The LIATE Rule The LIATE Rule The difficulty of integration by parts - - PowerPoint PPT Presentation

▶
the liate rule the liate rule
SMART_READER_LITE
LIVE PREVIEW

The LIATE Rule The LIATE Rule The difficulty of integration by parts - - PowerPoint PPT Presentation

Aug 22, 2022 23 likes •1.03k views

The LIATE Rule The LIATE Rule The difficulty of integration by parts is in choosing u ( x ) and v ( x ) correctly. The LIATE Rule The difficulty of integration by parts is in choosing u ( x ) and v ( x ) correctly. The LIATE rule is a

slide-1
SLIDE 1
slide-2
SLIDE 2
slide-3
SLIDE 3

The LIATE Rule

slide-4
SLIDE 4

The LIATE Rule

The difficulty of integration by parts is in choosing u(x) and v′(x) correctly.

slide-5
SLIDE 5

The LIATE Rule

The difficulty of integration by parts is in choosing u(x) and v′(x) correctly. The LIATE rule is a rule of thumb that tells you which function you should choose as u(x):

slide-6
SLIDE 6

The LIATE Rule

The difficulty of integration by parts is in choosing u(x) and v′(x) correctly. The LIATE rule is a rule of thumb that tells you which function you should choose as u(x):

LIATE

slide-7
SLIDE 7

The LIATE Rule

The difficulty of integration by parts is in choosing u(x) and v′(x) correctly. The LIATE rule is a rule of thumb that tells you which function you should choose as u(x):

L

  • Log

IATE

slide-8
SLIDE 8

The LIATE Rule

The difficulty of integration by parts is in choosing u(x) and v′(x) correctly. The LIATE rule is a rule of thumb that tells you which function you should choose as u(x):

L I

  • Inv. Trig

ATE

slide-9
SLIDE 9

The LIATE Rule

The difficulty of integration by parts is in choosing u(x) and v′(x) correctly. The LIATE rule is a rule of thumb that tells you which function you should choose as u(x):

LI A

  • Algebraic

TE

slide-10
SLIDE 10

The LIATE Rule

The difficulty of integration by parts is in choosing u(x) and v′(x) correctly. The LIATE rule is a rule of thumb that tells you which function you should choose as u(x):

LIA T

  • Trig

E

slide-11
SLIDE 11

The LIATE Rule

The difficulty of integration by parts is in choosing u(x) and v′(x) correctly. The LIATE rule is a rule of thumb that tells you which function you should choose as u(x):

LIAT E

  • Exp
slide-12
SLIDE 12

The LIATE Rule

The difficulty of integration by parts is in choosing u(x) and v′(x) correctly. The LIATE rule is a rule of thumb that tells you which function you should choose as u(x):

LIATE

slide-13
SLIDE 13

The LIATE Rule

The difficulty of integration by parts is in choosing u(x) and v′(x) correctly. The LIATE rule is a rule of thumb that tells you which function you should choose as u(x):

LIATE

The word itself tells you in which order of priority you should use u(x).

slide-14
SLIDE 14

Example 1

slide-15
SLIDE 15

Example 1

  • x sin xdx
slide-16
SLIDE 16

Example 1

  • x sin xdx

Here we have an algebraic and a trigonometric function:

slide-17
SLIDE 17

Example 1

  • x sin xdx

Here we have an algebraic and a trigonometric function:

  • x sin xdx
slide-18
SLIDE 18

Example 1

  • x sin xdx

Here we have an algebraic and a trigonometric function:

  • ✒

A x sin xdx

slide-19
SLIDE 19

Example 1

  • x sin xdx

Here we have an algebraic and a trigonometric function:

  • x✘✘

✘ ✿T

sin xdx

slide-20
SLIDE 20

Example 1

  • x sin xdx

Here we have an algebraic and a trigonometric function:

  • x sin xdx
slide-21
SLIDE 21

Example 1

  • x sin xdx

Here we have an algebraic and a trigonometric function:

  • x sin xdx

According to LIATE, the A comes before the T, so we choose u(x) = x.

slide-22
SLIDE 22

Example 2

slide-23
SLIDE 23

Example 2

  • ex sin xdx
slide-24
SLIDE 24

Example 2

  • ex sin xdx

In this case we have E and T:

slide-25
SLIDE 25

Example 2

  • ex sin xdx

In this case we have E and T:

  • ✚

✚ ❃E

ex sin xdx

slide-26
SLIDE 26

Example 2

  • ex sin xdx

In this case we have E and T:

  • ex✘✘

✘ ✿T

sin xdx

slide-27
SLIDE 27

Example 2

  • ex sin xdx

In this case we have E and T:

  • ex sin xdx
slide-28
SLIDE 28

Example 2

  • ex sin xdx

In this case we have E and T:

  • ex sin xdx

According to LIATE, the T comes before the E, so we choose u(x) = sin x.

slide-29
SLIDE 29

Example 2

  • ex sin xdx

In this case we have E and T:

  • ex sin xdx

According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works:

slide-30
SLIDE 30

Example 2

  • ex sin xdx

In this case we have E and T:

  • ex sin xdx

According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x

slide-31
SLIDE 31

Example 2

  • ex sin xdx

In this case we have E and T:

  • ex sin xdx

According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v′(x) = ex

slide-32
SLIDE 32

Example 2

  • ex sin xdx

In this case we have E and T:

  • ex sin xdx

According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v′(x) = ex u′(x) = cos x

slide-33
SLIDE 33

Example 2

  • ex sin xdx

In this case we have E and T:

  • ex sin xdx

According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v′(x) = ex u′(x) = cos x v(x) = ex

slide-34
SLIDE 34

Example 2

  • ex sin xdx

In this case we have E and T:

  • ex sin xdx

According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v′(x) = ex u′(x) = cos x v(x) = ex

  • ex sin xdx = ex sin x −
  • ex cos xdx
slide-35
SLIDE 35

Example 2

  • ex sin xdx

In this case we have E and T:

  • ex sin xdx

According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v′(x) = ex u′(x) = cos x v(x) = ex

  • ✚

✚ ❃v′(x)

ex sin xdx = ex sin x −

  • ex cos xdx
slide-36
SLIDE 36

Example 2

  • ex sin xdx

In this case we have E and T:

  • ex sin xdx

According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v′(x) = ex u′(x) = cos x v(x) = ex

  • ex✘✘

✘ ✿u(x)

sin xdx = ex sin x −

  • ex cos xdx
slide-37
SLIDE 37

Example 2

  • ex sin xdx

In this case we have E and T:

  • ex sin xdx

According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v′(x) = ex u′(x) = cos x v(x) = ex

  • ex sin xdx = ✚

✚ ❃v(x)

ex sin x −

  • ex cos xdx
slide-38
SLIDE 38

Example 2

  • ex sin xdx

In this case we have E and T:

  • ex sin xdx

According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v′(x) = ex u′(x) = cos x v(x) = ex

  • ex sin xdx = ex✘✘

✘ ✿u(x)

sin x −

  • ex cos xdx
slide-39
SLIDE 39

Example 2

  • ex sin xdx

In this case we have E and T:

  • ex sin xdx

According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v′(x) = ex u′(x) = cos x v(x) = ex

  • ex sin xdx = ex sin x −
  • ✚

✚ ❃v(x)

ex cos xdx

slide-40
SLIDE 40

Example 2

  • ex sin xdx

In this case we have E and T:

  • ex sin xdx

According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v′(x) = ex u′(x) = cos x v(x) = ex

  • ex sin xdx = ex sin x −
  • ex✘✘

✘ ✿u′(x)

cos xdx

slide-41
SLIDE 41

Example 2

  • ex sin xdx

In this case we have E and T:

  • ex sin xdx

According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v′(x) = ex u′(x) = cos x v(x) = ex

  • ex sin xdx = ex sin x −
  • ex cos xdx
slide-42
SLIDE 42

Example 2

slide-43
SLIDE 43

Example 2

Now we need to solve that integral:

slide-44
SLIDE 44

Example 2

Now we need to solve that integral:

  • ex sin xdx = ex sin x −
  • ex cos xdx
slide-45
SLIDE 45

Example 2

Now we need to solve that integral:

  • ex sin xdx = ex sin x −
  • ex cos xdx

We need to use integration by parts again.

slide-46
SLIDE 46

Example 2

Now we need to solve that integral:

  • ex sin xdx = ex sin x −
  • ex cos xdx

We need to use integration by parts again. By using LIATE, we choose u(x) = cos x:

slide-47
SLIDE 47

Example 2

Now we need to solve that integral:

  • ex sin xdx = ex sin x −
  • ex cos xdx

We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x

slide-48
SLIDE 48

Example 2

Now we need to solve that integral:

  • ex sin xdx = ex sin x −
  • ex cos xdx

We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v′(x) = ex

slide-49
SLIDE 49

Example 2

Now we need to solve that integral:

  • ex sin xdx = ex sin x −
  • ex cos xdx

We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v′(x) = ex u′(x) = − sin x

slide-50
SLIDE 50

Example 2

Now we need to solve that integral:

  • ex sin xdx = ex sin x −
  • ex cos xdx

We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v′(x) = ex u′(x) = − sin x v(x) = ex

slide-51
SLIDE 51

Example 2

Now we need to solve that integral:

  • ex sin xdx = ex sin x −
  • ex cos xdx

We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v′(x) = ex u′(x) = − sin x v(x) = ex

  • ex sin xdx = ex sin x −
  • ex cos xdx
slide-52
SLIDE 52

Example 2

Now we need to solve that integral:

  • ex sin xdx = ex sin x −
  • ex cos xdx

We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v′(x) = ex u′(x) = − sin x v(x) = ex

  • ex sin xdx = ex sin x −
  • ✚

✚ ❃v′(x)

ex cos xdx

slide-53
SLIDE 53

Example 2

Now we need to solve that integral:

  • ex sin xdx = ex sin x −
  • ex cos xdx

We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v′(x) = ex u′(x) = − sin x v(x) = ex

  • ex sin xdx = ex sin x −
  • ex✘✘

✘ ✿u(x)

cos xdx

slide-54
SLIDE 54

Example 2

Now we need to solve that integral:

  • ex sin xdx = ex sin x −
  • ex cos xdx

We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v′(x) = ex u′(x) = − sin x v(x) = ex

  • ex sin xdx = ex sin x −
  • ex cos xdx
slide-55
SLIDE 55

Example 2

Now we need to solve that integral:

  • ex sin xdx = ex sin x −
  • ex cos xdx

We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v′(x) = ex u′(x) = − sin x v(x) = ex

  • ex sin xdx = ex sin x −
  • ex cos xdx

= ex sin x −

  • ex cos x −
  • ex (− sin x) dx
slide-56
SLIDE 56

Example 2

Now we need to solve that integral:

  • ex sin xdx = ex sin x −
  • ex cos xdx

We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v′(x) = ex u′(x) = − sin x v(x) = ex

  • ex sin xdx = ex sin x −
  • ex cos xdx

= ex sin x −

  • ✚

✚ ❃v(x)

ex cos x −

  • ex (− sin x) dx
slide-57
SLIDE 57

Example 2

Now we need to solve that integral:

  • ex sin xdx = ex sin x −
  • ex cos xdx

We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v′(x) = ex u′(x) = − sin x v(x) = ex

  • ex sin xdx = ex sin x −
  • ex cos xdx

= ex sin x −

  • ex✘✘

✘ ✿u(x)

cos x −

  • ex (− sin x) dx
slide-58
SLIDE 58

Example 2

Now we need to solve that integral:

  • ex sin xdx = ex sin x −
  • ex cos xdx

We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v′(x) = ex u′(x) = − sin x v(x) = ex

  • ex sin xdx = ex sin x −
  • ex cos xdx

= ex sin x −

  • ex cos x −
  • ✚

✚ ❃v(x)

ex (− sin x) dx

slide-59
SLIDE 59

Example 2

Now we need to solve that integral:

  • ex sin xdx = ex sin x −
  • ex cos xdx

We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v′(x) = ex u′(x) = − sin x v(x) = ex

  • ex sin xdx = ex sin x −
  • ex cos xdx

= ex sin x −

  • ex cos x −
  • ex✘✘✘✘

✘ ✿u′(x)

(− sin x)dx

slide-60
SLIDE 60

Example 2

Now we need to solve that integral:

  • ex sin xdx = ex sin x −
  • ex cos xdx

We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v′(x) = ex u′(x) = − sin x v(x) = ex

  • ex sin xdx = ex sin x −
  • ex cos xdx

= ex sin x −

  • ex cos x −
  • ex (− sin x) dx
slide-61
SLIDE 61

Example 2

slide-62
SLIDE 62

Example 2

So, we have that:

slide-63
SLIDE 63

Example 2

So, we have that:

  • ex sin xdx = ex sin x − ex cos x −
  • ex sin xdx
slide-64
SLIDE 64

Example 2

So, we have that:

  • ex sin xdx
  • !!

= ex sin x − ex cos x −

  • ex sin xdx
  • !!
slide-65
SLIDE 65

Example 2

So, we have that:

  • ex sin xdx
  • !!

= ex sin x − ex cos x −

  • ex sin xdx
  • !!

So, we add

  • ex sin xdx to both sides of the equation:
slide-66
SLIDE 66

Example 2

So, we have that:

  • ex sin xdx
  • !!

= ex sin x − ex cos x −

  • ex sin xdx
  • !!

So, we add

  • ex sin xdx to both sides of the equation:

2

  • ex sin xdx = ex (sin x − cos x)
slide-67
SLIDE 67

Example 2

So, we have that:

  • ex sin xdx
  • !!

= ex sin x − ex cos x −

  • ex sin xdx
  • !!

So, we add

  • ex sin xdx to both sides of the equation:

2

  • ex sin xdx = ex (sin x − cos x)
  • ex sin xdx = ex

2 (sin x − cos x) + C

slide-68
SLIDE 68

Example 2

So, we have that:

  • ex sin xdx
  • !!

= ex sin x − ex cos x −

  • ex sin xdx
  • !!

So, we add

  • ex sin xdx to both sides of the equation:

2

  • ex sin xdx = ex (sin x − cos x)
  • ex sin xdx = ex

2 (sin x − cos x) + C This is in fact an example of Trick number 2.

slide-69
SLIDE 69

Example 3

slide-70
SLIDE 70

Example 3

Let’s do another example:

slide-71
SLIDE 71

Example 3

Let’s do another example:

  • x2exdx
slide-72
SLIDE 72

Example 3

Let’s do another example:

  • x2exdx

We have an A and an E:

slide-73
SLIDE 73

Example 3

Let’s do another example:

  • x2exdx

We have an A and an E:

  • ✒

A x2exdx

slide-74
SLIDE 74

Example 3

Let’s do another example:

  • x2exdx

We have an A and an E:

  • x2✚

✚ ❃E

exdx

slide-75
SLIDE 75

Example 3

Let’s do another example:

  • x2exdx

We have an A and an E:

  • x2exdx
slide-76
SLIDE 76

Example 3

Let’s do another example:

  • x2exdx

We have an A and an E:

  • x2exdx

LIATE says that the A comes before the E. So, we choose u(x) = x2:

slide-77
SLIDE 77

Example 3

Let’s do another example:

  • x2exdx

We have an A and an E:

  • x2exdx

LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2

slide-78
SLIDE 78

Example 3

Let’s do another example:

  • x2exdx

We have an A and an E:

  • x2exdx

LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v′(x) = ex

slide-79
SLIDE 79

Example 3

Let’s do another example:

  • x2exdx

We have an A and an E:

  • x2exdx

LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v′(x) = ex u′(x) = 2x

slide-80
SLIDE 80

Example 3

Let’s do another example:

  • x2exdx

We have an A and an E:

  • x2exdx

LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v′(x) = ex u′(x) = 2x v(x) = ex

slide-81
SLIDE 81

Example 3

Let’s do another example:

  • x2exdx

We have an A and an E:

  • x2exdx

LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v′(x) = ex u′(x) = 2x v(x) = ex

  • x2exdx =
slide-82
SLIDE 82

Example 3

Let’s do another example:

  • x2exdx

We have an A and an E:

  • x2exdx

LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v′(x) = ex u′(x) = 2x v(x) = ex

  • ✒

u(x) x2exdx =

slide-83
SLIDE 83

Example 3

Let’s do another example:

  • x2exdx

We have an A and an E:

  • x2exdx

LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v′(x) = ex u′(x) = 2x v(x) = ex

  • x2✚

✚ ❃v′(x)

exdx =

slide-84
SLIDE 84

Example 3

Let’s do another example:

  • x2exdx

We have an A and an E:

  • x2exdx

LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v′(x) = ex u′(x) = 2x v(x) = ex

  • x2exdx = x2ex −
  • 2xexdx
slide-85
SLIDE 85

Example 3

Let’s do another example:

  • x2exdx

We have an A and an E:

  • x2exdx

LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v′(x) = ex u′(x) = 2x v(x) = ex

  • x2exdx =
  • ✒

u(x) x2ex −

  • 2xexdx
slide-86
SLIDE 86

Example 3

Let’s do another example:

  • x2exdx

We have an A and an E:

  • x2exdx

LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v′(x) = ex u′(x) = 2x v(x) = ex

  • x2exdx = x2✚

✚ ❃v(x)

ex −

  • 2xexdx
slide-87
SLIDE 87

Example 3

Let’s do another example:

  • x2exdx

We have an A and an E:

  • x2exdx

LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v′(x) = ex u′(x) = 2x v(x) = ex

  • x2exdx = x2ex −
  • ✚

✚ ❃u′(x)

2xexdx

slide-88
SLIDE 88

Example 3

Let’s do another example:

  • x2exdx

We have an A and an E:

  • x2exdx

LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v′(x) = ex u′(x) = 2x v(x) = ex

  • x2exdx = x2ex −
  • 2x✚

✚ ❃v(x)

exdx

slide-89
SLIDE 89

Example 3

Let’s do another example:

  • x2exdx

We have an A and an E:

  • x2exdx

LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v′(x) = ex u′(x) = 2x v(x) = ex

  • x2exdx = x2ex −
  • 2xexdx
slide-90
SLIDE 90

Example 3

slide-91
SLIDE 91

Example 3

  • x2exdx = x2ex − 2
  • xexdx
slide-92
SLIDE 92

Example 3

  • x2exdx = x2ex − 2
  • xexdx

Now we need to solve that integral by integration by parts again:

slide-93
SLIDE 93

Example 3

  • x2exdx = x2ex − 2
  • xexdx

Now we need to solve that integral by integration by parts again: u(x) = x

slide-94
SLIDE 94

Example 3

  • x2exdx = x2ex − 2
  • xexdx

Now we need to solve that integral by integration by parts again: u(x) = x v′(x) = ex

slide-95
SLIDE 95

Example 3

  • x2exdx = x2ex − 2
  • xexdx

Now we need to solve that integral by integration by parts again: u(x) = x v′(x) = ex u′(x) = 1

slide-96
SLIDE 96

Example 3

  • x2exdx = x2ex − 2
  • xexdx

Now we need to solve that integral by integration by parts again: u(x) = x v′(x) = ex u′(x) = 1 v(x) = ex

slide-97
SLIDE 97

Example 3

  • x2exdx = x2ex − 2
  • xexdx

Now we need to solve that integral by integration by parts again: u(x) = x v′(x) = ex u′(x) = 1 v(x) = ex

  • x2exdx = x2ex − 2
  • xex −
  • 1.exdx
slide-98
SLIDE 98

Example 3

  • x2exdx = x2ex − 2
  • xexdx

Now we need to solve that integral by integration by parts again: u(x) = x v′(x) = ex u′(x) = 1 v(x) = ex

  • x2exdx = x2ex − 2
  • xex −
  • 1.exdx
  • = x2ex − 2xex + 2ex = ex

x2 − 2x + 2

slide-99
SLIDE 99

Example 3

  • x2exdx = x2ex − 2
  • xexdx

Now we need to solve that integral by integration by parts again: u(x) = x v′(x) = ex u′(x) = 1 v(x) = ex

  • x2exdx = x2ex − 2
  • xex −
  • 1.exdx
  • = x2ex − 2xex + 2ex = ex

x2 − 2x + 2

slide-100
SLIDE 100