```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
π΄πππ ≈ 4 β 2 + 5 β 2 + 6 β 2 + 7 β 2
y = 2x
12
1
π΄πππ ≈ (4 + 5 + 6 + 7) β 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
π΄πππ ≈ 5 β 2 + 6 β 2 + 7 β 2 + 8 β 2
y = 2x
12
1
π΄πππ ≈ (5 + 6 + 7 + 8) β 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: π΄πππ ≈ (4 + 4.5 + 5 + 5.5 + 6 + 6.5 + 7 + 7.5) β = 11.5
1
Right Sum: π΄πππ ≈ (4.5 + 5 + 5.5 + 6 + 6.5 + 7 + 7.5 + 8) β 4 = 12.5
Example 2. Find the approximate area between the graph of π¦ =
√−π₯ 2 + 2π₯ + 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 ππ 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:
π΄πππ ≈ 0 β 1 + 1.732051 β 1 + 2 β 1 + 1.732051 β 1 =(0 + 1.732051 + 2 + 1.732051) β 1 = 5.464
Using the right sum:
y = sqrt(4x - x2)
2
1
0
0
1
2
3
4
π΄πππ ≈ 1.732051 β 1 + 2 β 1 + 1.732051 β 1 + 0 β= (1.732051 + 2 + 1.732051 + 0) β 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
πΏπππ‘ ππ’π ≈ (0 + 1.32 + 1.73 + 1.94 + 2.00 + 1.94 + 1.73 + 1.32) β
1
= 5.991
2
πππβπ‘ ππ’π ≈ (1.32 + 1.73 + 1.94 + 2.00 + 1.94 + 1.73 + 1.32 + 0) β
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 widthβπ₯ =
π−π
.
π
These subintervals have
endpoints π₯0 , π₯1 , π₯2 , π₯3 , … , π₯π where π₯0 = π, π₯π = π, and π₯π − π₯π−1 = βπ₯ for any value of i from 1 to n. (This
implies that π₯π = π + πβπ₯ for any i from 1 to n.) If π₯π−1 ≤ ππ ≤ π₯π (i.e., ππ is in the ith subinterval) the Riemann
sum is given by:
π
ππ = ∑ π(ππ ) βπ₯
π=1
The left sum is a Riemann sum in which every ππ = π₯π−1 (the left end of the subinterval):
π
ππ = ∑ π(π₯π−1 ) βπ₯
π=1
The right sum is a Riemann sum in which every ππ = π₯π (the right end of the subinterval):
π
ππ = ∑ π(π₯π ) βπ₯
π=1
The midpoint sum is a Riemann sum in which every ππ =
π₯π −π₯π−1
2
(the midpoint of the subinterval):
π
π₯π − π₯π−1
ππ = ∑ π (
) βπ₯
2
π=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:
π
π
∫ π(π₯)ππ₯ = lim ππ = lim (∑ π(ππ ) βπ₯)
π
π→∞
π→∞
π=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.
```