4030 - Riemann’s Sums
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Transcript of 4030 - Riemann’s Sums
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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