Transcript Lesson 6-2

Lesson 5-2R
Riemann Sums
Objectives
• Understand Riemann Sums
Vocabulary
• Riemann Sum – a summation of n rectangles used to
estimate the area under curve; when used with a
limit as n approached infinity, then the Riemann sum
is the definite integral
• Definite Integral – the integral evaluated at an upper
limit (b) minus it evaluated at a lower limit (a); gives
the area under the curve (in two dimensions)
Example 2e
Use sums to describe the area of the region between
the graph of y = x² + 1 and the x-axis from x = 0 to x =
2. Partition [0,2] into n intervals, the width of the
intervals will be (2-0)/n = 2/n. Since the function is
increasing on this interval, the left-hand (inscribed)
heights will be f(xi-1) and the right-hand
(circumscribed) heights will be f(xi).
Rectangle Inscribed Area
Right -Hand
Circumscribed Area
1
(2/n) f(0)
(2/n) (1 + (2/n)²)
(2/n) f(0+2/n)
2
(2/n) f(0+1(2/n))
(2/n) (1 + (4/n)²)
(2/n) f(0+2(2/n))
3
(2/n) f(0+2(2/n))
(2/n) (1 + (6/n)²)
(2/n) f(0+3(2/n))
4
(2/n) f(0+3(2/n))
(2/n) f(0+4(2/n))
5
(2/n) f(0+4(2/n))
(2/n) (1 + (8/n)²)
(2/n) (1 + (10/n)²)
i
(2/n) f(0+(i-1)(2/n))
(2/n) (1 + (2i/n)²)
(2/n) f(0+(i)(2/n))
(2/n) f(0+5(2/n))
Example 3
Find the area bounded by the function f(x) = x² + 1 and
the
x-axis on the interval [0,2] using limits.
y
5
Lim ∑Ai = Lim ∑f(xi)∆x
n→∞
n→∞
∆x = (2-0)/n = 2/n
f(xi) = 1 + (2i/n)² = 1 + 4i²/n²
Ai = 2/n (1 + 4i²/n²)
Lim ∑ (2/n + 8i²/n³)
n→∞
x
0
0
2
Lim (2/n³) ∑ (n² + 4i²) = Lim (2/n³) (n³ + 4(n³/3 + n²/2 + n/6))
n→∞
n→∞
= Lim (2 + 8/3 + 4/n + 8/6n²) = 4.67
n→∞
Riemann Sums
Let f be a function that is defined on the closed interval [a,b]. If ∆
is a partition of [a,b] and ∆xi is the width of the ith interval, ci, is
any point in the subinterval, then the sum
Left-Edge Rectangles
Midpoint Rectangles
a
a
b
b
Right-Edge Rectangles
a
n
∑
i=1
f(ci)∆xi is called a Riemann Sum of f. Furthermore,
n
if exists, lim
∑ f(ci)∆xi
n→∞ i=1
b
The definite integral,
we say f is integrable on [a,b].
∫a f(x)dx , is the area under the curve
b
Definite Integral vs Riemann Sum
b
Area =
∫a f(x) dx
i=n
i=n
Area = Lim ∑Ai = Lim ∑f(xi) ∆x
n→∞
i=1
n→∞
i=1
∆x = (b – a) / n
5
Area =
∫2 (3x – 8) dx
a
5-2=3
xi
3i
3
Area = Lim ∑ [3(----- + 2) – 8] (---)
n→∞ i=1
n
n
i=n
[f(x)]
∆x
Sigma Notation
Operations:
C is a constant, n is a positive integer, and ai and bi are dependent on i
i=n
Σ
i=m
i=n
i=n
cai = c
Σ (a ± b ) = Σ a ± Σ b
Σa
i=m
i=n
i=n
i
i=m
i
i
i=m
i
i=m i
summations split across ±
constants factor out
Formulas:
C is a constant, n is a positive integer, and ai and bi are dependent on i
i=n
i=n
Σ1
Σ
= n
i=1
i=1
i=n
Σ
i =1
i=n
Σ
i=1
n(n + 1)
i = ------------- =
2
c = cn
n² + n
---------2
n(n + 1) ²
n² (n² + 2n + 1)
i³ = ----------- = --------------------2
4
i=n
Σ
i=1
n(n + 1)(2n + 1)
2n³ + 3n² + n
i² = -------------------- = -------------------6
6
Example 4
In the following summations, simplify in terms of n.
i=n
1.
Σ
i=1
5n
(5) =
2(n² + n)
------------- + n = n² + 2n
2
i=n
2.
Σ
(2i + 1) =
i=1
i=n
3.
Σ
i=1
6(2n³ + 3n² + n)
2(n² + n)
(6i² - 2i) = ---------------------- - ------------- = 2n³ + 2n²
6
2
i=n
4.
Σ
i=1
(4i³ - 6i²) =
4(n² (n² + 2n + 1))
6(2n³ + 3n² + n)
------------------------ - ---------------------4
6
= n4 + 2n³ + n² - 2n³ - 3n² - n = n4 - 2n² - n
Example 5
Rewrite following summations as definite integrals.
i=n
a) Lim
n→∞
Σ
i =1
3i 2 3
----n
n
i=n
b) Lim Σ
n→∞
i =1
2
3
∫0x³ dx
∫0x² dx
i=n
c) Lim
n→∞
Σ
i =1
πi
sin --n
π
--n
i=n
e) Lim
n→∞
π
i=n
n→∞
Σ
i =1
4i
2i 2 2
--1 + ---- + ---n
n
n
Σ
i =1
2i
2i 2 2
--1 + ---- + ---n
n
n
2
∫0 sin(x) dx
e) Lim
2i 3 2
----n
n
∫0(1 + x + x²) dx
2
∫0(1 + 2x + x²) dx
Summary & Homework
• Summary:
– Riemann Sums are Limits of Infinite sums
– Riemann Sums give exact areas under the curve
– Riemann Sums are the definite integral
• Homework:
– pg 390 - 393: 3, 5, 9, 17, 20, 33, 38