Calculus (I)

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Calculus (I)

W^EN-C^HING L^IEN

Department of Mathematics National Cheng Kung University

2008

WEN-CHINGLIEN Calculus (I)

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8.7 Numerical Integration

(1) The Midpoint Rule

Support that [x0,x1, . . . ,xn] is a partition of[a,b]into n subintervals of equal lengths. Assume that f is continuous on[a,b].

We approximate

Z b

a

f(x)dx

by

Mn= b−a n

n

X

k=1

f(ck),

where ck = ^x^k⁻¹₂^+x^k is the midpoint of [xk−1,xk].

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8.7 Numerical Integration

We approximate

Z b

a

f(x)dx

by

Mn= b−a n

n

X

k=1

f(ck),

(4)

We approximate

Z b

a

f(x)dx

by

Mn= b−a n

n

X

k=1

f(ck),

(5)

8.7 Numerical Integration

We approximate

Z b

a

f(x)dx

by

Mn= b−a n

n

X

k=1

f(ck),

(6)

We approximate

Z b

a

f(x)dx

by

Mn= b−a n

n

X

k=1

f(ck),

(7)

Error Estimate:

Suppose that |f

^′′

(x)| ≤ k ∀x ∈ [a, b]. Then the error in the midpoint rule is at most

Z

b

a

f ( x ) dx − M

_n

≤ K (b − a)

³

24n

²

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Error Estimate:

Suppose that |f

^′′

Z

b

a

f ( x ) dx − M

_n

≤ K (b − a)

³

24n

²

(9)

Error Estimate:

Suppose that |f

^′′

Z

b

a

f ( x ) dx − M

_n

≤ K (b − a)

³

24n

²

(10)

Error Estimate:

Suppose that |f

^′′

Z

b

a

f ( x ) dx − M

_n

≤ K (b − a)

³

24n

²

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Remark:

In the interval [x

k+1

, x

_k

], the tangent at the midpoint c

_k

is

φ(x ) = f (c

k

) + (x − c

_k

)f

^′

(c

k

)

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Remark:

In the interval [x

k+1

, x

_k

], the tangent at the midpoint c

_k

is

φ(x ) = f (c

k

) + (x − c

_k

)f

^′

(c

k

)

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Remark:

In the interval [x

k+1

, x

_k

], the tangent at the midpoint c

_k

is

φ(x ) = f (c

k

) + (x − c

_k

)f

^′

(c

k

)

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φ(x) =f(ck) + (x −ck)f^′(ck)

⇒ f(x)−φ(x) = 1

2(x−ck)²f^′′(ξ), ξ∈(x,ck)

Z x_k

xk−1

(f(x)−φ(x))dx

≤K · Z x_k

xk−1

1

2|x −ck|²dx

= K

24h³ (h= b−a n )

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φ(x) =f(ck) + (x −ck)f^′(ck)

⇒ f(x)−φ(x) = 1

2(x−ck)²f^′′(ξ), ξ∈(x,ck)

Z x_k

xk−1

(f(x)−φ(x))dx

≤K · Z x_k

xk−1

1

2|x −ck|²dx

= K

24h³ (h= b−a n )

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φ(x) =f(ck) + (x −ck)f^′(ck)

⇒ f(x)−φ(x) = 1

2(x−ck)²f^′′(ξ), ξ∈(x,ck)

Z x_k

xk−1

(f(x)−φ(x))dx

≤K · Z x_k

xk−1

1

2|x −ck|²dx

= K

24h³ (h= b−a n )

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φ(x) =f(ck) + (x −ck)f^′(ck)

⇒ f(x)−φ(x) = 1

2(x−ck)²f^′′(ξ), ξ∈(x,ck)

Z x_k

xk−1

(f(x)−φ(x))dx

≤K · Z x_k

xk−1

1

2|x −ck|²dx

= K

24h³ (h= b−a n )

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(2) The Trapezoidal Rule We approximate

Z

b

a

f (x )dx by

T

_n

= b − a n

f (x

0

)

2 + f ( x

₁

) + · · · + f ( x

_n−1

) + f (x

n

) 2

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Z

b

a

f (x )dx by

T

_n

= b − a n

f (x

0

)

2 + f ( x

₁

) + · · · + f ( x

_n−1

) + f (x

n

) 2

(20)

Z

b

a

f (x )dx

by

T

_n

= b − a n

f (x

0

)

2 + f ( x

₁

) + · · · + f ( x

_n−1

) + f (x

n

) 2

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Error Estimate:

Suppose that|f^′′(x)| ≤k ∀x ∈[a,b]. Then the error in the trapezoidal rule is at most

Z b

a

f(x)dx −Tn

≤K(b−a)³ 12n²

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Error Estimate:

Z b

a

f(x)dx −Tn

≤K(b−a)³ 12n²

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Error Estimate:

Z b

a

f(x)dx −Tn

≤K(b−a)³ 12n²

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Error Estimate:

Z b

a

f(x)dx −Tn

≤K(b−a)³ 12n²

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Examples:

f(x) = 1 x,

Z 4

1

1 xdx

(1)Midpoint Rule (2)Trapezoidal Rule (3)Error

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Examples:

f(x) = 1 x,

Z 4

1

1 xdx

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Examples:

f(x) = 1 x,

Z 4

1

1 xdx

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Examples:

f(x) = 1 x,

Z 4

1

1 xdx

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Remark:

The Midpoint Rule is the tangent formula So in this example,

M3 ≤ Z 4

1

xdx ≤T3

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Remark:

M3 ≤ Z 4

1

xdx ≤T3

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Remark:

M3 ≤ Z 4

1

xdx ≤T3

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Remark:

M3 ≤ Z 4

1

xdx ≤T3

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(3) The Simpon’s Rule

Step 1.

We consider g(x) =Ax²+Bx +C.

⇒ Z b

a

g(x)dx = b−a 6

g(a) +4g(a+b

2 ) +g(b)

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(3) The Simpon’s Rule

Step 1.

We consider g(x) =Ax²+Bx +C.

⇒ Z b

a

g(x)dx = b−a 6

g(a) +4g(a+b

2 ) +g(b)

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Step 2.

Sn =

b−a

6n f(x0) +f(xn) +2(f(x1) +· · ·+f(xn−1)) +4h

f(^x⁰^+x₂ ¹) +· · ·+f(^f(xⁿ⁻¹^)+f(x₂ ⁿ⁾)io

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Step 3.

Z b

a

f(x)dx −Sn

≤ 1

180(b−a)M4h⁴,

Here,

f⁽⁴⁾(x)

≤M4, h= b−a n

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Thank you.