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7- 1
Chapter 7 Differentiation and Integration
• Finite-difference differentiation
x
xxfxxf
dx
xdfx
xxfxf
dx
xdfx
xfxxf
dx
xdf
2
)()()( method step-two
)()()( difference backward
)()()( difference forward
7- 2
Table: Saturation Vapor Pressure (es) in
mm Hg as a Function of Temperature (T) in °C
T(°C) es(mm Hg)
20 17.53
21 18.65
22 19.82
23 21.05
24 22.37
25 23.75
Example: Evaporation Rates
7- 3
The slope of the saturation vapor pressure curve at 22°C (3 methods) :
The true value is 1.20 mm Hg/°C, so the two-step method provides the most accurate estimate.
CHg/ mm 20.12
65.1805.21
2123
)21()23( step-two
CHg/ mm 17.11
65.1882.19
2122
)21()22( backward
CHg/ mm 23.11
82.1905.21
2223
)22()23( forward
sss
sss
sss
ee
dT
de
ee
dT
de
ee
dT
de
7- 4
Example: Finite-difference Table for Specific Enthalpy (h) in Btu/lb and Temperature (T) in ºF
T h Δh Δ2h Δ3h Δ4h
800 1305
155
1000 1460 -30
125 25
1200 1585 -5 -20
120 5
1400 1705 0
120
1600 1825
Differentiation Using a Finite-difference Table
7- 5
• For example, at a temperature of 1200 ºF, the forward, backward, and two-step methods yield:
• The rate of change of cp at T= 1200 ºF is
FBtu/lb/ 6.0200
120
12001400
15851705
T
hcp
FBtu/lb/ 625.0200
125
10001200
14601585
T
hcp
FBtu/lb/ 6125.0400
245
10001400
14601705
T
hcp
22
2
F) Btu/lb/(025.0200
5
T
h
7- 6
Differentiating an Interpolating Polynomial
012
21
1 ...)( bxbxbxbxbxf nn
nn
nn
12
11 ...)1(
)(bxbnxnb
dx
xdf nn
nn
)])(())((
))([()]()[()(
2131
3241232
xxxxxxxx
xxxxaxxxxaadx
xdf
The derivative:
xxxxxxa
xxxxaxxaaf(x)
))()((
))(()(
3214
213121
Gregory-Newton interpolation polynomial:
It is more difficult to evaluate the derivative:
7- 7
Differentiation Using Taylor Series Expansion
.difference-finite forward theas same theisIt
)()()()('
get weThen,
)()()(
: sderivativehigher and derivative second the with termsTruncating
...!3
)()(
!2
)()()()()(
3
3
32
2
2
x
xfxxf
dx
xdfxf
xdx
xdfxfxxf
x
dx
xfdx
dx
xfdx
dx
xdfxfxxf
7- 8
• The second-order approximation:
The second-order approximation of the first derivative with forward difference:
2
2
2
2
2
)(
)()(2)2(
)(')(')(
:difference forward With!2
)()()()(
x
xfxxfxxfx
xfxxf
dx
xfd
x
dx
xfd
x
xfxxf
dx
xdf
x
xfxxfxxf
dx
xdf
2
)(3)(4)2()(
7- 9
• The second-order approximation of the second derivative with forward difference:
• The first-order and second-order approximation of the first derivative with backward difference:
22
2
)(
)(2)(5)2(4)3()(
x
xfxxfxxfxxf
dx
xfd
x
xxfxxfxf
dx
xdfx
xxfxf
dx
xdf
2
)2()(4)(3)(
)()()(
7- 10
• The first-order and second-order approximation of the second derivative with backward difference:
• How to derive
(for reference only)
22
2
22
2
)(
)3()2(4)(5)(2)(
)(
)2()(2)()(
x
xxfxxfxxfxf
dx
xfd
x
xxfxxfxf
dx
xfd
221
2
229
)()('')(')()(
)(2)(''2)(')()2(
)()(''3)(')()3(
xxfxxfxfxxf
xxfxxfxfxxf
xxfxxfxfxxf
)(''
)](2)('')()('5)(5-
)('')(8)('8)(4
)('')()('3)([)(
1
)](2)(5)2(4)3([)(
1
225
2
229
2
2
xf
xfxfxxxfxf
xfxxxfxf
xfxxxfxfx
xxfxxfxxfxxfx
7- 11
• The first-order and second-order approximation of the first derivative with the two-step method:
• The first-order and second-order approximation of second derivative with the two-step method:
x
xxfxxfxxfxxf
dx
xdfx
xxfxxf
dx
xdf
12
)2()(8)(8)2()(2
)()()(
22
2
22
2
)(12
)2()(16)(30)(16)2()(
)(
)()(2)()(
x
xxfxxfxfxxfxxf
dx
xfd
x
xxfxfxxf
dx
xfd
7- 12
Example: Evaporation Rates
The true value at T = 22ºC is 1.2 mm Hg/ºC
CHg/ mm 19667.1)1(12
)53.17()65.18(8)05.21(8)37.22(
)1(12
)20()21(8)23(8)24()(
CHg/ mm 195.1)1(2
53.17)65.18(4)82.19(3
)1(2
)20()21(4)22(3)(
CHg/ mm 185.1)1(2
)82.19(3)05.21(437.22
)1(2
)22(3)23(4)24()(
ssss
sss
sss
eeee
dx
xdf
eee
dx
xdf
eee
dx
xdf
Second-order with forward, backward, two-step:
7- 13
Numerical Integration
• Example: the volume rate of flow (Q) of water in a channel or through a pipe is the integral of the velocity (V) and the incremental area (dA):
b
adxxf )( integral
VdAQ
• The area under the curve f(x) between x=a and x=b:
7- 14
Interpolation Formula Approach
cxbxn
bx
n
bdxxf
bxbxbxf
n
aaa
xxxxxxa
xxxxaxxaaxf
nnn
nnn
n
nn
1211
11
21
121
211
213121
...1
)(
:follows asly analytical integrated be canit Then,
...)(
:polynomialorder -th an form it to
rearrange can we,determined are ,,After
))...()((
...))(()()(
The Gregory-Newton interpolation polynomial:
7- 15
Trapezoidal Rule
1
1
11 2
)()()()(
1
n
i
iiii
x
x
xfxfxxdxxf
n
7- 16
• Another way to get the trapezoidal formula The linear polynomial passing the data points:
)()()(
)()(1
1i
ii
iii xx
xx
xfxfxfxf
)]()())()((2
)([)(
get we, and points twobetween )( gIntegratin
1111
iiiiiiii
b
axfxxfxxfxf
ab
xx
abdxxf
baxf
)]()([2
)(
get we, and Let
11
1
1
iiii
x
x
ii
xfxfxx
dxxf
xa xb
i
i
7- 17
• The absolute value of the upper bound on the error for the Trapezoidal rule is:
)(''max)('' where
)(''12
)(error
1
31
xfxf
xfxx
ii xxxj
jii
7- 18
Example: Trapezoidal Rule for Integration
• n = 4
• The trapezoidal rule provides
345)( 23 xxxxf
10)2(1
)]3(3[2
34)]3(1[
2
23)]1(3[
2
12)(
4
1
dxxf
7- 19
7- 20
Example: Flow Rate• The flow rate (Q) of an incompressible fluid is given by th
e integral.
in which V is the velocity and A is the area.
• For a circular pipe of radius r, the incremental area dA is equal to 2πrdr.
VdAQ
0
02
rrdr)V(Q
7- 21
• Table 6: The Data for Estimating the Flow Rate of a Fluid in a Circular Piple
i ri (ft) Vi (fps)
1 0 10.000
2 1/12 9.722
3 1/6 8.889
4 1/4 7.500
5 1/3 5.556
6 5/12 3.056
7 1/2 0.
7- 22
• For a pipe of diameter of 1 ft
trapezoidal rule:
/secft 818.3
}2
10
12
5056.3
12
5056.3
3
1556.5
3
1556.5
4
15.7
4
15.7
6
1889.8
6
1889.8
12
1722.9
12
1722.9010{
12
π
][2
)121
(π2
3
6
111
Q
rVrVQi
iiii
6
111 )]π2()π2([
2 iiiii rVrV
rQ
7- 23
Simpson’s Rule
• Simpson’s rule:
where
• Simpson’s rule can only be applied when there are an even number of subintervals:
)]()(4)([3
)( bfxafafx
dxxfb
a
2
5,3,121
1 )]()(4)([3
)(1
n
iiii
iix
xxfxfxf
xxdxxf
n
2
abx
7- 24
• Proof of Simpson’s Rule
Using a second-order polynomial:
lhxkxxf 2)(
7- 25
Passing through the three data points:
Then, we can obtain the Simpson’s formula.
lhbkbbf
lxahxakxaf
lhakaaf
2
2
2
)(
)()()(
)(
)(2
)(
3
)(
)(
:polynomialorder -second thegIntegratin
2233
2
ablabhabk
dxlhxkxb
a
7- 26
• The absolute value of the upper bound on the error for the Simpson’s rule is estimated by
4
4
4
4
4
451
)(max
)( where
)(
90
)(error
1 dx
xfd
dx
xfd
dx
xfdxx
ii xxx
j
jii
7- 27
Example: Flow Rate Problem
• Applying Simpson’s rule to the data of Table.6
/secft 927.3)]2
1(0)
12
5)(056.3(4)
3
1(556.5[)]
3
1(556.5
)4
1)(5.7(4)
6
1(889.8[)]
6
1(889.8)
12
1)(722.9(4)0(10[
18
π
]4[3
)(π2π2
3
5
3,12211
12
1
0
0
r
iiiiii
rrVrVrVrdrVQ
7- 28
Romberg Integration
• Denoting the trapezoidal estimate as I01
where a and b are the start and end of an interval.
• A second estimate I11:
))()((201 bfaf
abI
22 where
))(2
1)()(
2
1(
211
aba
bam
bfmfafab
I
)]2
()([2
1 :rewritten be canIt 0111
abafabII
7- 29
• A third estimate I21 can be obtained using three equally spaced intermediate points m1, m2, and m3
and it can be rewritten as
)](4
1)(
2
1)(
2
1)(
2
1)(
4
1[
2 32121 bfmfmfmfafab
I
3
2
11121 )
4(
22
1
k
k
kab
afab
II
7- 30
• Continuing this subdividing of the interval leads to the following recursive relationship
• The general extrapolation formula in recursive form is
2,... 1, for )2
(22
1 12
5,3,111,11
ik
abaf
abII
i
kiiii
14
41
1,1,11
j
jijij
ij
III
7- 31
• The values of Iij can be presented in the following upper-triangular matrix form:
I01 I02 I03 I04 … I0,N-1 I0,N I0,N+1
I11 I12 I13 . … I1,N-1 I1,N
I21 I22 . . … I2,N-1
I31 . . .
. . .
. . IN-2,3
. IN-1,2
IN1
7- 32
Example: Romberg Method for Integration
• For the function f(u) = ueku, the integral is
• Let k = 2, we want to calculate
The true value of the integral is 2.097264 (seven significant digits).
ckuk
eduue
kuku )1(
2
duue u1
0
2
7- 33
20678.2
)]75.025.0(5.052683.2[5.0
))]4
3()4
()((5.0[5.0
2 For
52683.2)]5.0)(01(69453.3[5.0
)]2
()([5.0
1 For
69453.3)10)(01(5.0
))()()((5.0
0 For
75.0225.02
1121
5.02
0111
1202
01
ee
abaf
abafabII
i
e
abafabII
i
ee
bfafabI
iWe have a=0, b=1.
7- 34
12478.2
))]8
)(7()
8
)(5(
)8
)(3()
8()((25.0[5.0
3 For
2131
abaf
abaf
abaf
abafabII
i
10415.2
))]16
)(15()
16
)(13()
16
)(11(
)16
)(9()
16
)(7()
16
)(5(
)16
)(3()
16()((125.0[5.0
4 For
3141
abaf
abaf
abaf
abaf
abaf
abaf
abaf
abafabII
i
7- 35
Table.7 Computations According to the Romberg Method
i j = 1 2 3 4 5 6
0 3.69453 2.13760 2.09759 2.09727 2.09726 2.09726
1 2.52683 2.10009 2.09727 2.09726 2.09726
2 2.20678 2.09745 2.09726 2.09726
3 2.12478 2.09728 2.09726
4 2.10415 2.09727
5 2.09899
7- 36
• Similarly for i = 5, I51=2.09899
• I02 is computed as following:
• I03 is computed as following:
• The Romberg method yields an exact value to six significant digits for the integral of 2.09726
13760.214
69453.3)52683.2(412
12
02
I
09759.214
13760.2)10009.2(413
13
03
I
