The Definite Integral as an
Accumulator
Bob Arrigo
Scarsdale High School
Scarsdale, NY
r1arrigo@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”
n
lim
 f ( w )  x
k
n 
x  0 k 1
k
•Indefinite Integrals….a family of functions
•Integral functions….functions defined by an integral
x
F ( x)   g (t )dt
a
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.

t b
t a
(rate of change of F) dt  net change in F for a  t  b

t b
t a
RATE dt  Net Change

t b
t a
(rate of change of F) dt  net change in F for 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
0
6
12
18
24
R(t)
13
15
18
14
10

t b
t a
(rate of change of F) dt  net change in F over a  t  b

b
a
F '(t ) dt  F (b)  F (a)

24
0
R(t )dt  6   R(3)  R(9)  R(15)  R(21)  258.6
This is an approximation for the total flow
in gallons of water from the pipe in the
24-hour period.
6
 v(t )dt  (2)  (7.2)  (2)  (12.8)  2  (16.8)
0
TOTAL DISTANCE

Summing up lots of distances,
each of which equals the
product (rate)(time)
Method I to get the total distance traveled:
6
 v(t )dt  (2)  (7.2)  (2)  (12.8)  2  (16.8)
0
Break up the interval [0,6] into smaller and smaller subintervals.
To get the actual distance traveled, use more, smaller subintervals.
n
 v(t )dt  lim  v(t )  t
6
0
n 
k 1
k
t
v(t)
0 0
2 7.2
4 12.8
6 16.8
Next, reveal to your students
that the chart comes from
1 2
v(t )   t  4t
5
Method II
1 2
so, s '(t )   t  4t
5
1 3
 s(t )   t  2t 2  c
15
so, the change in position (displacement) is s(6)  s(0)
Since method I and method II, both yield total distance,
We get:
Answer method I
6
 v(t)dt
0
=
Answer method II

s(6)  s(0)
Since method I and method II both yield total distance,
We get:
Answer method I = Answer method II
6
 v(t )dt
0

t b
t a

s(6)  s(0)
RATE dt  Net Change
What is the minimum CO2 level of the pond?
The amount of CO 2 that has left the pond for
0  t  12 is aproximately
3  f '(3)  3  f '(6)  3  f '(9)  3  f '(12)
The EXACT amount of CO 2 that
entered is  
12
0
f '(t ) dt.
So, the actual amt of CO 2 that is in the pond at
t  12 is given by 
2.6 +

12
0
f '(t ) dt
So, to find the actual amt of CO 2 @ t  12 use
amt @ t  12 is given by
End Amt =
2.6 +

Start Amt 
f (12)  f (0) 

12
0
12
0
f '(t ) dt
NET CHANGE
f '(t ) dt
End Amt = Start Amt + NET CHANGE
f (12)  f (0)  
12
0
f '(t ) dt
In General,
b
f (b)  f (a )   f '(t ) dt
a
In General,
b
f (b)  f (a )   f '(t ) dt
a
OR,....
x
f ( x)  f (a)   f '(t ) dt
a
f ( x)

End Amt =
f (a)

Start Amt 
x
 a f '(t ) dt
NET CHANGE
4
f (4)  f (1)   f '( x) dx
1
end  start  net chg
2
x
f (4)  f (1)  
dx
5
1 1 x
4
f (4)  0  .376
x
f ( x)  f (a)   f '(t ) dt
a
end  start  net chg
For rectilinear particle motion, use
x
s ( x)

s (a )
  v(t ) dt
a
end position = start position  displacement
Ex A particle moves along the x  axis. Its velocity at time t is
t2
10
given by v(t )  2e . At time t  2, the particle is at s(2)  5. What is
the position of the particle at t=5?
Ex A particle moves along the x  axis. Its velocity at time t is
t2
10
given by v(t )  2e . At time t  2, the particle is at s(2)  5. What is
the position of the particle at t=5?
x
s ( x)  s (a )   v(t ) dt
a
s (5)
t2
10
5
 s (2)   2e dt
2
end position = start + net chg
End Amt =
Start Amt 
NET CHANGE
x
N ( x)  700   (r (t )  800) dt
10
N '( x)  r ( x)  800
Since is positive for and is negative for, the maximum value
for occurs at time.
13
At this time there are N (13)  700   (r (t )  800)) dt people, or
10
N (13)  700+3200 (from part a) 800  3  1500 people on line.
End Amt =
Start Amt 
x
NET CHANGE
N ( x)  700   (r (t )  800) dt
10
End Amt =
Start Amt 
NET CHANGE
x
N ( x)  700   (r (t )  800) dt
10
N '( x)  r ( x)  800
Since is positive for and is negative for, the maximum value
for occurs at time.
13
At this time there are N (13)  700   (r (t )  800)) dt people, or
10
N (13)  700+3200 (from part a) 800  3  1500 people on line.
Ex..A particle moves along a
straight line so that its acceleration
at any time t is given by
a (t )  4sin(et ). If its velocity
at time t  2 is 5, what is its velocity
at time t  4
Ex..A particle moves along a
straight line so that its acceleration
at any time t is given by
a (t )  4sin(et ). If its velocity
at time t  2 is 5, what is its velocity
at time t  4
End Amt =
Start Amt 

v(4) = v(2) 
v(4) =
5 
v ( 4) =
5 
4
2

4
2
NET CHANGE
a (t )dt
t t 4  t dt
60.9886

65.9886