The document discusses various numerical integration methods, including Newton-Cote's quadrature, the trapezoidal rule, Simpson's one-third, and three-eighths rules, along with examples demonstrating their application. It explains how to subdivide integration intervals and approximate integrals using tabulated values of functions. Additionally, the document covers Boole's rule for numerical integration with examples to illustrate accuracy and errors in approximations.

1. Dr. Deepa Chauhan
Numerical Integration
By
Dr. Deepa Chauhan
Associate Professor,
Applied Science & Humanities Department,
Axis Institute of Technology, Kanpur

2. Dr. Deepa Chauhan
Unit-III
Numerical Integration
Given a set of tabulated values of the integrand 𝑓(𝑥), to determine the value of ∫ 𝑓(𝑥)𝑑𝑥
𝑥𝑛
𝑥0
is called
numerical integration. We subdivide the given interval of integration into a large number of subintervals
of equal width h and replace the function tabulated at the points of subdivision by any one of the
interpolating polynomials.
Newton-Cote’s Quadrature Formula
Let 𝐼 = ∫ 𝑦𝑑𝑥
𝑏
𝑎
, where 𝑦 𝑡𝑎𝑘𝑒𝑠 𝑡ℎ𝑒 𝑣𝑎𝑢𝑒𝑠 𝑦0, 𝑦1,… … … , 𝑦𝑛 for 𝑥 = 𝑥0, 𝑥1, … … … , 𝑥𝑛
Let the interval of integration (a, b) be divided into n equal sub intervals, each of width ℎ =
𝑏−𝑎
𝑛
so that
𝑥0 = 𝑎, 𝑥1 = 𝑥0 + ℎ, 𝑥2 = 𝑥0 + 2ℎ, … … … , 𝑥𝑛 = 𝑥0 + 𝑛ℎ = 𝑏
𝐼 = ∫ 𝑓(𝑥)𝑑𝑥
𝑥0+𝑛ℎ
𝑥0
Since any 𝑥 is given by 𝑥 = 𝑥0 + 𝑟ℎ and 𝑑𝑥 = ℎ𝑑𝑟
𝐼 = ℎ ∫ 𝑓(𝑥0 + 𝑟ℎ)𝑑𝑟
𝑛
0
ℎ ∫ [𝑦0 + 𝑟∆𝑦0 +
𝑟 (𝑟 − 1)
2!
∆2
𝑦0 +
𝑟 (𝑟 − 1)(𝑟 − 2)
3!
∆3
𝑦0 + ⋯ … … … … … . ] 𝑑𝑟
𝑛
0
(By Newton’s forward interpolation formula)
= ℎ [𝑟𝑦0 +
𝑟2
2
∆𝑦0 +
1
2
(
𝑟3
3
−
𝑟2
2
) ∆2
𝑦0 +
1
6
(
𝑟4
4
− 𝑟3
+ 𝑟2
)∆3
𝑦0 + ⋯ … … … … … . ]
0
𝑛
= 𝑛ℎ [𝑦0 +
𝑛
2
∆𝑦0 +
𝑛 (2𝑛 − 3)
12
∆2
𝑦0 +
𝑛 (𝑛 − 2)2
24
∆3
𝑦0 + ⋯ … … … . ]
Then general quadrature formula is given by:
𝐼 = 𝑛ℎ [𝑦0 +
𝑛
2
∆𝑦0 +
𝑛 (2𝑛−3)
12
∆2
𝑦0 +
𝑛 (𝑛−2)2
24
∆3
𝑦0 + ⋯ … … … . ] (1)
This is also known as Newton-Cote’s Quadrature Formula

3. Dr. Deepa Chauhan
Trapezoidal Rule (n=1)
Putting n=1 in (1) and taking the curve through (𝑥0, 𝑦0) and (𝑥1, 𝑦1) as polynomial of degree one so that
differences of order higher than one vanish, we get
∫ 𝑓(𝑥)𝑑𝑥 =
ℎ
2
[(𝑦0 + 𝑦𝑛) + 2(𝑦1 + 𝑦2 + ⋯ … . . +𝑦𝑛−1)]
𝑥0+𝑛ℎ
𝑥0
Where ℎ =
𝑏−𝑎
𝑛
By increasing the number of subintervals, thereby making h very small, we can improve the accuracy of
the value of the given integral.
Example 1: Use Trapezoidal rule to evaluate ∫ 𝒙𝟑
𝒅𝒙
𝟏
𝟎
considering 5 sub-intervals.
Sol. Dividing the interval (0,1) into 5 equal parts, each of width ℎ =
𝑏−𝑎
𝑛
=
1−0
5
= 0.2
Here 𝑓(𝑥) = 𝑥3
𝑥: 0 0.2 0.4 0.6 0.8 1
𝑓(𝑥): 0 0.008 0.064 0.216 0.512 1
(𝑦0) 𝑦1 𝑦2 𝑦3 𝑦4 y5
By Trapezoidal rule
∫ 𝑥3
𝑑𝑥
1
0
=
ℎ
2
[(𝑦0 + 𝑦5) + 2(𝑦1 + 𝑦2 + 𝑦3 + 𝑦4)]
=
0.2
2
[(0 + 1) + 2(0.008 + 0.064 + 0.216 + 0.512)] = 0.26

4. Dr. Deepa Chauhan
Example 2: Use Trapezoidal rule to evaluate ∫
𝒅𝒙
𝟏+𝒙𝟐
𝟔
𝟎
.
Sol. Divide the interval (0, 6) into 6 equal parts each of width ℎ =
𝑏−𝑎
𝑛
=
6−0
6
= 1
Here 𝑓(𝑥) =
𝟏
𝟏+𝒙𝟐
𝑥: 0 1 2 3 4 5 6
𝑓(𝑥): 1 0.5 0.2 0.1 0.058 0.038 0.027
𝑦0 𝑦1 𝑦2 𝑦3 𝑦4 𝑦5
𝑦6
By Trapezoidal rule
∫
𝒅𝒙
𝟏 + 𝒙𝟐
𝟔
𝟎
=
ℎ
2
[(𝑦0 + 𝑦6) + 2(𝑦1 + 𝑦2 + 𝑦3 + 𝑦4 + 𝑦5)]
=
1
2
[(1 + 0.027) + 2(0.5 + 0.2 + 0.1 + 0.058 + 0.038] = 1.4095
Example 3: Evaluate using Trapezoidal rule ∫ 𝑡 𝑠𝑖𝑛𝑡 𝑑𝑡
𝜋
0
Sol. Divide the interval (0, 𝜋) into 4 equal parts each of width ℎ =
𝑏−𝑎
𝑛
=
𝜋−0
4
=
𝜋
4
= 0.785
Here 𝑓(𝑡) = 𝑡 𝑠𝑖𝑛𝑡
𝑡: 0
𝜋
4
𝜋
2
3𝜋
4
𝜋
In radian
𝑡: 0 0.785 1.57 2.355 3.14
𝑓(𝑡): 0 0.55485 1.56999 1.66722 0.005
∫ 𝑡 𝑠𝑖𝑛𝑡 𝑑𝑡
𝜋
0
=
ℎ
2
[(𝑦0 + 𝑦4) + 2(𝑦1 + 𝑦2 + 𝑦3)]
=
0.785
2
[(0 + 0.005) + 2(0.55485 + 1.56999 + 1.66722)] = 2.9787

5. Dr. Deepa Chauhan
Simpson’s One Third Rule
∫ 𝑓(𝑥)𝑑𝑥 =
ℎ
3
[(𝑦0 + 𝑦𝑛) + 4(𝑦1 + 𝑦3 + 𝑦5 … … . . +𝑦𝑛−1) + 2 (𝑦2 + 𝑦4 + 𝑦6 … … . . +𝑦𝑛−2)]
𝑥0+𝑛ℎ
𝑥0
While using this formula, the given interval of integration must be divided into even number of sub-
intervals.
Example 4: Evaluate ∫ 𝑓(𝑥)𝑑𝑥
11
1
where f(x) is given by the following table, using a suitable integration
formula
x: 1 2 3 4 5 6 7 8 9 10 11
f(x): 543 512 501 489 453 400 352 310 250 172 95
𝑦0 𝑦1 𝑦2 𝑦3 𝑦4 𝑦5 𝑦6 𝑦7 𝑦8 𝑦9 𝑦10
Sol. No of Sub-Intervals, n=10, ℎ =
𝑏−𝑎
𝑛
=
11−1
10
= 1, then by using Simpson’s 1/3rd
formula
∫ 𝑓(𝑥)𝑑𝑥 =
ℎ
3
[(𝑦0 + 𝑦𝑛) + 4(𝑦1 + 𝑦3 + 𝑦5 … … . . +𝑦𝑛−1) + 2 (𝑦2 + 𝑦4 + 𝑦6 … … . . +𝑦𝑛−2)]
𝑥0+𝑛ℎ
𝑥0
∫ 𝑓(𝑥)𝑑𝑥
11
1
=
ℎ
3
[(𝑦0 + 𝑦10) + 4 (𝑦1 + 𝑦3 + 𝑦5 + 𝑦7 + 𝑦9) + 2(𝑦2 + 𝑦4 + 𝑦6 + 𝑦8 )]
=
1
3
[(543 + 95) + 4 (512 + 489 + 400 + 310 + 172) + 2(501 + 453 + 352 + 250 )]
= 3760.67

6. Dr. Deepa Chauhan
Example 5: Evaluate ∫ 𝑒−𝑥/2
𝑑𝑥
2
1
using 4 intervals.
Sol. No of Sub-Intervals, n=4, ℎ =
𝑏−𝑎
𝑛
=
2−1
4
= 0.25, here 𝑓(𝑥) = 𝑒−𝑥/2
The table of values is:
x: 1 1.25 1.50 1.75 2
f(x): 0.6065 0.5352 0.4724 0.4168 0.3678
𝑦0 𝑦1 𝑦2 𝑦3 𝑦4
Since we have four (even) subintervals, we’ll use Simpson’s 1/3rd
rule
∫ 𝑓(𝑥)𝑑𝑥 =
ℎ
3
[(𝑦0 + 𝑦𝑛) + 4(𝑦1 + 𝑦3 + 𝑦5 … … . . +𝑦𝑛−1) + 2 (𝑦2 + 𝑦4 + 𝑦6 … … . . +𝑦𝑛−2)]
𝑥0+𝑛ℎ
𝑥0
∫ 𝑒−𝑥/2
𝑑𝑥
2
1
=
ℎ
3
[(𝑦0 + 𝑦4) + 4 (𝑦1 + 𝑦3) + 2(𝑦2 )]
=
0.25
3
[(0.6065 + 0.3678) + 4 (0.5352 + 0.4168) + 2(0.4724 )] = 0.4773
= 0.4773
Example 6: Evaluate ∫
𝑑𝑥
1+𝑥
1
0
by dividing the interval of integration into 8 equal parts.
Hence find 𝑙𝑜𝑔𝑒2 approximately.
Sol. No of Sub-Intervals, n=8, ℎ =
𝑏−𝑎
𝑛
=
1−0
8
=
1
8
= 0.125, 𝑓(𝑥) =
1
1+𝑥
x: 0 0.125 0.250 0.3750.5 0.625 0.75 0.875 1
f(x): 1 0.8888 0.8 0.7272 0.6666 0.6154 0.5714 0.5333 0.5
𝑦0 𝑦1 𝑦2 𝑦3 𝑦4 𝑦5 𝑦6 𝑦7 𝑦8
By Simpson’s 1/3rd
rule
∫ 𝑓(𝑥)𝑑𝑥 =
ℎ
3
[(𝑦0 + 𝑦𝑛) + 4(𝑦1 + 𝑦3 + 𝑦5 … … . . +𝑦𝑛−1) + 2 (𝑦2 + 𝑦4 + 𝑦6 … … . . +𝑦𝑛−2)]
𝑥0+𝑛ℎ
𝑥0

7. Dr. Deepa Chauhan
∫
𝑑𝑥
1 + 𝑥
1
0
= 0.6932
Now ∫
𝑑𝑥
1+𝑥
1
0
= [𝑙𝑜𝑔𝑒(1 + 𝑥)]0
1
= 𝑙𝑜𝑔𝑒2
Therefore, 𝑙𝑜𝑔𝑒2 = 0.6932

8. Dr. Deepa Chauhan
Simpson’s Three Eighth (3/8th
) Rule
∫ 𝑓(𝑥)𝑑𝑥 =
3ℎ
8
[(𝑦0 + 𝑦𝑛) + 3(𝑦1 + 𝑦2 + 𝑦4 + 𝑦5 … … . . +𝑦𝑛−1) + 2 (𝑦3 + 𝑦6 + 𝑦9 + 𝑦12
𝑥0+𝑛ℎ
𝑥0
+ ⋯ … . . +𝑦𝑛−3)]
While using this formula, the given interval of integration must be divided into number of sub-intervals n,
should be taken as multiple of 3.
Example 7. Evaluate ∫
𝑑𝑥
1+𝑥2
1
0
using Simpson’s rule taking ℎ =
1
6
Sol. Here (𝑥) =
1
1+𝑥2 , ℎ =
1
6
x: 0
1
6
2
6
3
6
4
6
5
6
1
f(x): 1
36
37
9
10
4
5
9
13
36
61
1
2
𝑦0 𝑦1 𝑦2 𝑦3 𝑦4 𝑦5 𝑦6
Here number of subintervals=6=multiple of 3
By Simpson’s 3/8th
rule:
∫ 𝑓(𝑥)𝑑𝑥 =
3ℎ
8
[(𝑦0 + 𝑦𝑛) + 3(𝑦1 + 𝑦2 + 𝑦4 + 𝑦5 … … . . +𝑦𝑛−1) + 2 (𝑦3 + 𝑦6 + 𝑦9 + 𝑦12
𝑥0+𝑛ℎ
𝑥0
+ ⋯ … . . +𝑦𝑛−3)]
∫
𝑑𝑥
1 + 𝑥2
1
0
=
3ℎ
8
[(𝑦0 + 𝑦6) + 3(𝑦1 + 𝑦2 + 𝑦4 + 𝑦5) + 2 (𝑦3)]
=
3 ×
1
6
8
[(1 +
1
2
) + 3 (
36
37
+
9
10
+
9
13
+
36
61
) + 2 (
4
5
)] = 0.78539375

9. Dr. Deepa Chauhan
Example 8. Use Simpson’s rule for evaluating ∫ 𝑓(𝑥)𝑑𝑥
0.3
−0.6
from the table given below
x: -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3
f(x): 4 2 5 3 -2 1 6 4 2 8
Sol. Here number of subinterval, n=9
By Using Simpson’s 3/8th
rule
∫ 𝑓(𝑥)𝑑𝑥
0.3
−0.6
= 2.475
Do Yourself.
Example 9. Find ∫
𝑒𝑥
1+𝑥
𝑑𝑥
6
0
approximately using Simpson’s 3/8th
rule on integration.
Sol. Here (𝑥) =
𝑒𝑥
1+𝑥
, Let number of subintervals, n=6, ℎ =
𝑏−𝑎
𝑛
=
6−0
6
= 1
x: 0 1 2 3 4 5 6
f(x): 1 1.3591 2.463 5.0213 10.9196 24.7355 57.6326
𝑦0 𝑦1 𝑦2 𝑦3 𝑦4 𝑦5 𝑦6
By Simpson’s 3/8th
rule:
∫ 𝑓(𝑥)𝑑𝑥 =
3ℎ
8
[(𝑦0 + 𝑦𝑛) + 3(𝑦1 + 𝑦2 + 𝑦4 + 𝑦5 … … . . +𝑦𝑛−1) + 2 (𝑦3 + 𝑦6 + 𝑦9 + 𝑦12
𝑥0+𝑛ℎ
𝑥0
+ ⋯ … . . +𝑦𝑛−3)]
∫
𝑒𝑥
1 + 𝑥
𝑑𝑥
6
0
=
3ℎ
8
[(𝑦0 + 𝑦6) + 3(𝑦1 + 𝑦2 + 𝑦4 + 𝑦5) + 2 (𝑦3)]
=
3
8
[(1 + 57.6326) + 3(1.3591 + 2.463 + 10.9196 + 24.7355) + 2 (5.0213)]
= 70.165205

10. Dr. Deepa Chauhan
Boole’ Rule
∫ 𝑓(𝑥)𝑑𝑥 =
2ℎ
45
[7𝑦0 + 32𝑦1 + 12𝑦2 + 32𝑦3 + 14𝑦4 + 32𝑦5 + 12𝑦6 + 32𝑦7 + 14𝑦8 + ⋯ … . ]
𝑥0+𝑛ℎ
𝑥0
While using this formula, the given interval of integration must be divided into multiple of 4.
Example 10: Evaluate ∫
𝑑𝑥
1+𝑥2
4
0
using Boole’s rule taking (a) h=1 (b) h=0.5. Compare the results with the
actual value and indicate the error in both.
Sol. (a) Dividing the given interval into 4 equal subintervals (h=1), the table is as below:
x: 0 1 2 3 4
f(x): 1 1/2 1/5 1/10 1/17
𝑦0 𝑦1 𝑦2 𝑦3 𝑦4
Using Boole’s rule
∫
𝑑𝑥
1 + 𝑥2
4
0
=
2ℎ
45
[7𝑦0 + 32𝑦1 + 12𝑦2 + 32𝑦3 + 14𝑦4 + ⋯ … . ] = 1.289412
(b) Dividing the given interval into 8 equal subintervals (h=0.5), the table is as below:
x: 0 0.5 1 1.5 2 2.5 3 3.5 4
f(x): 1 0.8 0.5 0.31 0.2 0.14 0.1 0.08 0.06
𝑦0 𝑦1 𝑦2 𝑦3 𝑦4 𝑦5 𝑦6 𝑦7 𝑦8
∫ 𝑓(𝑥)𝑑𝑥 =
2ℎ
45
[7𝑦0 + 32𝑦1 + 12𝑦2 + 32𝑦3 + 14𝑦4 + 32𝑦5 + 12𝑦6 + 32𝑦7 + 14𝑦8 + ⋯ … . ]
𝑥0+𝑛ℎ
𝑥0

11. Dr. Deepa Chauhan
∫
𝑑𝑥
1 + 𝑥2
4
0
=
2ℎ
45
[7𝑦0 + 32𝑦1 + 12𝑦2 + 32𝑦3 + 14𝑦4 + 32𝑦5 + 12𝑦6 + 32𝑦7 + 14𝑦8 + ⋯ … . ]
= 1.326373
Actual value is ∫
𝑑𝑥
1+𝑥2
4
0
= (𝑡𝑎𝑛−1
𝑥)0
4
= 𝑡𝑎𝑛−1
4 − 𝑡𝑎𝑛−1
1 = 0.5404195
Error in (a) result
= (
0.5404195 − 1.289412
0.5404195
) × 100 = −138.5946%
Error in (b) result
= (
0.5404195 − 1.326373
0.5404195
) × 100 = −145.415%

12. Dr. Deepa Chauhan
Weddle’s Rule
∫ 𝑓(𝑥)𝑑𝑥 =
3ℎ
10
[𝑦0 + 5𝑦1 + 𝑦2 + 6𝑦3 + 𝑦4 + 5𝑦5 + 2𝑦6 + 5𝑦7 + 𝑦8 + 6𝑦9 + 𝑦10 + 5𝑦11 + 𝑦12
𝑥0+𝑛ℎ
𝑥0
+ ⋯ … . ]
While using this formula, the given interval of integration must be divided into multiple of 6.
Example 11: Evaluate ∫ (𝑠𝑖𝑛𝑥 − log𝑒 𝑥 + 𝑒𝑥)
1.4
0.2
𝑑𝑥 taking n=12
Sol. Here 𝑓(𝑥) = 𝑠𝑖𝑛𝑥 − log𝑒 𝑥 + 𝑒𝑥
, ℎ =
𝑏−𝑎
𝑛
=
1.4−0.2
12
= 0.1
We shall divide the given interval of integration into 12 equal parts so that
x f(x)
0.2 3.0295 y0
0.3 2.8495 y1
0.4 2.7975 y2
0.5 2.8213 y3
0.6 2.8976 y4
0.7 3.0147 y5
0.8 3.1661 y6
0.9 3.3483 y7
1 3.5598 y8
1.1 3.8001 y9
1.2 4.0698 y10
1.3 4.3705 y11
1.4 4.7042 y12
By Weddle’s rule:
∫ 𝑓(𝑥)𝑑𝑥 =
3ℎ
10
[𝑦0 + 5𝑦1 + 𝑦2 + 6𝑦3 + 𝑦4 + 5𝑦5 + 2𝑦6 + 5𝑦7 + 𝑦8 + 6𝑦9 + 𝑦10 + 5𝑦11 + 𝑦12
𝑥0+𝑛ℎ
𝑥0
+ ⋯ … . ]
∫ (𝑠𝑖𝑛𝑥 − log𝑒 𝑥 + 𝑒𝑥)
1.4
0.2
𝑑𝑥 = 4.050

13. Dr. Deepa Chauhan
Example 12. A train is moving at the speed of 30m/sec. suddenly breaks are applied. The speed of the
train per second after t seconds is given by
Time (t): 0 5 10 15 20 25 30 35 40 45
Speed (v): 30 24 19 16 13 11 10 8 7 5
𝑦0 𝑦1 𝑦2 𝑦3 𝑦4 𝑦5 𝑦6 𝑦7 𝑦8 𝑦9
Apply Simpson’s rule to determine the distance covered by the train in 45 seconds.
Sol. If s meters is the distance covered in t seconds, then
𝑑𝑠
𝑑𝑡
= 𝑣
𝑑𝑠 = 𝑣 𝑑𝑡
𝑠 = ∫ 𝑣 𝑑𝑡
𝑡=45
𝑡=0
Since number of subintervals is 9 hence by using Simpson’s 3/8th
rule:
∫ 𝑣 𝑑𝑡
𝑡=45
𝑡=0
=
3ℎ
8
[(𝑦0 + 𝑦𝑛) + 3(𝑦1 + 𝑦2 + 𝑦4 + 𝑦5 … … . . +𝑦𝑛−1) + 2 (𝑦3 + 𝑦6 + 𝑦9 + 𝑦12
+ ⋯ … . . +𝑦𝑛−3)]
𝑠 = ∫ 𝑣 𝑑𝑡
𝑡=45
𝑡=0
=
3ℎ
8
[(𝑦0 + 𝑦9) + 3(𝑦1 + 𝑦2 + 𝑦4 + 𝑦5 + 𝑦7 + 𝑦8 … ) + 2 (𝑦3 + 𝑦6)]
=
3×5
8
[(30 + 5) + 3(24 + 19 + 13 + 11 + 8 + 7) + 2 (16 + 10)]
= 624.375 𝑚𝑒𝑡𝑟𝑒𝑠
Example 13. . 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:
x: 0 10 20 30 40 50 60 70 80
y: 0 4 7 9 12 15 14 8 3

14. Dr. Deepa Chauhan
Find the approximate area of the cross-section of the river using Simpson’s rule.
Sol. The required area of cross-section of the river= ∫ 𝑦 𝑑𝑥
80
0
Here the number of subintervals is 8, then by using Simpson’s 1/3rd
rule
∫ 𝑓(𝑥)𝑑𝑥 =
ℎ
3
[(𝑦0 + 𝑦𝑛) + 4(𝑦1 + 𝑦3 + 𝑦5 … … . . +𝑦𝑛−1) + 2 (𝑦2 + 𝑦4 + 𝑦6 … … . . +𝑦𝑛−2)]
𝑥0+𝑛ℎ
𝑥0
𝐴 = ∫ 𝑦 𝑑𝑥 =
80
0
ℎ
3
[(𝑦0 + 𝑦8) + 4(𝑦1 + 𝑦3 + 𝑦5 + 𝑦7) + 2 (𝑦2 + 𝑦4 + 𝑦6)]
=
10
3
[(0 + 3) + 4(4 + 9 + 15 + 8) + 2 (7 + 12 + 14)] = 710 𝑚2
Example 14. A solid is formed by rotating about x-axis, the lines x=0 and x=1 and a curve through the
points with the following coordinates
x: 0 0.25 0.5 0.75 1
y: 1 0.9896 0.9589 0.9089 0.8415
Estimate the volume of solid formed using Simpson’s rule.
Sol. If V is the volume of the solid formed, then
𝑉 = 𝜋 ∫ 𝑦2
𝑑𝑥
1
0
x: 0 0.25 0.5 0.75 1
y2
: 0 0.9793 0.9195 0.8261 0.7081
Here the number of subintervals is 4, then by using Simpson’s 1/3rd
rule
∫ 𝑓(𝑥)𝑑𝑥 =
ℎ
3
[(𝑦0 + 𝑦𝑛) + 4(𝑦1 + 𝑦3 + 𝑦5 … … . . +𝑦𝑛−1) + 2 (𝑦2 + 𝑦4 + 𝑦6 … … . . +𝑦𝑛−2)]
𝑥0+𝑛ℎ
𝑥0
∫ 𝑦2
𝑑𝑥 =
1
0
ℎ
3
[(𝑦0 + 𝑦4) + 4(𝑦1 + 𝑦3) + 2 (𝑦2)]
=
0.25
3
[(0 + 0.7081) + 4(0.9793 + 0.8261) + 2(0.9195)] = 0.8140
𝑉 = 𝜋 ∫ 𝑦2
𝑑𝑥 = 𝜋 × 0.8140 = 2.5561 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠
1
0