1
Section 4.4 Indefinite Integrals and the Net Change
The notation is traditionally used for an
antiderivative of f and called indefinite integral.
means
∫ f (x)dx
∫ f (x)dx=F (x) F ' (x)=f (x)
2
Indefinite Integrals and the Net Change
The notation is traditionally used for an
antiderivative of f and called indefinite integral.
means
Example: because
∫ f (x)dx
∫ f (x)dx=F (x) F ' (x)=f (x)
∫ x3dx=14x4+C ( 14 x4+C)'=x3
3
The notation is traditionally used for an
antiderivative of f and called indefinite integral.
means
Example: because
Therefore, we can look at the indefinite integral as representing an entire family of functions (one antiderivative for each value of constant C).
∫ f (x)dx
∫ f (x)dx=F (x) F ' (x)=f (x)
∫ x3dx=14x4+C ( 14 x4+C)'=x3
Indefinite Integrals and the Net Change
4
The notation is traditionally used for an
antiderivative of f and called indefinite integral.
means
Example: because
Recall that the most general antiderivative is given on an interval – we adopt the convention that when an antiderivative is obtained (indefinite integral), it is valid on an interval.
∫ f (x)dx
∫ f (x)dx=F (x) F ' (x)=f (x)
∫ x3dx=14x4+C ( 14 x4+C)'=x3
Indefinite Integrals and the Net Change
5
Table of Indefinite Integrals
(1)∫ c f (x )dx=c∫ f (x )dx
Indefinite Integrals and the Net Change
(3)∫( f (x)+g (x))dx=∫ f (x )dx+∫ g (x)dx
(2)∫ k dx=kx+C
(4)∫ xndx= xn+1
n+1+C (n≠−1) (5)∫ cos x dx=sin x+C
(6)∫sin x dx=−cos x+C (7)∫ sec2 x dx=tan x+C
(8)∫ csc2 x dx=−cot x+C
(10)∫ csc x cot x dx=−csc x+C
(9)∫ sec x tan x dx=sec x+C
6
Example 1: Verify by differentiation that the formula is correct:
∫cos2 x dx=12 x+14sin (2 x)+C
Indefinite Integrals and the Net Change
7
Example 2: Find general indefinite integral
∫(√x3+ 3√x2)dx
Indefinite Integrals and the Net Change
(4)∫ xndx= xn+1
n+1+C (n≠−1)
8
Example 3: Find general indefinite integral
∫ v (v2+2)2dv
Indefinite Integrals and the Net Change
(4)∫ xndx= xn+1
n+1+C (n≠−1)
9
Example 4: Find general indefinite integral ∫ sin (2 x)sin (x)dx
Indefinite Integrals and the Net Change
cos(2α )=cos2α−sin2α=2cos2α−1=1−2sin2αsin(2α )=2sinα cosα
10
Example 4: Find general indefinite integral
Solution: let’s re-write sin(2x) as 2 sinx cosx:
∫ sin (2 x)sin (x)dx
Indefinite Integrals and the Net Change
cos(2α )=cos2α−sin2α=2cos2α−1=1−2sin2αsin(2α )=2sinα cosα
∫ sin (2 x)sin (x)dx∫ sin (2 x)sin (x)dx=∫ 2sin (x )cos(x )sin (x )
dx
11
Example 4: Find general indefinite integral
Solution: let’s re-write sin(2x) as 2 sinx cosx:note that we should not cancel sin(x)
∫ sin (2 x)sin (x)dx
Indefinite Integrals and the Net Change
cos(2α )=cos2α−sin2α=2cos2α−1=1−2sin2αsin(2α )=2sinα cosα
∫ sin (2 x)sin (x)dx∫ sin (2 x)sin (x)dx=∫ 2sin (x )cos(x )sin (x )
dx
12
Example 4: Find general indefinite integral
Solution: let’s re-write sin(2x) as 2 sinx cosx:note that we should not cancel sin(x)
Otherwise we will get this:
∫ sin (2 x)sin (x)dx
Indefinite Integrals and the Net Change
cos(2α )=cos2α−sin2α=2cos2α−1=1−2sin2αsin(2α )=2sinα cosα
∫ sin (2 x)sin (x)dx∫ sin (2 x)sin (x)dx=∫ 2sin (x )cos(x )sin (x )
dx
∫ sin (2 x)sin (x)dx=∫ 2sin (x )cos(x )sin (x )
dx=∫2cos(x )dx=2sin x
13
Example 4: Find general indefinite integral
Solution: let’s re-write sin(2x) as 2 sinx cosx:note that we should not cancel sin(x)
Otherwise we will get this:
and
∫ sin (2 x)sin (x)dx
Indefinite Integrals and the Net Change
cos(2α )=cos2α−sin2α=2cos2α−1=1−2sin2αsin(2α )=2sinα cosα
∫ sin (2 x)sin (x)dx∫ sin (2 x)sin (x)dx=∫ 2sin (x )cos(x )sin (x )
dx
∫ sin (2 x)sin (x)dx=∫ 2sin (x )cos(x )sin (x )
dx=∫2cos(x )dx=2sin x
(2sin x) '=2cos x≠sin (2 x )sin (x)
14
Example 4: Find general indefinite integral
Solution: let’s re-write sin(2x) as 2 sinx cosx:note that we should not cancel sin(x)
Otherwise we will get this:
and
∫ sin (2 x)sin (x)dx
Indefinite Integrals and the Net Change
cos(2α )=cos2α−sin2α=2cos2α−1=1−2sin2αsin(2α )=2sinα cosα
∫ sin (2 x)sin (x)dx∫ sin (2 x)sin (x)dx=∫ 2sin (x )cos(x )sin (x )
dx
∫ sin (2 x)sin (x)dx=∫ 2sin (x )cos(x )sin (x )
dx=∫2cos(x )dx=2sin x
(2sin x) '=2cos x≠sin (2 x )sin (x)
15
Example 4: Find general indefinite integral
Solution: let’s re-write sin(2x) as 2 sinx cosx:note that we should not cancel sin(x)
we are looking for F(x) such that
∫ sin (2 x)sin (x)dx
Indefinite Integrals and the Net Change
cos(2α )=cos2α−sin2α=2cos2α−1=1−2sin2αsin(2α )=2sinα cosα
∫ sin (2 x)sin (x)dx∫ sin (2 x)sin (x)dx=∫ 2sin (x )cos(x )sin (x )
dx
F ' (x)=2sin (x)cos(x)
sin(x)F ' (x)=
2sin (x)cos(x)sin(x)
F (x)=cos2(x )sin (x)
?
16
Example 4: Find general indefinite integral
Solution: let’s re-write sin(2x) as 2 sinx cosx:note that we should not cancel sin(x)
we are looking for F(x) such that
∫ sin (2 x)sin (x)dx
Indefinite Integrals and the Net Change
cos(2α )=cos2α−sin2α=2cos2α−1=1−2sin2αsin(2α )=2sinα cosα
∫ sin (2 x)sin (x)dx∫ sin (2 x)sin (x)dx=∫ 2sin (x )cos(x )sin (x )
dx
F ' (x)=2sin (x)cos(x)
sin(x)F ' (x)=
2sin (x)cos(x)sin(x)
F (x)=cos2(x )sin (x)
?
F ' (x)=( cos2 xsin x )'=2cos x (−sin x )sin x−cos2 x cos x
sin2 x=
17
Example 4: Find general indefinite integral
Solution: let’s re-write sin(2x) as 2 sinx cosx:note that we should not cancel sin(x)
we are looking for F(x) such that
∫ sin (2 x)sin (x)dx
Indefinite Integrals and the Net Change
cos(2α )=cos2α−sin2α=2cos2α−1=1−2sin2αsin(2α )=2sinα cosα
∫ sin (2 x)sin (x)dx∫ sin (2 x)sin (x)dx=∫ 2sin (x )cos(x )sin (x )
dx
F ' (x)=2sin (x)cos(x)
sin(x)F ' (x)=
2sin (x)cos(x)sin(x)
F (x)=cos2(x )sin (x)
?
F ' (x)=( cos2 xsin x )'=2cos x (−sin x )sin x−cos2 x cos x
sin2 x=
=−2cos x sin2 x−cos3 xsin2 x
=
18
Example 4: Find general indefinite integral
Solution: let’s re-write sin(2x) as 2 sinx cosx:note that we should not cancel sin(x)
we are looking for F(x) such that
∫ sin (2 x)sin (x)dx
Indefinite Integrals and the Net Change
cos(2α )=cos2α−sin2α=2cos2α−1=1−2sin2αsin(2α )=2sinα cosα
∫ sin (2 x)sin (x)dx∫ sin (2 x)sin (x)dx=∫ 2sin (x )cos(x )sin (x )
dx
F ' (x)=2sin (x)cos(x)
sin(x)F ' (x)=
2sin (x)cos(x)sin(x)
F (x)=cos2(x )sin (x)
?
F ' (x)=( cos2 xsin x )'=2cos x (−sin x )sin x−cos2 x cos x
sin2 x=
=−2cos x sin2 x−cos3 xsin2 x
=
=−cos x (2sin2 x+cos2 x)
sin2 x
19
Example 4: Find general indefinite integral
Solution: let’s re-write sin(2x) as 2 sinx cosx:note that we should not cancel sin(x)
we are looking for F(x) such that
∫ sin (2 x)sin (x)dx
Indefinite Integrals and the Net Change
cos(2α )=cos2α−sin2α=2cos2α−1=1−2sin2αsin(2α )=2sinα cosα
∫ sin (2 x)sin (x)dx∫ sin (2 x)sin (x)dx=∫ 2sin (x )cos(x )sin (x )
dx
F ' (x)=2sin (x)cos(x)
sin(x)F ' (x)=
2sin (x)cos(x)sin(x)
F (x)=cos2(x )sin (x)
?
F ' (x)=( cos2 xsin x )'=2cos x (−sin x )sin x−cos2 x cos x
sin2 x=
=−2cos x sin2 x−cos3 xsin2 x
=
=−cos x (2sin2 x+cos2 x)
sin2 x=
−cos x (sin2 x+1)sin2 x
=…
20
Example 4: Find general indefinite integral
Solution: let’s re-write sin(2x) as 2 sinx cosx:note that we should not cancel sin(x)
we are looking for F(x) such that
∫ sin (2 x)sin (x)dx
Indefinite Integrals and the Net Change
cos(2α )=cos2α−sin2α=2cos2α−1=1−2sin2αsin(2α )=2sinα cosα
∫ sin (2 x)sin (x)dx∫ sin (2 x)sin (x)dx=∫ 2sin (x )cos(x )sin (x )
dx
F ' (x)=2sin (x)cos(x)
sin(x)F ' (x)=
2sin (x)cos(x)sin(x)
F (x)=sin2(x )sin (x)
?
21
Example 4: Find general indefinite integral
Solution: let’s re-write sin(2x) as 2 sinx cosx:note that we should not cancel sin(x)
we are looking for F(x) such that
∫ sin (2 x)sin (x)dx
Indefinite Integrals and the Net Change
cos(2α )=cos2α−sin2α=2cos2α−1=1−2sin2αsin(2α )=2sinα cosα
∫ sin (2 x)sin (x)dx∫ sin (2 x)sin (x)dx=∫ 2sin (x )cos(x )sin (x )
dx
F ' (x)=2sin (x)cos(x)
sin(x)F ' (x)=
2sin (x)cos(x)sin(x)
F (x)=sin2(x )sin (x)
?
F ' (x)=( sin2 xsin x )'=2sin x cos x sin x−sin2 x cos x
sin2 x=
22
Example 4: Find general indefinite integral
Solution: let’s re-write sin(2x) as 2 sinx cosx:note that we should not cancel sin(x)
we are looking for F(x) such that
∫ sin (2 x)sin (x)dx
Indefinite Integrals and the Net Change
cos(2α )=cos2α−sin2α=2cos2α−1=1−2sin2αsin(2α )=2sinα cosα
∫ sin (2 x)sin (x)dx∫ sin (2 x)sin (x)dx=∫ 2sin (x )cos(x )sin (x )
dx
F ' (x)=2sin (x)cos(x)
sin(x)F ' (x)=
2sin (x)cos(x)sin(x)
F (x)=sin2(x )sin (x)
?
F ' (x)=( sin2 xsin x )'=2sin x cos x sin x−sin2 x cos x
sin2 x=
=cos x sin2 x
sin2 x
23
Example 4: Find general indefinite integral
Solution: let’s re-write sin(2x) as 2 sinx cosx:note that we should not cancel sin(x)
we are looking for F(x) such that
∫ sin (2 x)sin (x)dx
Indefinite Integrals and the Net Change
cos(2α )=cos2α−sin2α=2cos2α−1=1−2sin2αsin(2α )=2sinα cosα
∫ sin (2 x)sin (x)dx∫ sin (2 x)sin (x)dx=∫ 2sin (x )cos(x )sin (x )
dx
F ' (x)=2sin (x)cos(x)
sin(x)F ' (x)=
2sin (x)cos(x)sin(x)
F (x)=sin2(x )sin (x)
?
F ' (x)=( sin2 xsin x )'=2sin x cos x sin x−sin2 x cos x
sin2 x=
=cos x sin2 x
sin2 x=cos x sin x
sin x
24
Example 4: Find general indefinite integral
Solution: let’s re-write sin(2x) as 2 sinx cosx:note that we should not cancel sin(x)
we are looking for F(x) such that
∫ sin (2 x)sin (x)dx
Indefinite Integrals and the Net Change
cos(2α )=cos2α−sin2α=2cos2α−1=1−2sin2αsin(2α )=2sinα cosα
∫ sin (2 x)sin (x)dx∫ sin (2 x)sin (x)dx=∫ 2sin (x )cos(x )sin (x )
dx
F ' (x)=2sin (x)cos(x)
sin(x)F ' (x)=
2sin (x)cos(x)sin(x)
F (x)=2sin2(x )sin (x)
?
F ' (x)=( sin2 xsin x )'=2sin x cos x sin x−sin2 x cos x
sin2 x=
=cos x sin2 x
sin2 x=cos x sin x
sin x
25
Example 4: Find general indefinite integral
Solution: let’s re-write sin(2x) as 2 sinx cosx:note that we should not cancel sin(x)
we are looking for F(x) such that
∫ sin (2 x)sin (x)dx
Indefinite Integrals and the Net Change
cos(2α )=cos2α−sin2α=2cos2α−1=1−2sin2αsin(2α )=2sinα cosα
∫ sin (2 x)sin (x)dx∫ sin (2 x)sin (x)dx=∫ 2sin (x )cos(x )sin (x )
dx
F ' (x)=2sin (x)cos(x)
sin(x)F ' (x)=
2sin (x)cos(x)sin(x)
F (x)=2sin2(x )sin (x)
?
F ' (x)=(2 sin2 xsin x )'=2 2 sin x cos x sin x−sin2 x cos x
sin2 x=
=2 cos x sin2 x
sin2 x=2 cos x sin x
sin x= sin 2 xsin x
26
Example 5: Evaluate the integral ∫−1
1
t (1−t2)dt
Indefinite Integrals and the Net Change
(4)∫ xndx= xn+1
n+1+C (n≠−1)
27
Example 6: Evaluate the integral ∫1
93 x−2√ x
dx
Indefinite Integrals and the Net Change
(4)∫ xndx= xn+1
n+1+C (n≠−1)
28
Example 7: Evaluate the integral∫0
3π2
|sin x|dx
Indefinite Integrals and the Net Change
29
Applications: New Change Theorem
Recall the Fundamental Theorem of Calculus, part 2:
If f is continuous on [a,b], then ,
where is any antiderivative of f , i.e. .
F'(x) represents the rate of change of y = F(x) with respect to x, and F(b) – F(a) is the change in y, when x changes from a to b, i.e. net change in y.
So we can re-formulate FTC2 as follows:
Indefinite Integrals and the Net Change
∫a
b
f (x )dx=F (b)−F (a)
F F '=f
30
New Change Theorem
The integral of a rate of change is the net change:
Indefinite Integrals and the Net Change
∫a
b
F ' (x)dx=F (b)−F (a)
31
New Change Theorem
The integral of a rate of change is the net change:
This principal can be applied to all of the rates of change in natural and social studies.
Check page 333 in the textbook for a list of instances of this idea.
Meanwhile, we will consider some examples.
Indefinite Integrals and the Net Change
∫a
b
F ' (x)dx=F (b)−F (a)
32
Example: A particle moves along a straight line so that its velocity at a time t is v(t) = t2-2t-8, 1 t 6,measured in meters per second.
(a) Find the displacement of the particle during the time period 1 t 6.
(b) Find the distance traveled during this time period.
Indefinite Integrals and the Net Change
33
If an object is moving along a straight line with position function s(t), then its velocity v(t) = s'(t), so
is the net change in its position, or displacement of the particle.
If we want to calculate the distance an object traveled,
then = the total distance traveled
displacement = A1 - A
2 + A
3
distance = A1 + A
2 + A
3
Indefinite Integrals and the Net Change
∫t1
t2
v (t )dt=s (t 2)−s(t 1)
∫t1
t2
|v (t )|dt
A1
A2
A3
t1 t
2
34
Example: A particle moves along a straight line so that its velocity at a time t is v(t) = t2-2t-8, 1 t 6,measured in meters per second.
(a) Find the displacement of the particle during the time period 1 t 6.
(b) Find the distance traveled during this time period.
Indefinite Integrals and the Net Change
∫t1
t2
v (t )dt=s(t 2)−s (t 1)
∫t1
t2
|v (t )|dt
35
Example: A particle moves along a straight line so that its velocity at a time t is v(t) = t2-2t-8, 1 t 6,measured in meters per second.
(a) Find the displacement of the particle during the time period 1 t 6.
Indefinite Integrals and the Net Change
∫t1
t2
v (t )dt=∫1
6
(t 2−2 t−8)dt=
36
Example: A particle moves along a straight line so that its velocity at a time t is v(t) = t2-2t-8, 1 t 6,measured in meters per second.
(a) Find the displacement of the particle during the time period 1 t 6.
Indefinite Integrals and the Net Change
∫t1
t2
v (t )dt=∫1
6
(t 2−2 t−8)dt= 13t3]
1
6
−t2 ]16−8 t ]1
6=
37
Example: A particle moves along a straight line so that its velocity at a time t is v(t) = t2-2t-8, 1 t 6,measured in meters per second.
(a) Find the displacement of the particle during the time period 1 t 6.
Indefinite Integrals and the Net Change
∫t1
t2
v (t )dt=∫1
6
(t 2−2 t−8)dt= 13t3]
1
6
−t2 ]16−8 t ]1
6=
=13(216−1)−(36−1)−8(6−1)=215
3−35−40=215−225
3=
=−103meters
38
Example: A particle moves along a straight line so that its velocity at a time t is v(t) = t2-2t-8, 1 t 6,measured in meters per second.
(b) Find the distance traveled during this time period.
Indefinite Integrals and the Net Change
∫t1
t2
|v (t )|dt=∫1
6
|t2−2 t−8|dt=
39
Example: A particle moves along a straight line so that its velocity at a time t is v(t) = t2-2t-8, 1 t 6,measured in meters per second.
(b) Find the distance traveled during this time period.
Indefinite Integrals and the Net Change
∫t1
t2
|v (t )|dt=∫1
6
|t2−2 t−8|dt=
t2−2 t−8=(t−4)(t+2)=0t2−2 t−8=(t−4)(t+2)=0
40
Example: A particle moves along a straight line so that its velocity at a time t is v(t) = t2-2t-8, 1 t 6,measured in meters per second.
(b) Find the distance traveled during this time period.
Indefinite Integrals and the Net Change
∫t1
t2
|v (t )|dt=∫1
6
|t2−2 t−8|dt=
t2−2 t−8=(t−4)(t+2)=0t2−2 t−8=(t−4)(t+2)=0
t−4=0 t+2=0or
t=−2,4-2 4
41
Example: A particle moves along a straight line so that its velocity at a time t is v(t) = t2-2t-8, 1 t 6,measured in meters per second.
(b) Find the distance traveled during this time period.
Indefinite Integrals and the Net Change
∫t1
t2
|v (t )|dt=∫1
6
|t2−2 t−8|dt=−∫1
4
(t2−2 t−8)dt+∫4
6
(t2−2 t−8)dt=
t2−2 t−8=(t−4)(t+2)=0t2−2 t−8=(t−4)(t+2)=0
t−4=0 t+2=0or
t=−2,4-2 4
42
Example: A particle moves along a straight line so that its velocity at a time t is v(t) = t2-2t-8, 1 t 6,measured in meters per second.
(b) Find the distance traveled during this time period.
Indefinite Integrals and the Net Change
∫t1
t2
|v (t )|dt=∫1
6
|t2−2 t−8|dt=−∫1
4
(t2−2 t−8)dt+∫4
6
(t2−2 t−8)dt=
t2−2 t−8=(t−4)(t+2)=0t2−2 t−8=(t−4)(t+2)=0
t−4=0 t+2=0or
t=−2,4-2 4
=− 13t3]
1
4
+ t2 ]14+8 t ]1
4+
13t3]
4
5
−t2 ]46−8 t ]4
6
43
Example: A particle moves along a straight line so that its velocity at a time t is v(t) = t2-2t-8, 1 t 6,measured in meters per second.
(b) Find the distance traveled during this time period.
Indefinite Integrals and the Net Change
∫t1
t2
|v (t )|dt=∫1
6
|t2−2 t−8|dt=−∫1
4
(t2−2 t−8)dt+∫4
6
(t2−2 t−8)dt=
t2−2 t−8=(t−4)(t+2)=0t2−2 t−8=(t−4)(t+2)=0
t−4=0 t+2=0or
t=−2,4-2 4
=− 13t3]
1
4
+ t2 ]14+8 t ]1
4+
13t3]
4
5
− t2 ]46−8 t ]4
6=…=32 23meters