Section 4.4 Indefinite Integrals and the Net Change...1 4 x4+C)'=x3 Indefinite Integrals and the Net Change 4 The notation is traditionally used for an antiderivative of f and called

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



means

Example: because

∫ f (x)dx

∫ f (x)dx=F (x) F ' (x)=f (x)

∫ x3dx=14x4+C ( 14 x4+C)'=x3

3



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


4



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


5

Table of Indefinite Integrals

(1)∫ c f (x )dx=c∫ f (x )dx


(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


7

Example 2: Find general indefinite integral

∫(√x3+ 3√x2)dx


(4)∫ xndx= xn+1

n+1+C (n≠−1)

8


∫ v (v2+2)2dv


(4)∫ xndx= xn+1

n+1+C (n≠−1)

9

Example 4: Find general indefinite integral ∫ sin (2 x)sin (x)dx


cos(2α )=cos2α−sin2α=2cos2α−1=1−2sin2αsin(2α )=2sinα cosα

10


Solution: let’s re-write sin(2x) as 2 sinx cosx:

∫ sin (2 x)sin (x)dx



∫ sin (2 x)sin (x)dx∫ sin (2 x)sin (x)dx=∫ 2sin (x )cos(x )sin (x )

dx

11


Solution: let’s re-write sin(2x) as 2 sinx cosx:note that we should not cancel sin(x)





dx

12



Otherwise we will get this:





dx

∫ sin (2 x)sin (x)dx=∫ 2sin (x )cos(x )sin (x )

dx=∫2cos(x )dx=2sin x

13




and





dx



(2sin x) '=2cos x≠sin (2 x )sin (x)

14




and





dx



(2sin x) '=2cos x≠sin (2 x )sin (x)

15



we are looking for F(x) such that





dx

F ' (x)=2sin (x)cos(x)

sin(x)F ' (x)=

2sin (x)cos(x)sin(x)

F (x)=cos2(x )sin (x)

?

16








dx


sin(x)F ' (x)=



?

F ' (x)=( cos2 xsin x )'=2cos x (−sin x )sin x−cos2 x cos x

sin2 x=

17








dx


sin(x)F ' (x)=



?


sin2 x=

=−2cos x sin2 x−cos3 xsin2 x

=

18








dx


sin(x)F ' (x)=



?


sin2 x=


=

=−cos x (2sin2 x+cos2 x)

sin2 x

19








dx


sin(x)F ' (x)=



?


sin2 x=


=

=−cos x (2sin2 x+cos2 x)

sin2 x=

−cos x (sin2 x+1)sin2 x

=…

20








dx


sin(x)F ' (x)=


F (x)=sin2(x )sin (x)

?

21








dx


sin(x)F ' (x)=



?

F ' (x)=( sin2 xsin x )'=2sin x cos x sin x−sin2 x cos x

sin2 x=

22








dx


sin(x)F ' (x)=



?


sin2 x=

=cos x sin2 x

sin2 x

23








dx


sin(x)F ' (x)=



?


sin2 x=

=cos x sin2 x

sin2 x=cos x sin x

sin x

24








dx


sin(x)F ' (x)=


F (x)=2sin2(x )sin (x)

?


sin2 x=

=cos x sin2 x

sin2 x=cos x sin x

sin x

25








dx


sin(x)F ' (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


(4)∫ xndx= xn+1

n+1+C (n≠−1)

27

Example 6: Evaluate the integral ∫1

93 x−2√ x

dx


(4)∫ xndx= xn+1

n+1+C (n≠−1)

28

Example 7: Evaluate the integral∫0

3π2

|sin x|dx


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:


∫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:


∫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.


∫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.


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


∫t1

t2

v (t )dt=s (t 2)−s(t 1)

∫t1

t2

|v (t )|dt

A1

A2

A3

t1 t

2

34





∫t1

t2

v (t )dt=s(t 2)−s (t 1)

∫t1

t2

|v (t )|dt

35




∫t1

t2

v (t )dt=∫1

6

(t 2−2 t−8)dt=

36




∫t1

t2

v (t )dt=∫1

6

(t 2−2 t−8)dt= 13t3]

1

6

−t2 ]16−8 t ]1

6=

37




∫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




∫t1

t2

|v (t )|dt=∫1

6

|t2−2 t−8|dt=

39




∫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




∫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




∫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




∫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




∫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

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Section 4.4 Indefinite Integrals and the Net Change...1 4 x4+C)'=x3 Indefinite Integrals and the Net Change 4 The notation is traditionally used for an antiderivative of f and called