Numerical Integration

Trapezoidal Rule

Q1- Using Trapezoidal rule, calculate the value of integral ∫ logx dx given that:

x      4.0     4.2     4.4     4.6     4.8     5.0     5.2
y   1.3863  1.4251  1.4816  1.5260  1.5686  1.6094  1.6486

Solution:

x₀ = 4.0 which is the first value of x and x₆ = 5.2 which is last value of x
These becomes upper and lower limit of ∫logx dx

Calculate y which equal to logx and which is already given in that question

x0 = 4.0 and y0 = 1.3863
x1 = 4.2 and y1 = 1.4351
x2 = 4.4 and y2 = 1.4816
x3 = 4.6 and y3 - 1.5266
x4 = 4.8 and y4 = 1.5686
x5 = 5.0 and y5 = 1.6094
x6 = 5.2 and y6 = 1.6486

Trapezoidal Rule Formula:
∫ f(x) dx = $\frac{h}{2}$ [(y0 + y6) + 2 (y1 + y2 + ... + y5)]
∫ f(x) dx = $\frac{h}{2}$ [(first y value + last y value) + 2 (remaining y value in between first and last)]
h is the difference between two x , 4.2 - 4.0 = 0.2
∫ logx dx = $\frac{0.2}{2}$ [(1.3863 + 1.6486) + 2 (1.4351 + 1.4816 + 1.5266 + 1.5686 + 1.6094)]
∫ logx dx = 0.1[(3.0349) + 2 (7.6213)]
∫ logx dx = 0.1[3.0349 + 15.2426]
∫ logx dx = 0.1[18.2775]
∫ logx dx = 1.8278 (answer)

Q2- $ \int_{0}^{6} \frac{1}{1+x^2} $ using Trapezoidal rule.

Solution:

x₀ = 0 and x_n = 6
Now for h we can assume it whatever like 1, 2 etc. or we can calculate it by assuming n
as x₀ + nh = x_n
0 + nh = 6
taking n = 6
h = 1

Now create the table
x₁ = x₀ + h
y₀ = $\frac{1}{1+x0^2}$

x0 = 0 and y0 = 1
x1 = 1 and y1 = 0.5
x2 = 2 and y2 = 0.2
x3 = 3 and y3 = 0.1
x4 = 4 and y4 = 0.0588
x5 = 5 and y5 = 0.0385
x6 = 6 and y6 = 0.0270

Now putting values into the formula:
∫ f(x) dx = $\frac{h}{2}$ [(y0 + y6) + 2 (y1 + y2 + ... + y5)]
$ \int_{0}^{6} \frac{1}{1+x^2}\ dx $ = $\frac{h}{2}$ [(1 + 0.0270) + 2 (0.5 + 0.2 + 0.1 + 0.0588 + 0.0385)]
$ \int_{0}^{6} \frac{1}{1+x^2}\ dx $ = $\frac{h}{2}$ [(1.0270) + 2 (0.8973)]
$ \int_{0}^{6} \frac{1}{1+x^2}\ dx $ = $\frac{h}{2}$ [(1.0270) + 1.7946]
$ \int_{0}^{6} \frac{1}{1+x^2}\ dx $ = $\frac{1}{2}$ [2.8216]
$ \int_{0}^{6} \frac{1}{1+x^2}\ dx $ = 0.5 * 2.8216
$ \int_{0}^{6} \frac{1}{1+x^2}\ dx $ = 1.4108 (answer)

Q3 - $ \int_{0}^{1} \frac{1}{1+x^2}\ dx $ taking h = 1/4 by Trapezoidal rule

Solution:

x₀ = 0 and x_n = 1
h = 0.25

Now we know that x_n = x_n-1 + h
x₁ = x₀ + h
y₀ = $\frac{1}{1+x0^2}$ , so using this we can create the table

x0 = 0      and y0 = 1
x1 = 0.25   and y1 = 0.9412
x2 = 0.5    and y2 = 0.8
x3 = 0.75   and y3 = 0.64
x4 = 1      and y4 = 0.5

Now putting values into the formula:
∫ f(x) dx = $\frac{h}{2}$ [(y0 + y6) + 2 (y1 + y2 + ... + y5)]
$ \int_{0}^{6} \frac{1}{1+x^2}\ dx $ = $\frac{h}{2}$ [(1 + 0.5) + 2 (0.9412 + 0.8 + 0.64)]
$ \int_{0}^{6} \frac{1}{1+x^2}\ dx $ = $\frac{h}{2}$ [(1.5) + 2 (2.3812)]
$ \int_{0}^{6} \frac{1}{1+x^2}\ dx $ = $\frac{h}{2}$ [1.5 + 4.7624]
$ \int_{0}^{6} \frac{1}{1+x^2}\ dx $ = $\frac{0.25}{2}$ [6.2624]
$ \int_{0}^{6} \frac{1}{1+x^2}\ dx $ = 0.125 * 6.2624
$ \int_{0}^{6} \frac{1}{1+x^2}\ dx $ = 0.7828 (answer)

Q4- Using Trapezoidal Rule evaluate $ \int_{0}^{1} x^3 dx $ & taking 5 sub intervals.

Solution:

As sub intervals is given as 5 so n = 5
x₀ = 0 and x_n = 1
h = x_n - x₀ / n
h = 1/5
h = 0.2

x₁ = x₀ + h
y₀ = $ \int_{0}^{1} x^3 dx $

x0 = 0   and y0 = 0
x1 = 0.2 and y1 = 0.008
x2 = 0.4 and y2 = 0.064
x3 = 0.6 and y3 = 0.216
x4 = 0.8 and y4 = 0.512
x5 = 1   and y5 = 1

Formula:
∫ f(x) dx = $\frac{h}{2}$ [(y0 + y6) + 2 (y1 + y2 + ... + y5)]
$ \int_{0}^{1} x^3 dx $ = $\frac{h}{2}$ [(0 + 1) + 2 (0.008 + 0.064 + 0.216 + 0.512)]
$ \int_{0}^{1} x^3 dx $ = $\frac{h}{2}$ [1 + 2 (0.8)]
$ \int_{0}^{1} x^3 dx $ = $\frac{h}{2}$ [1 + 1.6]
$ \int_{0}^{1} x^3 dx $ = $\frac{0.2}{2}$ [2.6]
$ \int_{0}^{1} x^3 dx $ = 0.1 * 2.6
$ \int_{0}^{1} x^3 dx $ = 0.26 (answer)

Simpson's 1/3rd Rule

Q1- Using Simpson's 1/3rd Rule, calculate the value of integral $ \int_{0}^{1} \frac{1}{1+x^2}\ dx $ and h = 1/4

Solution:

h = 0.25

x₁ = x₀ + h
y₀ = $\frac{1}{1+x0^2}$

x0 = 0    and y0 = 1
x1 = 0.25 and y1 = 0.9412
x2 = 0.5  and y2 = 0.8
x3 = 0.75 and y3 = 0.64
x4 = 1    and y4 = 0.5

Formula:
∫ f(x) dx = $\frac{h}{3}$ [(y0 + y4) + 4 (y1 + y3) + 2(y2)]
∫ f(x) dx = $\frac{h}{3}$ [(sum of first and last y) + 4 (sum of odd y) + 2(sum of even y)]
$ \int_{0}^{1} \frac{1}{1+x^2}\ dx $ = $\frac{0.25}{3}$ [(1 + 0.5) + 4(0.9412 + 0.64) + 2(0.8)]
$ \int_{0}^{1} \frac{1}{1+x^2}\ dx $ = 0.7860 (approx)

Q2- Evaluate $ \int_{0}^{1} e^{-x^2} \, dx $ by Simpson's 1/3rd Rule.

x₀ = 0 and x_n = 1
taking n = 4 (we try to take even sub intervals)
x₀ + nh = x_n
0 + 4h = 1
h = 0.25

x₁ = x₀ + h
y₀ = $ \int_{0}^{1} e^{-x0^2} \, dx $

x0 = 0    and y0 = 1
x1 = 0.25 and y1 = 0.9394
x2 = 0.5  and y2 = 0.7788
x3 = 0.75 and y3 = 0.5697
x4 = 1    and y4 = 0.3679

$ \int_{0}^{1} e^{-x^2} \, dx $ = $\frac{0.25}{3}$ [(1 + 0.3679) + 4(0.9394 + 0.5697) + 2(0.7788)]
$ \int_{0}^{1} e^{-x^2} \, dx $ = 0.7468 (approx)

Q3- Evaluate $ \int_{1}^{2} e^{x^3} \, dx $ by Simpson's 1/3rd Rule.

Simpson's 3/8th Rule

Q1- Using Simpson's 3/8th Rule, calculate the value of integral $\int_{0}^{6} \frac{1}{1+x^2}\ dx$

Solution:

x0 = 0 and xn = 6
x0 + nh = 6
taking n = 6
0 + 6h = 6
h = 1

x₁ = x₀ + h
y₀ = $ \frac{1}{1+x0^2}\ $

x0 = 0 and y0 = 1
x1 = 1 and y1 = 0.5
x2 = 2 and y2 = 0.2
x3 = 3 and y3 = 0.1
x4 = 4 and y4 = 0.0588
x5 = 5 and y3 = 0.0385
x6 = 6 and y4 = 0.027

Formula:
∫ f(x) dx = $\frac{3h}{8}$ [(y0 + y6) + 2(y3+...) + 3(y1 + y2 + y4 + y5 + ...)]
∫ f(x) dx = $\frac{3h}{8}$ [(sum of first and last y) + 2(y, which are multiple of 3) + 3(rest of y)]
$ \int_{0}^{6} \frac{1}{1+x^2}\ dx $ = $\frac{3}{8}$ [(1 + 0.027) + 2(0.1) + 3(0.5 + 0.2 + 0.0588 + 0.0385)]
$ \int_{0}^{6} \frac{1}{1+x^2}\ dx $ = 1.3571 (approx)

Q2- Using Simpson's 3/8th Rule, calculate the value of integral $ \int_{0}^{3} \frac{1}{1+x^5}\ dx $ and n = 6

Solution:

x0 = 0 and xn = 3
x0 + nh = 3
0 + 6h = 3
h = 0.5

x₁ = x₀ + h
y₀ = $\frac{1}{1+x0^2}$

x0 = 0   and y0 = 1
x1 = 0.5 and y1 = 0.9697
x2 = 1   and y2 = 0.5
x3 = 1.5 and y3 = 0.11636
x4 = 2   and y4 = 0.03031
x5 = 2.5 and y3 = 0.01014
x6 = 3   and y4 = 0.00410

$ \int_{0}^{3} \frac{1}{1+x^5}\ dx $ = $\frac{3*05}{8}$ [(1 + 0.00410) + 2(0.11636) + 3(1.5102)]
$ \int_{0}^{3} \frac{1}{1+x^5}\ dx $ = 1.0814 (approx)

Boole's Rule

Formula:

$ \int_{x0}^{xn} f(x) dx $ = $\frac{2h}{45}$ [(7y0 + 32y1 + 12y2 + 32y3 + 7y4 + 7y4 + 32y5 + 12 y6 + ....)]

7 - 32 - 12 - 32 - 7 again repeat
y4 is repeated twice because this method works with instance of 4

Q- Evaluate $ \int_{0}^{4} \frac{1}{1+x^2}\ dx $ using Boole's Rule taking h = 0.5

x0 = 0   and y0 = 1
x1 = 0.5 and y1 = 0.8
x2 = 1   and y2 = 0.5
x3 = 1.5 and y3 = 0.3077
x4 = 2   and y4 = 0.2
x5 = 2.5 and y3 = 0.1379
x6 = 3   and y4 = 0.1
x7 = 3.5 and y3 = 0.0755
x8 = 4   and y4 = 0.0588

$ \int_{0}^{4} \frac{1}{1+x^2}\ dx $ = $\frac{2 * 0.5}{45}$ [7(1) + 32(0.8) + 12(0.5) + 32(0.3077) + 7(0.2) + 7(0.2) + 32(0.1379) + 12(0.1) + 32(0.0755) + 7(0.0588)]
$ \int_{0}^{4} \frac{1}{1+x^2}\ dx $ = 1.3250

Weedle's Rule

Note:

We divide the interval into an even number of subintervals in the 1/3rd Simpson Method
We divide the interval into multiples of 3 subintervals in the 3/8th Simpson Method
We divide the interval into multiples of 4 subintervals in Boole's Rule
We divide the interval into multiples of 6 subintervals in Weddle's Method

Formula:$ \int_{x0}^{xn} f(x) dx $ = $\frac{3h}{10}$ [1(y0) + 5(y1) + 1(y2) + 6(y3) + 1(y4) + 5(y5) + 1(y6)]
the sequence goes like: 1,5,1,6,1,5,1,5,1,6,1,5,2,5,1,6,1,5