MAT 1235
Calculus II
4.4 Part II
Indefinite Integrals and the
Net Change Theorem
http://myhome.spu.edu/lauw
Homework
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WebAssign HW 4.4 II
Preview
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We will discuss the Net Change Theorem
The NCT explains the formulas for computing
displacement and distance traveled you have
learned in physics.
b
From t  a to t  b, displacement   v(t ) dt
a
b
distance traveled   v(t ) dt
a
Recall: Fundamental Theorem of
Calculus Part 2

b
a
f (t )dt  F (b)  F (a)
Where F (t ) is any anitderivatives of f (t ).
Thus, F (t )  f (t )
Recall: Fundamental Theorem of
Calculus Part 2

b
a
f (t )dt  F (b)  F (a)
Where F (t ) is any anitderivatives of f (t ).
Thus, F (t )  f (t )
Recall: Fundamental Theorem of
Calculus Part 2

b
a
f (t )dt  F (b)  F (a)
Where F (t ) is any anitderivatives of f (t ).
Thus, F (t )  f (t )
So , we have

b
a
F (t )dt  F (b)  F (a )
The Net Change Theorem

b
a
Rate of change
of 𝐹(𝑡)
F (t )dt  F (b)  F (a)
Final value –Initial Value
= Net change of 𝐹(𝑡) from 𝑡 = 𝑎 to 𝑡 = 𝑏
The integral of rate of change is the net
change
Example 1:
Solution leaking from a tank
Example 1
𝑉(𝑡) = volume of the solution in the tank at time 𝑡
𝑉′(𝑡) = rate of change of the volume

100
0
V (t )dt  V (100)  V (0)

Example 2:
Linear motion of a particle
Reference
Frame
particle
0
positive direction
negative direction
x
particle
Example 2:
ta
t b
s (t )  position function
v(t )  s(t )  velocity function
b
b
a
a
 v(t )dt   s(t )dt
 change of position from t  a to t  b
 displaceme nt of the particle from t  a to t  b
x
Formulas
From t  a to t  b
b
displacement   v(t )dt
a
distance traveled  ?
Recall: Section 2.7
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displacement ≠ distance traveled
displacement ≠ distance traveled
t 1
t 0
3
0
0
t 4
t 1
2
3
displacement =
distance traveled =
x
s (0)  0
s (1)  3
x
s (4)  2
Observations
t 0
0
t 1
3
x
v  t   0 distance traveled = displacement =
Observations
0
t 4
t 1
2
3
x
v  t   0 distance traveled = -displacement =
Observations: A Unified Formula
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v(t ) if v  t   0
v(t )  

v(t ) if v  t   0
v  t   0 distance traveled = displacement =
v  t   0 distance traveled = -displacement =
Formulas
From t  a to t  b
b
displaceme nt   v(t )dt
a
b
distance traveled   v(t ) dt
a
Example 3
The velocity function (in meters per second) is
given for a particle moving along a line.
v(t )  t  t  6; 1  t  4
2
(a) Find the displacement of the particle.
(b) Find the distance traveled.
Step 1: Find the interval where v(t)>0 and <0
9
 m
2
Example 3
v(t )  t  t  6; 1  t  4
2
Step 2: Use the distance formula
4
distance traveled   v(t ) dt
1
61
m
6
Expectations
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Please pay attention to the solutions
steps.
Do not skip steps.
Do not do multiple things at one step.
It is very helpful to have a diagram.