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Lecture 11Section 8.7 Numerical Integration
Jiwen He
1 Riemann Sums
1.1 Area Problem
Area Problem
Partition of [a, b]
Take a partition P = {x 0 , x 1 , · · · , xn} of [a, b]. Then P splits up the interval
[a, b] into a finite number of subintervals [x 0 , x 1 ], · · · , [xn− 1 , xn] with a = x 0 <
x 1 < · · · < xn = b. We have [a, b] = [x 0 , x 1 ] ∪ · · · ∪ [xi− 1 , xi] ∪ · · · ∪ [xn− 1 , xn]
Remark
This breaks up the region Ω into n subregions Ω 1 ,· · · ,Ωn: Ω = Ω 1 ∪ · · · ∪ Ωi ∪ · · · ∪ Ωn
We can estimate the total area of Ω by estimating the area of each subregion Ωi
and adding up the results.
1.2 Lower and Upper Sums
Lower and Upper Sums
Let ∆xi = xi−xi− 1 , mi = min x∈[xi− 1 ,xi]
f (x), Mi = max x∈[xi− 1 ,xi]
f (x) mi∆xi = area of ri ≤ area of Ωi ≤
Lf (P ) :=
n ∑
i=
mi∆xi ≤ I =
∫ (^) b
a
f (x) dx ≤
n ∑
i=
Mi∆xi =: Uf (P )
1.3 Riemann Sum
Riemann Sum
Riemann Sum
S ∗ (P ) = f (x ∗ 1 )∆x 1 + f (x ∗ 2 )∆x 2 + · · · + f (x ∗ n )∆xn
where x ∗ i is any point picked in [xi− 1 , xi] for i = 1 , · · · , n. We have
Lf (P ) :=
n ∑
i=
mi∆xi ≤ S
∗ (P ) :=
n ∑
i=
f (x
∗ i )∆xi ≤
n ∑
i=
Mi∆xi =: Uf (P )
Limit of Riemann Sums ∫ b
a
f (x) dx = lim |P |→ 0
[f (x
∗ 1 )∆x^1 +^ f^ (x
∗ 2 )∆x^2 +^ · · ·^ +^ f^ (x
∗ n)∆xn]
2 Rectangle Approximations
2.1 Regular Partition
Regular Partition
n ∑
i=
f (xi− 1 )∆x,
n ∑
i=
f (xi)∆x,
n ∑
i=
f
xi− 1 + xi
∆x
left-endpoints right-endpoints midpoints
Rectangle Rules
Rectangle Rule for Approximating
∫ (^) b
a f (x) dx
b − a
n
[f (x 0 ) + f (x 1 ) + · · · + f (xn− 1 )]
- Right-endpoint rule: Rn =
b − a
n
[f (x 1 ) + f (x 2 ) + · · · + f (xn)]
b − a
n
[
f
x 0 + x 1
xn− 1 + xn
)]
Problem
Find the approximate value of ln 2 =
1
dx x using only the values of f (x) =
1 x
at 1, 6 5
7 5
8 5
9 5
Approximating ln 2 = 0. 69314718 · · ·
L 5 = 1 5
5 6
5 7
5 8
5 9
R 5 =
1 5
5 6
5 7
5 8
5 9
1 2
M 5 =
1 5
10 11
10 13
10 15
10 17
10 19
2.3 Error Estimates
Error Estimates
b
a
f (x) dx − Ln
b − a
n
[f (b) − f (a)]
Error Estimates
L n =
b
a
f (x) dx − Ln =
(b − a) 2
n
f
′ (c) = O(∆x)
R n =
b
a
f (x) dx − Rn =
(b − a) 2
n
f
′ (c) = O(∆x)
M n =
b
a
f (x) dx − Mn =
(b − a) 3
n 2
f
′′ (c) = O((∆x)
2 )
Error Estimates: Example
Error Estimates
L n
∫ (^) b
a
f (x) dx − Ln =
(b − a) 2
n
f
′ (c) = O(∆x)
R n
∫ (^) b
a
f (x) dx − Rn =
(b − a) 2
n
f
′ (c) = O(∆x)
M n
∫ (^) b
a
f (x) dx − Mn =
(b − a) 3
n 2
f
′′ (c) = O((∆x)
2 )
Trapeziodal and Simpson’s Rules
Trapeziodal and Simpson’s Rules for Approximating
∫ (^) b
a f (x) dx
b − a
2 n
[f (x 0 ) + 2f (x 1 ) + · · · + 2f (xn− 1 ) + f (xn)]
- Simpson’s Rule (Parabolic): Sn =
b − a
6 n
f (x 0 ) + f (xn) + 2
[
f (x 1 ) + · · · + 2f (xn− 1 )
]
[
f
x 0 + x 1
xn− 1 + xn
)]}
Problem
Find the approximate value of ln 2 =
1
dx x using only the values of f (x) =
1 x
at 1,
6 5
7 5
8 5
9 5
Approximating ln 2 = 0. 69314718 · · ·
T 5 =
1 10
10 6
10 7
10 8
10 9
1 2
S 5 =
1 30
1 2
5 6
5 7
5 8
5 9
[
10 11
10 13
10 15
10 17
10 19
]
3.2 Error Estimates
Error Estimates
b
a
f (x) dx − Tn
b − a
n
[f (b) − f (a)]
Error Estimates
T n =
b
a
f (x) dx−Tn = −
(b − a) 3
n 2
f
′′ (c) = O((∆x)
2 )
S n =
b
a
f (x) dx−Sn = −
(b − a) 5
n 4
f
(4) (c) = O((∆x)
4 )
Error Estimates: Example
Error Estimates
T n
∫ (^) b
a
f (x) dx−Tn = −
(b − a) 3
n 2
f
′′ (c) = O((∆x)
2 )
S n
∫ (^) b
a
f (x) dx−Sn = −
(b − a) 5
n 4
f
(4) (c) = O((∆x)
4 )
