the definite integral as an accumulator bob arrigo scarsdale high school scarsdale, ny...
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The Definite Integral as an Accumulator
Bob ArrigoScarsdale High School
Scarsdale, [email protected]
www.BCCalculus.com
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Traditional applications of the Definite Integral prior to the Calculus reform movement
• Area, volume, total distance traveled.. (AB)• Arc length, work.. (BC)• Mass, fluid pressure.. (Some college
Calculus courses)
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Calculus Reform in the early 90’s brought in “broader”, more robust applications of the
definite integral……
most prominently, use of the definite integral to calculate “net change”, or “accumulated
change.”
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Types of Integrals
•Definite Integrals…limits of Riemann sums
…”summing up infinitely many infinitesimally small products”
•Indefinite Integrals….a family of functions
•Integral functions….functions defined by an integral
10
lim ( )n
k kn
kx
f w x
( ) ( )x
aF x g t dt
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The definite integral provides net change in a quantity over time.
The definite integral of a rate function yields accumulated change of the
associated function over some interval.
(rate of change of F) net change in F for t b
t adt a t b
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Net Changet b
t aRATE dt
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(rate of change of F) net change in F for t b
t adt a t b
Motivate with a water flow problem:
The rate at which water flows into a tank, in gallons per hour, is given by a differentiable function R of time t. Values of R are given at various times t during a 24 hour period. Approximate the number of gallons of water that flowed into the tank over the 24 hour period.
t R(t)0 136 1512 1818 1424 10
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(rate of change of F) net change in F over t b
t adt a t b
'( ) ( ) ( )b
aF t dt F b F a
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24
0( ) 6 (3) (9) (15) (21) 258.6R t dt R R R R
This is an approximation for the total flow in gallons of water from the pipe in the 24-hour period.
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6
0( ) (2) (7.2) (2) (12.8) 2 (16.8)v t dt
TOTAL DISTANCE
Summing up lots of distances, each of which equals the product (rate)(time)
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6
0( ) (2) (7.2) (2) (12.8) 2 (16.8)v t dt
6
01
( ) lim ( )n
kn
k
v t dt v t t
Method I to get the total distance traveled:
Break up the interval [0,6] into smaller and smaller subintervals.To get the actual distance traveled, use more, smaller subintervals.
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t v(t)
0 0
2 7.2
4 12.8
6 16.8
2
Next, reveal to your students
that the chart comes from
1( ) 4
5v t t t
21so, '( ) 4
5s t t t
3 21( ) 2
15s t t t c
so, the change in position (displacement) is (6) (0)s s
Method II
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Since method I and method II, both yield total distance,
We get:
Answer method I = Answer method II
6
0( ) (6) (0)v t dt s s
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Since method I and method II both yield total distance,
We get:
Answer method I = Answer method II
6
0( ) (6) (0)v t dt s s
Net Changet b
t aRATE dt
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2What is the minimum CO level of the pond?
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2
2
12
0
2
The amount of CO that has left the pond for
0 12 is aproximately
3 '(3) 3 '(6) 3 '(9) 3 '(12)
The EXACT amount of CO that
entered is '( ) .
So, the actual amt of CO that is in the pond at
t
f f f f
f t dt
12
0 12 is given by 2.6 + '( )t f t dt
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1
0
2
2
So, to find the actual amt of CO @ 12 use
is given by 2 amt @ 12 ' ( ).6 + f tt dt
t
12
0(02 (( ) )1 ' )f tf dtf
Start A= End Amt NET m Et CHANG
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End Amt = Start Amt + NET CHANGE
12
0(12) (0) '( )f f f t dt
In General,
( ) ( ) '( )b
f b f a f t dta
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In General,
( ) ( ) '( )b
f b f a f t dta
,....
( ) ( ) '( )
ORx
f x f a f t dta
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End A
( ) ( ) '( )
= Start A NET CHANGmt m t E
xf x f a f t dta
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4
1
24
51
(4) (1) '( )
(4) 0 .376
(4) (11
)
net chend g
xdxf
star
f f f x dx
f
t
xf
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( ) ( ) '( )x
f x f a f t dta
For rectilinear particle motion, use
( ) ( ) (
st
)
=end positio ar displacementt posn ition
xs x s a v t dta
start neend t chg
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2
10
Ex A particle moves along the x axis. Its velocity at time is
given by ( ) 2 . At time 2, the particle is at (2) 5. What is
the position of the particle at t=5?
t
t
v t e t s
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2
10
Ex A particle moves along the x axis. Its velocity at time is
given by ( ) 2 . At time 2, the particle is at (2) 5. What is
the position of the particle at t=5?
t
t
v t e t s
2
105
22(2)
sta
(5)
end positi
( ) ( ) ( )
rt= + o c gn net h
t
e ds t
xs x s a v t d
s
ta
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10( ) 700 ( ( ) 800)
xN x r t dt
'( ) ( ) 800N x r x
At this time there are 13
10(13) 700 ( ( ) 800))N r t dt people, or
(13) 700+3200N (from part a) 800 3 1500 people on line.
Since is positive for and is negative for, the maximum value for occurs at time.
Start A= End Amt NET m Et CHANG
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Start A= End Amt NET m Et CHANG
10( ( ) 807( 0 00 ))
xr tN x dt
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'( ) ( ) 800N x r x
At this time there are 13
10(13) 700 ( ( ) 800))N r t dt people, or
(13) 700+3200N (from part a) 800 3 1500 people on line.
Since is positive for and is negative for, the maximum value for occurs at time.
Start A= End Amt NET m Et CHANG
10( ( ) 807( 0 00 ))
xr tN x dt
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..A particle moves along a
straight line so that its acceleration
at any time is given by
( ) 4sin( ). If its velocity
at time 2 is 5, what is its velocity
at time 4
t
Ex
t
a t e
t
t
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Start A= End Amt NET m Et CHANG
..A particle moves along a
straight line so that its acceleration
at any time is given by
( ) 4sin( ). If its velocity
at time 2 is 5, what is its velocity
at time 4
t
Ex
t
a t e
t
t
4
2
4 4
2
(2)
5
( )
65.98860.
(4) =
(4
988 6
=
6
)
4 = 5( )
a t dt
t t t
v
v
v
dtv
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