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 2eUse 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 Circumscribed Area1

2

3

4

5

i

(2/n) f(0) (2/n) f(0+2/n)

(2/n) f(0+1(2/n)) (2/n) f(0+2(2/n))(2/n) f(0+2(2/n)) (2/n) f(0+3(2/n))(2/n) f(0+3(2/n)) (2/n) f(0+4(2/n))(2/n) f(0+4(2/n)) (2/n) f(0+5(2/n))(2/n) f(0+(i-1)(2/n)) (2/n) f(0+(i)(2/n))

(2/n) (1 + (2/n)²)

(2/n) (1 + (4/n)²)(2/n) (1 + (6/n)²)(2/n) (1 + (8/n)²)(2/n) (1 + (10/n)²)(2/n) (1 + (2i/n)²)

Right -Hand

Example 3Find the area bounded by the function f(x) = x² + 1 and the x-axis on the interval [0,2] using limits.y

x2

5

00

∆x = (2-0)/n = 2/nf(xi) = 1 + (2i/n)² = 1 + 4i²/n²

Ai = 2/n (1 + 4i²/n²)

Lim ∑Ai = Lim ∑f(xi)∆xn→∞ n→∞

Lim ∑ (2/n + 8i²/n³)n→∞

Lim (2/n³) ∑ (n² + 4i²)n→∞

= Lim (2/n³) (n³ + 4(n³/3 + n²/2 + n/6))n→∞

= Lim (2 + 8/3 + 4/n + 8/6n²) = 4.67n→∞

Riemann SumsLet 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

f(ci)∆xi is called a Riemann Sum of f. Furthermore,

if exists, lim f(ci)∆xi we say f is integrable on [a,b].

The definite integral, f(x)dx , is the area under the curve

∑i=1

n

n→∞

∑i=1

n

∫b

a

Definite Integral vs Riemann Sum

Area = f(x) dx∫b

aArea = Lim ∑Ai = Lim ∑f(xi) ∆x

n→∞ n→∞i=1

i=n

i=1

i=n

∆x = (b – a) / n

Area = (3x – 8) dx∫5

2

3i 3Area = Lim ∑ [3(----- + 2) – 8] (---) n nn→∞ i=1

i=n

5-2=3

∆x[f(x)]

xi

a

Σ cai = c Σ aii = m

i = n

Operations:

i = m

i = n

Σ (ai ± bi) = Σ ai ± Σ bii = m

i = n

i = m

i = n

i = m

i = n

constants factor out summations split across ±

C is a constant, n is a positive integer, and ai and bi are dependent on i

Formulas:C is a constant, n is a positive integer, and ai and bi are dependent on i

Σi = 1

i = n1 = n

Σi =1

i = n

Σi = 1

i = n

Σi = 1

i = n

Σi = 1

i = n

c = cn

n(n + 1) n² + ni = ------------- = ---------- 2 2

n(n + 1)(2n + 1) 2n³ + 3n² + ni² = -------------------- = -------------------- 6 6

n(n + 1) ² n² (n² + 2n + 1)i³ = ----------- = --------------------- 2 4

Sigma Notation

Example 4In the following summations, simplify in terms of n.

1. (5) =

2. (2i + 1) =

3. (6i² - 2i) =

4. (4i³ - 6i²) =

Σi = 1

i = n

Σi = 1

i = n

Σi = 1

i = n

Σi = 1

i = n

5n

2(n² + n)------------- + n = n² + 2n 2

6(2n³ + 3n² + n) 2(n² + n)---------------------- - ------------- = 2n³ + 2n² 6 2

4(n² (n² + 2n + 1)) 6(2n³ + 3n² + n) ------------------------ - ---------------------- 4 6

= n4 + 2n³ + n² - 2n³ - 3n² - n = n4 - 2n² - n

Example 5Rewrite following summations as definite integrals.

Σi =1

i = n 3i--- n2 3--- na) Lim

n→∞Σi =1

i = n 2i--- n3 2--- nb) Lim

n→∞

Σi =1

i = n 4i 2i1 + ---- + ---- n n2 2--- ne) Lim

n→∞

∫3

0x² dx ∫

2

0x³ dx

∫2

0(1 + 2x + x²) dx

Σ i =1

i = n πisin --- n π--- nc) Lim

n→∞Σi =1

i = n 2i 2i1 + ---- + ---- n n2 2--- ne) Lim

n→∞

∫2

0(1 + x + x²) dx∫

π

0 sin(x) 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