ 𝑓(𝑥𝑖 ) ∆𝑥
𝑖=1
The midpoint sum is a Riemann sum in which every 𝑐𝑖 =
𝑥𝑖 −𝑥𝑖−1
2
(the midpoint of the subinterval):
𝑛
𝑥𝑖 − 𝑥𝑖−1
𝑆𝑛 = ∑ 𝑓 (
) ∆𝑥
2
𝑖=1
The Definite Integral
If f(x) is continuous on [a,b] the definite integral of f(x) from a to b is the limit of any Riemann sum as n
approaches infinity:
𝑛
𝑏
∫ 𝑓(𝑥)𝑑𝑥 = lim 𝑆𝑛 = lim (∑ 𝑓(𝑐𝑖 ) ∆𝑥)
𝑎
𝑛→∞
𝑛→∞
𝑖=1
The definite integral represents the signed area between the graph of f(x) and the x-axis on the interval [a,b].
The area above the x-axis is positive and the area below the x-axis is negative.
```