The Definite Integral as an Accumulator

Bob ArrigoScarsdale High School

Scarsdale, NYr1arrigo@gmail.com

www.BCCalculus.com

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)

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.”

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

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

Net Changet b

t aRATE dt

(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

(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

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.

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)

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.

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

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

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

2What is the minimum CO level of the pond?

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

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

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

In General,

( ) ( ) '( )b

f b f a f t dta

,....

( ) ( ) '( )

ORx

f x f a f t dta

End A

( ) ( ) '( )

= Start A NET CHANGmt m t E

xf x f a f t dta

4

1

24

51

(4) (1) '( )

(4) 0 .376

(4) (11

)

net chend g

xdxf

star

f f f x dx

f

t

xf

( ) ( ) '( )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

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

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

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

Start A= End Amt NET m Et CHANG

10( ( ) 807( 0 00 ))

xr tN x 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

10( ( ) 807( 0 00 ))

xr tN x dt

..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

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