4030 - Riemann’s Sums

AP Calculus

A). AccumulationRate * Time = Distance

T

R

a bArea under the curve represents the Accumulated Distance

Big Umbrella: Two Models:

50 mphD=r*t

50(3)=150

accumulation Area under the curveDistance traveled or money

y = (x+5)(x^2-x+7)*.1

a b

R

T

AccumulationWe don’t drive

at a constant rate due to lights,

traffic, and etc. But the distance must be the same

Method of Exhaustion:

Archimedes approximated π by trapping the area of a circle between inscribed and circumscribed polygons

up to a 96-gon

Rectangular Approximations y = (x+5)(x^2-x+7)*.1

Rectangular Approximations - Development

y = (x+5)(x^2-x+7)*.1

1 2 3 4

( )k

TA A A A ATA ATA f x x

y = (x+5)(x^2-x+7)*.1

A1 A2 A3A4

b=x

h=f (x)

*( )

K

k

A b hA f x x

Under Approximations

y = (x+5)(x^2-x+7)*.1

Under Approximation: Inscribed Rectangles

( )ih f x

( )iTA f x x

Over Approximations

y = (x+5)(x^2-x+7)*.1

Over Approximation: Circumscribed Rectangles

( )ch f x

( )cTA f x x

Under Approximation Total Area Over Approximation

( ) ( )i cf x x TA f x x

MethodA). Sketch the graph and the partitions on x.

B). Sketch the height at each partition and find its height.

C) Choose the value required for the approximation.

xc = Circumscribed rectangle for OVER Approximation

xi = Inscribed rectangle for UNDER Approximation

and multiply by the base ( x ) for the Area of each rectangle.

D) Add the areas.

Ex 1:Find the Under and Over Approximations of the area under the

graph of from 0 to 3 with x = 1. 2( ) 16f x x

A). Sketch the graph and the partitions on x.

B). Sketch the height at each partition and find its height.

C). Aunder =

C). Aover =

under over

0 1 2 3 0 1 2 3𝑓 (0 )=16𝑓 (1 )=15𝑓 (2 )=12𝑓 (3 )=7

1 (15 )+1 (12 )+1 (7 )=34

1 (16 )+1 (15 )+1 (12 )=43

34 ≤𝑇𝐴 ≤43

Ex 2:Find the Under and Over Approximations of the area under the

graph of from to with partitions at the friendly numbers

( ) sin( )f x x

Aover =

Aunder =

4 5

6

𝜋4𝜋3𝜋2

2𝜋3

3𝜋4

5𝜋6

𝜋4𝜋3𝜋2

2𝜋3

3𝜋4

5𝜋6

𝑓 ( 5𝜋61 )=1

2

𝑓 ( 3𝜋4 )=√2

2

𝑓 ( 2𝜋3 )=√3

2

=1

𝑓 ( 𝜋3 )=√32

𝑓 ( 𝜋4 )=√22

𝜋12 (√2

2 )+𝜋6 (√32 )+ 𝜋6 ( √3

2 )+ 𝜋12 (√2

2 )+ 𝜋12 ( 1

2 )=¿

𝜋12 (√3

2 )+ 𝜋6 (1 )+ 𝜋6

(1 )+ 𝜋12 (√3

2 )+ 𝜋12 (√2

2 )=¿

𝜋 √212

+ 𝜋 √36

+ 𝜋24

𝜋√312

+ 𝜋√224

+ 𝜋3

Event Time(s) Velocity ft/s

Launch 0 0

Begin Roll maneuver 10 185

End Roll maneuver 15 319

Throttle to 89% 20 447

Throttle to 67% 32 742

Throttle to 104% 59 1325

Maximum dynamic pressure 62 1445

Endeavor, Lift-off May 7,1992

Find the Over and Under Approximations for the altitude of Endeavor at 62 sec.

B). Riemann’s Sumsy = (x+5)(x^2-x+7)*.1

Xa Xb

f (xa )

f (x1)

f (x2)

f (xb )

Riemann showed that any height in the subinterval could be used to approximate the accumulation.

Definition : Riemann’s Sum

11

( )n

i i i i ii

f c x x c x

is the length of the subinterval.

is any point in the subintervalix thi

thiic

Riemann’s LeftWe will use three of Riemann’s sums.

LEFT Riemann:

using the LEFT partition always

1 2 3 4 5

Riemann’s RightWe will use three of Riemann’s sums.

RIGHT Riemann:

using the RIGHT partition always

1 2 3 4 5

Riemann’s MidpointWe will use three of Riemann’s sums.

MIDPOINT Riemann:

using the MIDPOINT of the partition always

NOTE: x does not change. 1 2 3 4 5

Ex 3: Regular Partitions

Find the Left Riemann’s, Right Riemann’s, and Midpoint Riemann’s approximations for the accumulation.

2( ) [0,2]f x x on x = ½.

𝑓 (0 )=0𝑓 ( .5 )= .25

𝑓 (1 )=1𝑓 (1.5 )=2.25𝑓 (2 )=4

𝑓 ( .25 )= .0625𝑓 ( .75 )=5625

𝑓 (1.25 )=1.5625𝑓 (1.75 )=3.0625

𝑅𝐿=12 [ 0+.25+1+2.25 ]=1.75

𝑅𝑅=12 [ .25+1+2.25+4 ]=3.75

𝑅𝑀=12 [ .0635+ .5625+1.5625+3.0625 ]=2.625

Ex 4: Convenient Partitions

Find the Left Riemann’s, Right Riemann’s, and Midpoint Riemann’s approximations for the accumulation.

( ) 1 sin( ) [0,2 ]f x x on 2 40, , , ,23 3

p

1

𝑓 ( 2𝜋3 )=1.8660

𝑓 (𝜋 )=1𝑓 ( 4𝜋

3 )= .1340

𝑓 (2𝜋 )=1

𝑓 ( 𝜋3 )=1.8660

𝑓 ( 5𝜋6 )=1.5

.5

𝑓 ( 5𝜋3 )= .1340

𝑅𝐿=2𝜋3

(1 )+ 𝜋3 (1.866 )+ 𝜋3 (1 )+ 2𝜋3

( .134 )=5.6763

𝑅𝑅=2𝜋3

(1.866 )+𝜋3 (1 )+ 𝜋3 ( .134 )+ 2𝜋3

(1 )=7.19

𝑅𝑀=2𝜋3

(1.866 )+ 𝜋3 (1.5 )+𝜋3 (.5 )+ 2𝜋3

( .134 )=6.2832

Pollution ControlThe TABLE shows the rate of emissions of pollutants from a plant from 12 midnight to 6 am. The EPA regulates the quantity of pollutants and assesses a fine if the quantity is over 10,000.

0 0:30 1 1:30 2 2:30 3 3:30 4 4:30 5 5:30 6

1814 1735 1686 1646 1637 1609 1604 1611 1621 1666 1745 1886 2052

T

ppi

Find the Over and Under approximations for the quantity with t =½ hr.

The plants officials use the under approximation to argue that their emissions are with in the standards. The environmental advocates use the over approximation to argue for sanctions. the EPA, required to make a decision, wants a better estimate. Use the Left and Right Riemann’s and determine if sanctions are required.

𝐴𝑂=12 [1814+1735+1686+1646+1637+1609+1611+1621+1666+1745+1886+2052 ]=10354

𝐴𝑈=12 [1735+1686+1646+1637+1609+1604+1604+1611+1621+1666+1745+1886 ]=10025

𝑅𝐿=12 [1814+1735+1686+1646+1637+1609+1604+1611+1621+1666+1745+1886 ]=10130

𝑅𝑅=12 [1735+1686+1646+1637+1609+1604+1611+1621+1666+1745+1886+2052 ]=10354

Last Update:

• 01/03/07

• Assignment : worksheet

Volume:

Text #24 p. 272

Volume:

Riemann’s Sums 2We will use three of Riemann’s sums.

LEFT Riemann RIGHT Riemann MIDPOINT Riemann

y = (x+5)(x^2-x+7)*.1

The rate of exchange of the dollar versus the Euro over time graph is given. Find the net value of the investment during the 8 months using four subintervals and a) Left Riemanns,(b) Right Riemanns,and (c) Midpoint Riemanns

Graphical

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