Note: Using the formula, the given interval of integration must be divided into an even number of sub-intervals... Note: Using this formula, the given interval of integration must be div
Trang 16.5 TRAPEZOIDAL RULE
Putting n =1 in equation (2) and taking the curve y = f(x) through (x0, y0) and (x0, y0) as a polynomial
of degree one so that differences of order higher than one vanish, we get
0
0
1
x h
x
f x dx h y y y y y y y
+
∫ Similarly, for the next sub interval (x0+h x, 0+2h),we get
2
( 1)
x nh
f x dx y y f x dx y y
+ +
−
Adding the above integrals, we get
0
0
2
x nh
x
h
f x dx y y y y y
+
−
∫ which is known as Trapezoidal rule
Putting n = 2 in equation (2 ) and taking the curve through ( ,x y0 0),( ,x y1 1) and ( ,x y2 2) as a polynomial of degree two so that differences of order higher than two vanish, we get
0
0
2
2
1
6
x
+
∫
2
y y y y y y y y y
0
4
2
3
h
f x dx y y y
+
+
∫
0
0
( 2)
3
x nh
h
f x dx y y y
+
+ −
∫ Adding the above integrals, we get
0
0
3
x nh
x
h
f x dx y y y y y y y y
+
∫ which is known as Simpson’s one-third rule
Note: Using the formula, the given interval of integration must be divided into an even number of sub-intervals.
Trang 26.7 SIMPSON’S THREE-EIGHT RULE
Putting n = 3 in equation (2) and taking the curve through (x0, y0), (x1, y1), (x2, y2) and (x3, y3) as a polynomial of degree three so that differences of order higher than three vanish, we get
0
0
3
( ) 3
x
f x dx h y y y y
+
∫
3
8
h
y y y y y y y y y y
( 0 1 2 3)
3
8
h
y y y y
0
6
3
3
8
h
f x dx y y y y
+
+
∫
0
0
6
( 3)
3
8
h
+
+ −
∫
Adding the above integrals, we get
0
0
3
8
x nh
x
h
f x dx y y y y y y y y y y y
+
∫
which is known as Simpson’s three-eighth rule
Note: Using this formula, the given interval of integration must be divided into sub-intervals whose
number n is a multiple of 3.
Putting n = 4 in equation (2) and neglecting all differences of order higher than four, we get
0
0
4
0
( )
x
+
(By Newton’s forward interpolation formula)
4 4
0
y
h y y y y y
2
45
h
2
45
h
x x
h
+
Trang 3Adding all these integrals from x0 to x0 + nh, where n is a multiple of 4, we get
0
2
45
x nh
x
h
+
∫
This is known as Boole’s rule
Note: Using Boole’s rule, the number of sub-intervals should be taken as a multiple of 4.
Putting n = 6 in equation (2) and neglecting all differences of order higher than six, we get
0
0
0
( )
x
r r r r r
+
( 1)( 2)( 3)( 4)( 5)
6!
r r r r r r
y dr
h ry y y r r y
6 6 0 0
720 7r 2r r 4r 3r r y
= 6 3 9
41 20
11 20
41 840
h yL + y + y + y + y + y + y
h
3
10
h
+41(y4−4y3+6y2−4y1+y0) 11(+ y5−5y4+10y3−10y2+5y1−y0)
+(y6−6y5+15y4−20y3+15y2−6y1+y0)] 341
42≈1
L
NM O QP
( 0 1 2 3 4 5 6)
3
10
h
y y y y y y y
Trang 4Similarly, 0 ( ) ( )
0
12
6
3
10
h
+
∫ 0
( 6)
3
10
x nh
h
+
∫ Adding the above integrals, we get
0
3
10
x nh
x
h
+
∫
which is known as Weddle’s rule Here n must be a multiple of 6.
Example 1 Use Trapezoidal rule to evaluate 10 1
1+x dx
∫
Sol Let h = 0.125 and y = f(x) = 1
1 x+ , then the values of y are given for the arguments which
are obtained by dividing the interval [0,1] into eight equal parts are given below:
1
1
y
x
=
+ 1.0 0.8889 0.8000 0.7273 0.6667 0.6154 0.5714 0.5333 0.5
Now by Trapezoidal rule
0
1
h
+
∫
0.125 [1 2(0.8889 0.800 0.7273 0.6667 0.6154 0.5714 0.5333) 0.5] 2
0.125 [1.5 2(4.803)]
2
Example 2 Evaluate 01 1 2
dx x
+
(i) Simpson’s 1
3 rule taking
1 4
h=
(ii) Simpson’s 3
8 rule taking
1 6
h=
(iii) Weddle’s rule taking 1
6
h=
Hence compute an approximate value of π in each case.
Trang 5Sol (i) The values of 2
1 ( ) 1
f x
x
= + at
1 2 3
0, , , , 1
4 4 4
( )
16
17
x
f x
By Simpon’s 1
3 rule
1
2
3 1
+
∫
1 { }16
Also,
1
4 1
dx
x x
+
∫
4
(ii) The values of ( ) 1 2
1
f x
x
= + at x = 0,
1 2 3 4 5 , , , , ,1
( )
1
x
f x
By Simpson’s 3
8 rule
1
2 0
3
8 1
+
∫ =
6
1
36 37
9 10
9 13
36
4 5
FH IK L FHG IKJ+ + RST + + + UVW+ FHG IKJ
Also,
1 2
dx x
π
= +
∫
4
Trang 6(iii) By Weddle’s rule
1
2 0
3
10 1
+
∫
1 3
6
Since
1 2
dx x
π
= +
∫
4
π =
Example 3 Evaluate 06 1 2
dx x
+
(i) Trapezoidal Rule
(ii) Simpson’s one-third rule
(iii) Simpson’s three-eighth rule
(iv) Weddle’s rule.
Sol Divide the interval (0, 6) into six parts each of width h = 1.
The value of ( ) 1 2
1
f x
x
= + are given below:
17
1 26
1 37
(i) By Trapezoidal rule,
2 1
+
∫
1 1 1 2 0.5 0.2 0.1 1 1
= + + + + + +
= 1.410798581 Ans
(ii) By Simpson’s one-third rule,
2
3 1
+
∫
1 1 1 4 0.5 0.1 1 2 0.2 1
= + + + + + +
Trang 7(iii) By Simpson’s three-eighth rule,
6
2 0
3
8 1
+
∫ = 3 1 1 3 0.5 0.2 1 1 2(0.1)
= 1.357080836 Ans
(iv) By Weddle’s rule,
6
2 0
3
10 1
+
∫
3 1 5(0.5) 0.2 6(0.1) 1 5 1 1
=1.373447475 Ans
Example 4 Using Simpson’s one-third rule, find 6 2
0 (1 )
dx x
+
Sol Divide the interval [0,6] into 6 equal parts with 6 0
6
h= −
= 1 The values of 1 2
(1 )
y x
=
each points of sub-divisions are given by
2
1
y
x
=
By Simpson’s one-third rule, we get
6
2
3 (1 )
dx h
y y y y y y y
x = + + + + + + +
∫
1 4(0.25 0.0625 0.02778) 2(0.11111 0.04) 0.02041
= 1[1.02041 4(0.34028) 2(0.15111)]
= 0.89458 Ans
Example 5 Evaluate 4 2
0 1
dx x
x
+
(i) h = 1 (ii) h = 0.5 Compare the rusults with the actual value and indicate the error in both.
Trang 8Sol (i) Dividing the given interval into 4 equal subintervals (i.e h = 1), the table is as below:
2
1 5
1 10
1 17
Using Boole’s rule,
1
4
0
2
45
h
∫
= 2 1
45 7 1 32
1
2 12
1
5 32
1
10 7
1 17
( ) ( )+ FH IK+ FH IK+ FH IK+ FH IK
L
∴ 4 2
1
dx
x = +
(ii) Dividing the given interval into 8 equal subintervals (i.e h = 0.5), the table is as below:
53
1 17
Using Boole’s rule,
0
2
45
h
∫
= 1.326373
x
1 2
0
4
+
But the actual value is
2
1
dx
x x
+
∫
Error in I result = 1.325818 1.289412 100 2.746%
1.325818
−
Error in II result = 1.325818 1.326373 100 0.0419%
1.325818
−
Trang 9Example 6 Evaluate the integral 6 3
0 1
dx x
+
Sol Divide the interval [0,6] into 6 equal parts each of width 6 0 1
6
h= − =
The value of 1 3
1
y x
= +
at each points of sub-divisions are given below:
y 1.0000 0.5000 0.1111 0.0357 0.0153 0.0079 0.0046
By Weddle’s Rule, we get
3 0
3
10 1
+
∫
3 1.0000 5(0.5000 0.0079) 0.1111 0.0153 6(0.357) 0.0046 10
3 1.131 5 0.5079 6 0.0357 10
3 1.131 2.5395 0.2142 10
3 (3.8847) 1.1654
10
Example 7 Evaluate the integral 1.5 3
x dx
e −
Sol Dividing the interval [0, 1, 5] into 6 equal parts of each of width 1.5 0 0.25
6
h= − =
and the values of
3 1
x
x y
e
=
− at each points of sub-interval are given by
Now by Weddle’s rule, we get
0
3
10 1
x
dx y y y y y y y
−
∫ 3(0.25[0 5(0.0549 0.7843) 0.1927 0.5820 0.9694 6(0.3777)]
10
= 0.075[1.7441 + 5(0.8392) +6(0.3777)]
Trang 10= 0.075[1.7441 + 4.196 + 2.2662] = 0.075(8.2063) = 0.6155 Ans
Example 8 Evaluate the integral 5.2
4 logxdx,
Sol Divide the interval [4, 5.2] into 6 equal sub-interval of each width 5.2 4 0.2
6
−
values of y = log x are given below:
y 1.3862 1.4350 1.4816 1.5261 1.5686 1.6094 1.6486
By Weddle’s Rule, we get
3
10
h xdx= y + y +y +y +y + y +y
∫ =3(0.2)10 [1.3862 5(1.4350 1.6094) 1.4816 1.5686 6(1.5261) 1.6486]+ + + + + +
0.6[1.3862 5(3.0444) 1.4816 1.5686 6(1.526) 1.6486]
10
0.6[6.085 5(3.0444) 6(1.5261)]
10
0.6[6.085 15.222 9.1566]
10
=0.6(30.4636) 1.8278=
Example 9 A river is 80 m wide The depth y of the river at a distance ‘x’ from one bank is given by the
following table:
Find the approximate area of cross section of the river using
(i) Boole’s rule.
(ii) Simpson’s one-third rule.
Sol The required area of the cross-section of the river
Here no of sub intervals is 8
(i) Boole’s rule,
∫ 80
0
2
45
h