Numerical Integration: Multi-Segment Trapezoidal Rule

1. What is Numerical Integration? (2 marks)

Numerical Integration (also called numerical quadrature) is the process of computing the approximate value of a definite integral using numerical methods when:

• The integrand f(x) is too complex to integrate analytically, or
• The function is given only as a set of discrete data points (tabular data), or
• The antiderivative does not exist in closed form.

The general goal is to approximate:

$$I = \int_a^b f(x), dx$$

by replacing the integrand with a simpler approximating function (such as polynomials) and summing the areas of geometric shapes under the curve.

2. Application of Numerical Integration (1 mark)

Engineering Application: Computing the area under a velocity-time curve to find displacement, or calculating the work done by a variable force:

$$W = \int_{x_1}^{x_2} F(x), dx$$

where F(x) may be given only as experimental data points, making analytical integration impossible.

3. Multi-Segment Trapezoidal Rule (7 marks)

Concept

The interval [a, b] is divided into n equal sub-intervals (segments), each of width:

$$h = \frac{b - a}{n}$$

The nodes are: $x_0 = a,; x_1 = a+h,; x_2 = a+2h,; \ldots,; x_n = b$

Each sub-interval is approximated by a trapezoid. The formula is:

$$\boxed{I \approx \frac{h}{2}\left[f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_n)\right]}$$

Or equivalently:

$$I \approx \frac{h}{2}\left[f(x_0) + f(x_n) + 2\sum_{i=1}^{n-1} f(x_i)\right]$$

Algorithm

Algorithm: Multi-Segment Trapezoidal Rule
Input : a (lower limit), b (upper limit), n (number of segments)
Output: Approximate value of integral I

Step 1: START
Step 2: Read a, b, n
Step 3: Compute h = (b - a) / n
Step 4: Set sum = f(a) + f(b)          [first and last terms]
Step 5: Set i = 1
Step 6: WHILE i <= n-1 DO
            x = a + i * h
            sum = sum + 2 * f(x)       [middle terms multiplied by 2]
            i = i + 1
        END WHILE
Step 7: I = (h / 2) * sum
Step 8: Print I
Step 9: STOP

C Program

#include <stdio.h>
#include <math.h>

/* Define the function to be integrated */
/* Example: f(x) = x^2, change as needed */
double f(double x) {
    return x * x;   /* Replace with desired function */
}

int main() {
    double a, b, h, sum, x, result;
    int n, i;

    /* Input */
    printf("Enter lower limit (a): ");
    scanf("%lf", &a);

    printf("Enter upper limit (b): ");
    scanf("%lf", &b);

    printf("Enter number of segments (n): ");
    scanf("%d", &n);

    /* Step size */
    h = (b - a) / n;

    /* Apply Multi-Segment Trapezoidal Rule */
    /* First and last terms */
    sum = f(a) + f(b);

    /* Middle terms */
    for (i = 1; i <= n - 1; i++) {
        x = a + i * h;
        sum = sum + 2 * f(x);
    }

    /* Final result */
    result = (h / 2) * sum;

    printf("\nApproximate value of integral = %.6lf\n", result);

    return 0;
}

Sample Output

For $\int_0^1 x^2, dx$ with n = 4:

$$h = \frac{1-0}{4} = 0.25$$

$$I \approx \frac{0.25}{2}\left[0 + 1 + 2(0.0625 + 0.25 + 0.5625)\right]$$

$$I \approx 0.125 \times [1 + 2(0.875)] = 0.125 \times 2.75 = \mathbf{0.34375}$$

Exact value = 1/3 = 0.33333... (small error due to finite segments; error decreases as n increases)

Key Points

Least Squares Method and Second Order Polynomial Fitting

Definition

The least squares method finds the best-fit curve for a set of data points by minimizing the sum of squares of residuals (differences between observed values $y_i$ and fitted values $f(x_i)$):

$$S = \sum_{i=1}^{n} [y_i - f(x_i)]^2 \to \text{minimum}$$

Given Data

$$x: 1, 2, 3, 4, 5, 6, 7$$ $$y: 2, 6, 7, 8, 10, 12, 15$$ $$n = 7$$

Model and Normal Equations

Fit $y = a_0 + a_1 x + a_2 x^2$. The normal equations are:

$$\sum y = na_0 + a_1\sum x + a_2\sum x^2$$ $$\sum xy = a_0\sum x + a_1\sum x^2 + a_2\sum x^3$$ $$\sum x^2y = a_0\sum x^2 + a_1\sum x^3 + a_2\sum x^4$$

Calculation Table

Also $\sum x = 28$.

Substituting

$$60 = 7a_0 + 28a_1 + 140a_2 \quad (1)$$ $$294 = 28a_0 + 140a_1 + 784a_2 \quad (2)$$ $$1634 = 140a_0 + 784a_1 + 4676a_2 \quad (3)$$

Note: The coefficient of $a_0$ in equation (1) must be $\sum x = 28$, while in equation (3) the $a_0$ coefficient is $\sum x^2 = 140$; mixing the two is an easy slip to make.

Solving

Eliminate $a_0$. Multiply (1) by 4 and subtract from (2):

$(2) - 4\times(1)$: $$294 - 240 = (28-28)a_0 + (140-112)a_1 + (784-560)a_2$$ $$54 = 28a_1 + 224a_2 \quad (4)$$

Multiply (1) by 20 and subtract from (3):

$(3) - 20\times(1)$: $$1634 - 1200 = (140-140)a_0 + (784-560)a_1 + (4676-2800)a_2$$ $$434 = 224a_1 + 1876a_2 \quad (5)$$

Eliminate $a_1$. Multiply (4) by 8: $$432 = 224a_1 + 1792a_2 \quad (4')$$

$(5) - (4')$: $$434 - 432 = (1876 - 1792)a_2$$ $$2 = 84a_2 \implies a_2 = 0.02381$$

From (4): $$54 = 28a_1 + 224(0.02381) = 28a_1 + 5.333$$ $$28a_1 = 48.667 \implies a_1 = 1.7381$$

From (1): $$60 = 7a_0 + 28(1.7381) + 140(0.02381)$$ $$60 = 7a_0 + 48.667 + 3.333$$ $$7a_0 = 8.0 \implies a_0 = 1.1429$$

Result

$$\boxed{y = 1.1429 + 1.7381x + 0.0238x^2}$$

Verification (at $x=4$): $1.1429 + 6.9524 + 0.381 = 8.476 \approx 8$ ✓ (at $x=7$): $1.1429 + 12.167 + 1.167 = 14.48 \approx 15$ ✓