Matematika

(1)

Section 5.7

Numerical Integration

Math 1a

Introduction to Calculus

April 25, 2008

Announcements

◮ Midterm III is Wednesday 4/30 in class (covers §4.9–5.6) ◮ Friday 5/2 is Movie Day!

(2)

Announcements

◮ Midterm III is Wednesday 4/30 in class (covers §4.9–5.6)

◮ Friday 5/2 is Movie Day!

◮ Final (tentative) 5/23 9:15am

◮ Problem Sessions Sunday, Thursday, 7pm, SC 310

(3)

Outline

Last Time: Integration by Parts

The Trapezoidal Rule

Trapezoids instead of rectangles

The Midpoint Rule

Simpson’s Rule

Error estimates for the Trapezoidal and Midpoint Rules

(4)

Theorem (Integration by Parts)

Let u and v be differentiable functions. Then

∫

u(x)v′₍_x₎_dx₌_u₍_x₎_v₍_x₎

− ∫

v(x)u′₍_x₎_dx .

Succinctly,

∫

u dv=uv− ∫

v du.

Theorem (Integration by Parts, deﬁnite form)

b

a u dv uv b

(5)

∫

u(x)v′₍_x₎_dx₌_u₍_x₎_v₍_x₎

− ∫

v(x)u′₍_x₎_dx .

Succinctly,

∫

u dv=uv− ∫

v du.

∫ b

a u dv= uv

¯ ¯ ¯ ¯

b

a−

∫ b

(6)

Outline

The Midpoint Rule

Simpson’s Rule

(7)

Why estimate an integral when we have the FTC?

◮ Antidifferentiation is

“hard”

◮ Sometimes

antidifferentiation is

impossible

◮ Sometimes all we need

is an approximation

◮ These methods actually

(8)

Trapezoids instead of rectangles

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

◮ In the case of a simple

decreasing function like this one, clearlyLnis too

much andRnis not

enough.

Why not average them out and make a trapezoid?

A

f xi 1 f xi

2 x

The smaller the

(9)

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

much andRnis not

enough.

A

f xi 1 f xi

2 x

The smaller the

(10)

Trapezoids instead of rectangles

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

much andRnis not

enough.

◮ Why not average them

out and make a trapezoid?

∆A=

f xi 1 f xi

2 x

The smaller the

(11)

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

much andRnis not

enough.

∆A= f(xi−1) +f(xi)

2 ∆x

The smaller the

(12)

Trapezoids instead of rectangles

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

much andRnis not

enough.

∆A= f(xi−1) +f(xi)

2 ∆x

◮ The smaller the

(13)

Deﬁnition

Divide the interval[a,b]up intonpieces. Let∆x=

b−a n , xi=a+i∆x, andyi =f(xi)for eachifrom 1 ton.

The(n+1)-pointTrapezoidal Ruleapproximation to

∫ b

a f(x)dx

is given by

Tn(f) = n

∑

i=1

f(xi−1) +f(xi)

2 ∆x

= ∆x

[y 0+y1

2 +

y₁+y₂

2 +· · ·+

y_n₋₁+y_n

2

]

= ∆x

2

(

y₀+2y₁+2y₂+· · ·+2yn−1+yn

(14)

Example

Example Estimate

∫ 1

0 x

2_dx_{with the Trapezoidal Rule and}_n₌_4.

Solution

The set of values is

y₀ y₁ y₂ y₃ y₄ 0 1 4

2 ₂ 4

2 ₃ 4

2 ₄ 4

2

So

Tn f _{2 4}1 0 2 ₁₆1 2 ₁₆4 2 ₁₆9 16₁₆

1 8

2 8 18 16

16 44

(15)

Example

Example Estimate

∫ 1

0 x

Solution

{

y₀,y1,y2,y3,y4

}

={0,(1

4 )2 , (2 4 )2 , (3 4 )2 , (4 4

)2}

So

Tn(f) =₂1

·4 (

0+2·

1

16 +2·

4

16+2·

9 16+ 16 16 ) = 1 8 (₂

+8+18+16

16

)

(16)

Your Turn

Example Estimate

ln 2=

∫ 2

1

xdx

with the Trapezoidal Rule andn=8.

(17)

Your Turn

Example Estimate

ln 2=

∫ 2

1

xdx

(18)

Outline

The Midpoint Rule

Simpson’s Rule

(19)

Midpoints to cancel out the slop

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

◮ The Trapezoidal Rule

averages the values

Another possibility would be to average the

points.

A

f xi 1 xi

2 x

(20)

Midpoints to cancel out the slop

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

averages the values

points.

A

f xi 1 xi

2 x

(21)

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

averages the values

points.

A

f xi 1 xi

2 x

(22)

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

averages the values

◮ Another possibility

would be to average the

points.

∆A=

f xi 1 xi

2 x

(23)

Midpoints to cancel out the slop

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

averages the values

points.

∆A=f

(_x

i−1+xi 2

)

∆x

(24)

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

averages the values

points.

∆A=f

(_x

i−1+xi 2

)

∆x

◮ Again, the smaller the

(25)

The Midpoint Rule

Deﬁnition

b−a

n , and

xi=a+i∆xfor eachifrom 1 ton.

The(n+1)-pointMidpoint Ruleapproximation to

∫ b

a f(x)dxis

given by

Mn(f) = n

∑

i=1

f (x

i−1+xi 2

)

∆x

= ∆x

( f

(_x 0+x1

2

)

+f

(_x 1+x2

2

)

+_{· · ·}+f (_x

n−1+xn 2

(26)

Example

Example Estimate

∫ 1

0 x

2_dx_{with the Midpoint Rule and}_n₌_4.

Solution

The set of midpoints is

0 1₈ 3₈ 5₈ 7₈

So

Mn f 1₄ 0 1₈ 2 3₈ 2 5₈ 2 7₈ 2

1 4

1 9 25 49

64 84

(27)

Example

Example Estimate

∫ 1

0 x

Solution

{

0,1₈,

3 8, 5 8, 7 8 } So

Mn(f) =1₄

(

0+(1 8

)2

+(3

8 )2

+(5

8 )2

+(7

8 )2)

= 1

4

(₁

+9+25+49

64

)

(28)

Your Turn

Example Estimate

ln 2=

∫ 2

1

xdx

with the Midpoint Rule andn=8.

(29)

Your Turn

Example Estimate

ln 2=

∫ 2

1

xdx

(30)

Outline

The Midpoint Rule

Simpson’s Rule

(31)

Parabolas instead of lines

.

◮ Trapezoidal Rule

approximates the function with a line

◮ Why not use a parabola?

◮ A section of a parabola

passing through equally spaced points is

∆x

3 (y0+4y1+y2)

◮ need an even number of

(32)

Simpson’s Rule

Deﬁnition

Divide the interval[a,b]up intonpieces, wherenis even. Let

∆x= b−a

n ,xi=a+i∆x, andyi =f(xi)for eachifrom 1 ton.

The(n+1)-pointSimpson’s Ruleapproximation to

∫ b

a f(x)dxis

given by

Sn(f) = n/2 ∑

i=1

∆x

3

(

y₂_i+4y₂_i+1+y2i+2

)

= b−a

3n (

y₀+4y₁+2y₂+4y₃+· · ·+2yn−2+4yn−1+yn

(33)

Meet the Simpsons

Thomas Simpson

(34)

Examples

Example

Use Simpson’s rule withn=4 to estimate

∫ 1

0 x 2_dx

Solution

y₀ y₁ y₂ y₃ y₄ 0 1₄ 2 2₄ 2 3₄ 2 4₄ 2

So

S4 f _{3 4}1 0 4 ₁₆1 2 ₁₆4 4 ₁₆9 16₁₆

1 12

4 8 36 16

16

1 12

64 16

1 3

(35)

Examples

Example

∫ 1

0 x 2_dx

Solution

{

y₀,y1,y2,y3,y4

}

={0,(1

4 )2 , (2 4 )2 , (3 4 )2 , (4 4

)2}

So

S4(f) =₃1 ·4

(

0+4·

1

16 +2·

4

16 +4·

9 16+ 16 16 ) = 1 12 (₄

+8+36+16

16 ) = 1 12 · 64 16 = 1 3

(36)

Examples

Example

∫ 1

0 x 2_dx

Solution

{

y₀,y1,y2,y3,y4

}

={0,(1

4 )2 , (2 4 )2 , (3 4 )2 , (4 4

)2}

So

S4(f) =₃1 ·4

(

0+4·

1

16 +2·

4

16 +4·

9 16+ 16 16 ) = 1 12 (₄

+8+36+16

(37)

Your turn

Example

Use Simpson’s Rule withn=8 to estimate ln 2.

(38)

Your turn

Example

(39)

Section 5.7

Math 1a

April 25, 2008

Announcements

(40)

Announcements

(41)

Outline

The Midpoint Rule

Simpson’s Rule

(42)

∫

u(x)v′₍_x₎_dx₌_u₍_x₎_v₍_x₎

− ∫

v(x)u′₍_x₎_dx .

Succinctly,

∫

u dv=uv− ∫

v du.

b

a u dv uv b

(43)

∫

u(x)v′₍_x₎_dx₌_u₍_x₎_v₍_x₎

− ∫

v(x)u′₍_x₎_dx .

Succinctly,

∫

u dv=uv− ∫

v du.

∫ b

a u dv= uv

¯ ¯ ¯ ¯

b

a−

∫ b

(44)

Outline

The Midpoint Rule

Simpson’s Rule

(45)

“hard”

◮ Sometimes

impossible

is an approximation

(46)

Trapezoids instead of rectangles

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

much andRnis not

enough.

A

f xi 1 f xi

2 x

The smaller the

(47)

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

much andRnis not

enough.

A

f xi 1 f xi

2 x

The smaller the

(48)

Trapezoids instead of rectangles

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

much andRnis not

enough.

∆A=

f xi 1 f xi

2 x

The smaller the

(49)

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

much andRnis not

enough.

∆A= f(xi−1) +f(xi)

2 ∆x

The smaller the

(50)

Trapezoids instead of rectangles

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

much andRnis not

enough.

∆A= f(xi−1) +f(xi)

2 ∆x

◮ The smaller the

(51)

Deﬁnition

∫ b

a f(x)dx

is given by

Tn(f) = n

∑

i=1

f(xi−1) +f(xi)

2 ∆x

= ∆x

[y 0+y1

2 +

y₁+y₂

2 +· · ·+

y_n₋₁+y_n

2

]

= ∆x

2

(

y₀+2y₁+2y₂+· · ·+2yn−1+yn

(52)

Example

Example Estimate

∫ 1

0 x

Solution

y₀ y₁ y₂ y₃ y₄ 0 1 4

2 ₂ 4

2 ₃ 4

2 ₄ 4

2

So

Tn f _{2 4}1 0 2 ₁₆1 2 ₁₆4 2 ₁₆9 16₁₆

1 8

2 8 18 16

16 44

(53)

Example

Example Estimate

∫ 1

0 x

Solution

{

y₀,y1,y2,y3,y4

}

={0,(1

4 )2 , (2 4 )2 , (3 4 )2 , (4 4

)2}

So

Tn(f) =₂1

·4 (

0+2·

1

16 +2·

4

16+2·

9 16+ 16 16 ) = 1 8 (₂

+8+18+16

16

)

(54)

Your Turn

Example Estimate

ln 2=

∫ 2

1

xdx

(55)

Your Turn

Example Estimate

ln 2=

∫ 2

1

xdx

(56)

Outline

The Midpoint Rule

Simpson’s Rule

(57)

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

averages the values

points.

A

f xi 1 xi

2 x

(58)

Midpoints to cancel out the slop

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

averages the values

points.

A

f xi 1 xi

2 x

(59)

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

averages the values

points.

A

f xi 1 xi

2 x

(60)

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

averages the values

points.

∆A=

f xi 1 xi

2 x

(61)

Midpoints to cancel out the slop

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

averages the values

points.

∆A=f

(_x

i−1+xi 2

)

∆x

(62)

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

averages the values

points.

∆A=f

(_x

i−1+xi 2

)

∆x

(63)

The Midpoint Rule

Deﬁnition

b−a

n , and

∫ b

a f(x)dxis

given by

Mn(f) = n

∑

i=1

f (x

i−1+xi 2

)

∆x

= ∆x

( f

(_x 0+x1

2

)

+f

(_x 1+x2

2

)

+_{· · ·}+f (_x

n−1+xn 2

(64)

Example

Example Estimate

∫ 1

0 x

Solution

0 1₈ 3₈ 5₈ 7₈

So

Mn f 1₄ 0 1₈ 2 3₈ 2 5₈ 2 7₈ 2

1 4

1 9 25 49

64 84

(65)

Example

Example Estimate

∫ 1

0 x

Solution

{

0,1₈,

3 8, 5 8, 7 8 } So

Mn(f) =1₄

(

0+(1 8

)2

+(3

8 )2

+(5

8 )2

+(7

8 )2)

= 1

4

(₁

+9+25+49

64

)

(66)

Your Turn

Example Estimate

ln 2=

∫ 2

1

xdx

(67)

Your Turn

Example Estimate

ln 2=

∫ 2

1

xdx

(68)

Outline

The Midpoint Rule

Simpson’s Rule

(69)

Parabolas instead of lines

.

∆x

3 (y0+4y1+y2)

(70)

Simpson’s Rule

Deﬁnition

∆x= b−a

∫ b

a f(x)dxis

given by

Sn(f) = n/2 ∑

i=1

∆x

3

(

y₂_i+4y₂_i+1+y2i+2

)

= b−a

3n (

y₀+4y₁+2y₂+4y₃+· · ·+2yn−2+4yn−1+yn

(71)

Meet the Simpsons

Thomas Simpson

(72)

Examples

Example

∫ 1

0 x 2_dx

Solution

y₀ y₁ y₂ y₃ y₄ 0 1₄ 2 2₄ 2 3₄ 2 4₄ 2

So

S4 f _{3 4}1 0 4 ₁₆1 2 ₁₆4 4 ₁₆9 16₁₆

1 12

4 8 36 16

16

1 12

64 16

1 3

(73)

Examples

Example

∫ 1

0 x 2_dx

Solution

{

y₀,y1,y2,y3,y4

}

={0,(1

4 )2 , (2 4 )2 , (3 4 )2 , (4 4

)2}

So

S4(f) =₃1 ·4

(

0+4·

1

16 +2·

4

16 +4·

9 16+ 16 16 ) = 1 12 (₄

+8+36+16

16 ) = 1 12 · 64 16 = 1 3

(74)

Examples

Example

∫ 1

0 x 2_dx

Solution

{

y₀,y1,y2,y3,y4

}

={0,(1

4 )2 , (2 4 )2 , (3 4 )2 , (4 4

)2}

So

S4(f) =₃1 ·4

(

0+4·

1

16 +2·

4

16 +4·

9 16+ 16 16 ) = 1 12 (₄

+8+36+16

(75)

Your turn

Example

(76)

Your turn

Example

(77)

Section 5.7

Math 1a

April 25, 2008

Announcements

(78)

Announcements

(79)

Outline

The Midpoint Rule

Simpson’s Rule

(80)

∫

u(x)v′₍_x₎_dx₌_u₍_x₎_v₍_x₎

− ∫

v(x)u′₍_x₎_dx .

Succinctly,

∫

u dv=uv− ∫

v du.

b

a u dv uv b

(81)

∫

u(x)v′₍_x₎_dx₌_u₍_x₎_v₍_x₎

− ∫

v(x)u′₍_x₎_dx .

Succinctly,

∫

u dv=uv− ∫

v du.

∫ b

a u dv= uv

¯ ¯ ¯ ¯

b

a−

∫ b

(82)

Outline

The Midpoint Rule

Simpson’s Rule

(83)

“hard”

◮ Sometimes

impossible

is an approximation

(84)

Trapezoids instead of rectangles

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

much andRnis not

enough.

A

f xi 1 f xi

2 x

The smaller the

(85)

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

much andRnis not

enough.

A

f xi 1 f xi

2 x

The smaller the

(86)

Trapezoids instead of rectangles

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

much andRnis not

enough.

∆A=

f xi 1 f xi

2 x

The smaller the

(87)

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

much andRnis not

enough.

∆A= f(xi−1) +f(xi)

2 ∆x

The smaller the

(88)

Trapezoids instead of rectangles

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

much andRnis not

enough.

∆A= f(xi−1) +f(xi)

2 ∆x

◮ The smaller the

(89)

Deﬁnition

∫ b

a f(x)dx

is given by

Tn(f) = n

∑

i=1

f(xi−1) +f(xi)

2 ∆x

= ∆x

[y 0+y1

2 +

y₁+y₂

2 +· · ·+

y_n₋₁+y_n

2

]

= ∆x

2

(

y₀+2y₁+2y₂+· · ·+2yn−1+yn

(90)

Example

Example Estimate

∫ 1

0 x

Solution

y₀ y₁ y₂ y₃ y₄ 0 1 4

2 ₂ 4

2 ₃ 4

2 ₄ 4

2

So

Tn f _{2 4}1 0 2 ₁₆1 2 ₁₆4 2 ₁₆9 16₁₆

1 8

2 8 18 16

16 44

(91)

Example

Example Estimate

∫ 1

0 x

Solution

{

y₀,y1,y2,y3,y4

}

={0,(1

4 )2 , (2 4 )2 , (3 4 )2 , (4 4

)2}

So

Tn(f) =₂1

·4 (

0+2·

1

16 +2·

4

16+2·

9 16+ 16 16 ) = 1 8 (₂

+8+18+16

16

)

(92)

Your Turn

Example Estimate

ln 2=

∫ 2

1

xdx

(93)

Your Turn

Example Estimate

ln 2=

∫ 2

1

xdx

(94)

Outline

The Midpoint Rule

Simpson’s Rule

(95)

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

averages the values

points.

A

f xi 1 xi

2 x

(96)

Midpoints to cancel out the slop

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

averages the values

points.

A

f xi 1 xi

2 x

(97)

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

averages the values

points.

A

f xi 1 xi

2 x

(98)

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

averages the values

points.

∆A=

f xi 1 xi

2 x

(99)

Midpoints to cancel out the slop

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

averages the values

points.

∆A=f

(_x

i−1+xi 2

)

∆x

(100)

. .

xi−1 .xi .

f(xi−1)

.

f(xi)

averages the values

points.

∆A=f

(_x

i−1+xi 2

)

∆x

(101)

The Midpoint Rule

Deﬁnition

b−a

n , and

∫ b

a f(x)dxis

given by

Mn(f) = n

∑

i=1

f (x

i−1+xi 2

)

∆x

= ∆x

( f

(_x 0+x1

2

)

+f

(_x 1+x2

2

)

+_{· · ·}+f (_x

n−1+xn 2

(102)

Example

Example Estimate

∫ 1

0 x

Solution

0 1₈ 3₈ 5₈ 7₈

So

Mn f 1₄ 0 1₈ 2 3₈ 2 5₈ 2 7₈ 2

1 4

1 9 25 49

64 84

(103)

Example

Example Estimate

∫ 1

0 x

Solution

{

0,1₈,

3 8, 5 8, 7 8 } So

Mn(f) =1₄

(

0+(1 8

)2

+(3

8 )2

+(5

8 )2

+(7

8 )2)

= 1

4

(₁

+9+25+49

64

)

(104)

Your Turn

Example Estimate

ln 2=

∫ 2

1

xdx

(105)

Your Turn

Example Estimate

ln 2=

∫ 2

1

xdx

(106)

Outline

The Midpoint Rule

Simpson’s Rule

(107)

Parabolas instead of lines

.

∆x

3 (y0+4y1+y2)

(108)

Simpson’s Rule

Deﬁnition

∆x= b−a

∫ b

a f(x)dxis

given by

Sn(f) = n/2 ∑

i=1

∆x

3

(

y₂_i+4y₂_i+1+y2i+2

)

= b−a

3n (

y₀+4y₁+2y₂+4y₃+· · ·+2yn−2+4yn−1+yn

(109)

Meet the Simpsons

Thomas Simpson

(110)

Examples

Example

∫ 1

0 x 2_dx

Solution

y₀ y₁ y₂ y₃ y₄ 0 1₄ 2 2₄ 2 3₄ 2 4₄ 2

So

S4 f _{3 4}1 0 4 ₁₆1 2 ₁₆4 4 ₁₆9 16₁₆

1 12

4 8 36 16

16

1 12

64 16

1 3

(111)

Examples

Example

∫ 1

0 x 2_dx

Solution

{

y₀,y1,y2,y3,y4

}

={0,(1

4 )2 , (2 4 )2 , (3 4 )2 , (4 4

)2}

So

S4(f) =₃1 ·4

(

0+4·

1

16 +2·

4

16 +4·

9 16+ 16 16 ) = 1 12 (₄

+8+36+16

16 ) = 1 12 · 64 16 = 1 3

(112)

Examples

Example

∫ 1

0 x 2_dx

Solution

{

y₀,y1,y2,y3,y4

}

={0,(1

4 )2 , (2 4 )2 , (3 4 )2 , (4 4

)2}

So

S4(f) =₃1 ·4

(

0+4·

1

16 +2·

4

16 +4·

9 16+ 16 16 ) = 1 12 (₄

+8+36+16

(113)

Your turn

Example

(114)

Your turn

Example