Numerical Methods -- Integration -- Module 3

#Numerical-Methods #Semester-6

Index

  1. Numerical Integration
  2. 1. The Trapezoidal Rule of Numerical Integration
  3. 2. Simpson's 1/3 rule
  4. Solved Examples

Numerical Integration

Overview of Numerical Integration

Numerical integration methods allow us to estimate the area under a curve, that is, the integral of a function, using discrete data points. The two most common techniques are:

Each method has its own advantages, assumptions, and error characteristics.

https://www.youtube.com/watch?v=iviiGB5vxLA&list=PLU6SqdYcYsfLrTna7UuaVfGZYkNo0cpVC&index=5

Suppose we have a definite integral like this:

Pasted image 20250403121555.png

Now suppose this is the area defined by this integral

Pasted image 20250403122215.png

To solve this, we can divide the interval (a,b) into n-equal intervals of length h , which is called the step-size.

Now depending on how the interval (a,b) is divided, there are two rules which can be applied to solve the integral.

1. The Trapezoidal Rule of Numerical Integration

Concept

The trapezoidal rule approximates the area under a curve by dividing it into a series of trapezoids rather than rectangles. The area of each trapezoid is computed and summed up.

For a function f(x) on the interval [a,b] divided into n equal subintervals of width h=b−an​, the composite trapezoidal rule is:

∫abf(x)dx ≈ h2[f(x0​)+2f(x1​)+2f(x2​)+ ⋯ + 2f(xn−1​) + f(xn​)]

Do note that the end points don't have any coefficients so x0 and xn don't have any coefficients attached.

or

Pasted image 20250403123030.png

where y0 and yn are the first and last terms respectively, followed by the remaining terms.

Error Estimate

The error ET​ for the trapezoidal rule is given by:

E_T = -\frac{(b-a)h^2}{12} f''(\xi)$$, for some $\xi \in [a,b]$. The error depends on the second derivative of $f(x)$; if $f(x)$ is nearly linear, the error will be small. --- # 2. Simpson's 1/3 rule ### Concept Simpson’s 1/3 rule approximates the area under the curve by fitting quadratic polynomials (parabolas) to the function over pairs of subintervals. It provides better accuracy than the trapezoidal rule when the function is smooth. ### Requirements - The number of subintervals $n$ must be even. - The interval is again $[a,b]$ with $h = \frac{b-a}{n}$ ### Formula The composite Simpson's 1/3 rule is: $$\int_{a}^{b}{f(x) \ dx} \ \approx \ \frac{h}{3}[f(x_0​)+4f(x_1​)+2f(x_2​)+4f(x_3​)+⋯+4f(x_{n−1}​)+f(x_n​)]

or

Pasted image 20250403130020.png

where y1 , y3 are the odd terms and y2 , y4 are the even terms.

This means that 4 is multiplied to the odd terms and 2 is multiplied to the even terms.

Do note that the end points don't have any coefficients so y0 and yn don't have any coefficients attached.

Error Estimate

The error ES for Simpson's rule is:

E_S​\ = \ \frac{-(b−a)h^4}{180} ​f^{(4)}(\xi)$$, for some $\xi \in [a,b]$. Notice the higher power of $h$ in the error term, which explains why Simpson's rule is generally more accurate for smooth functions. --- # Solved Examples Let's go through a simple example to illustrate these methods. ## Example 1 Approximate $$\int_{0}^{2}{(x^3 \ + \ x) \ dx}

with n=4 subintervals.

Before trying out any rule, we first find a few data points.

1. Find h

h=2−04 = 12 = 0.5

2. Calculate xi values

Since the step-size is of 0.5 then for 4 subintervals :

x0 =0x1=0+h =0.5x2 = 0+2h = 1x3 = 0+3h =1.5x4 = 0+4h =2.0

3. Evaluate the values for f

  1. f(0) = 03+0 =0
  2. f(0.5) = 0.53+0.5 = 0.625
  3. f(1.0) = 13+1 =2
  4. f(1.5) = 1.53 + 1.5 = 4.875
  5. f(2.0) = 23+2 =10

Now, we will solve this integral using both the rules.

1. Trapezoidal rule solving

∫02(x3+x)dx ≈ h2[f(x0​)+2f(x1​)+2f(x2​)+2f(x3)+f(x4)]= 0.52[0 + 2 ×0.625 + 2×2 + 2×4.875 + 10]= 0.25[1.25 + 4 + 9.75 + 10]= 0.25 [25]=6.25

2. Simpson's 1/3 rule solving

∫abf(x) dx ≈ h3[f(x0​) + 4f(x1​) + 2f(x2​) + 4f(x3​)+ f(x4)]=0.53 [0 + 4×0.625 + 2×2+ 4×4.875 + 10]= 0.53 [0+2.5+4+19.5 +10]=0.53 ×36= 0.5 ×12= 6.0

Comparison

Depending on the function's curvature, Simpson's rule often gives a better approximation. In this case, Simpson's 1/3 rule approximates the integral as 6.

Example 2

Pasted image 20250403132014.png

We are not going to do the 3/8 rule as it's not in our syllabus.

So we have :

∫01dx1+x2

Let's solve this for 6 sub intervals (n=6)

So firstly we will find the value of h

h = 1−06 =16 = 0.16667

So with a step size of 0.16667,

x0=0x1=0 + h =0.16667x2 =0 + 2h =0.33333x3 = 0 + 3h =0.50001x4 = 0 + 4h = 0.66667x5 = 0 + 5h = 0.83333x6 = 0+6h = 1

Now we find the values of f(x0) till f(x6)

  1. f(0) = 11 + 02 =1
  2. f(0.16667) = 11 + 0.166672 ≈ 11.02778 ≈ 0.97297
  3. f(0.33333) = 11 + 0.333332 ≈ 11.11111 ≈ 0.9
  4. f(0.50001) =11 + 0.500012 ≈ 11.25001 ≈0.8
  5. f(0.66667) = 11 +0.666672 ≈ 11.44444 ≈ 0.6923
  6. f(0.83333) = 11 + 0.833332 ≈ 11.69444 ≈ 0.5909
  7. f(1) = 11 + 12 = 12 = 0.5

1. Solving using Trapezoidal Rule

∫02(11+x2)dx ≈ h2[f(x0​)+2f(x1​)+2f(x2​)+2f(x3)+2f(x4) +2f(x5) +f(x6)]= 0.166672 [1 + 1.94594 + 1.8 + 1.6 + 1.3846 + 1.1818 + 0.5]= 0.16667 ×9.412342= 1.5687547082≈ 0.784377353

2. Solving using Simpson's 1/3 rule.

∫02(11+x2)dx ≈ h3[f(x0​) + 4f(x1​) + 2f(x2​) + 4f(x3​)+ 2f(x4) + 4f(x5) +f(x6)]= 0.166673 [1 +3.89188 +1.8 + 3.2 + 1.3846 + 2.3636 + 0.5]= 0.166673 [14.14008]≈ 0.785575711

As we can see, Simpson's 1/3 rule resulted in a better precision than the trapezoidal rule.