🎯 Unit Learning Objectives:
- Derive Numerical Differentiation formulas for first and second derivatives from Newton's forward and backward differences.
- Derive the General Quadrature Formula for equidistant ordinates from first principles.
- Deduce the Trapezoidal Rule ($n=1$), Simpson's 1/3rd Rule ($n=2$), and Simpson's 3/8th Rule ($n=3$) from the General Quadrature formula.
- Evaluate the mandatory subinterval constraints for each numerical integration rule.
- Compare truncation error bounds and degrees of precision across rules.
- Solve university numerical integrals (e.g. $\int_0^6 \frac{dx}{1+x^2}$) using all rules and compute error relative to exact calculus values.
💡 Conceptual Intuition: Why Do We Need Numerical Integration?
In analytical calculus, we learn integration tricks (substitution, integration by parts, partial fractions). But many real-world functions have no elementary anti-derivative: $$\int e^{-x^2}\,dx \quad \text{or} \quad \int \frac{\sin x}{x}\,dx \quad \text{or} \quad \int \sqrt{1 + x^4}\,dx$$ Furthermore, in engineering experiments, we often don't have a mathematical function at all—we only have a table of discrete sensor readings (e.g. velocity over time)!
**Numerical Integration (Quadrature)** solves this by calculating the **definite area under the curve** by slicing the area into simple geometric strips (trapezoids or parabolas) whose areas can be added with basic arithmetic!
In analytical calculus, we learn integration tricks (substitution, integration by parts, partial fractions). But many real-world functions have no elementary anti-derivative: $$\int e^{-x^2}\,dx \quad \text{or} \quad \int \frac{\sin x}{x}\,dx \quad \text{or} \quad \int \sqrt{1 + x^4}\,dx$$ Furthermore, in engineering experiments, we often don't have a mathematical function at all—we only have a table of discrete sensor readings (e.g. velocity over time)!
**Numerical Integration (Quadrature)** solves this by calculating the **definite area under the curve** by slicing the area into simple geometric strips (trapezoids or parabolas) whose areas can be added with basic arithmetic!
Numerical Differentiation Formulas
⭐ CSJMU PYQs:
2024–25, 2022–23 (Derive first and second derivatives from forward difference formula)
Numerical differentiation is the process of computing the value of the derivative $\frac{dy}{dx}$ of a function at a given point using a discrete table of values.
Derivation from Newton's Forward Difference Formula:
Newton's forward formula states: $y(x) = y_0 + p \Delta y_0 + \frac{p(p-1)}{2} \Delta^2 y_0 + \frac{p(p-1)(p-2)}{6} \Delta^3 y_0 + \dots$ where $p = \frac{x - x_0}{h} \implies \frac{dp}{dx} = \frac{1}{h}$.
By the chain rule of differentiation: $$\frac{dy}{dx} = \frac{dy}{dp} \cdot \frac{dp}{dx} = \frac{1}{h} \left[\frac{d}{dp} y(x)\right]$$ Differentiating term-by-term with respect to $p$: $$\mathbf{\left(\frac{dy}{dx}\right)_{x = x_0} = \frac{1}{h}\left[\Delta y_0 - \frac{1}{2}\Delta^2 y_0 + \frac{1}{3}\Delta^3 y_0 - \frac{1}{4}\Delta^4 y_0 + \dots\right]}$$ Differentiating a second time yields the second derivative: $$\mathbf{\left(\frac{d^2y}{dx^2}\right)_{x = x_0} = \frac{1}{h^2}\left[\Delta^2 y_0 - \Delta^3 y_0 + \frac{11}{12}\Delta^4 y_0 - \frac{5}{6}\Delta^5 y_0 + \dots\right]}$$
Newton's forward formula states: $y(x) = y_0 + p \Delta y_0 + \frac{p(p-1)}{2} \Delta^2 y_0 + \frac{p(p-1)(p-2)}{6} \Delta^3 y_0 + \dots$ where $p = \frac{x - x_0}{h} \implies \frac{dp}{dx} = \frac{1}{h}$.
By the chain rule of differentiation: $$\frac{dy}{dx} = \frac{dy}{dp} \cdot \frac{dp}{dx} = \frac{1}{h} \left[\frac{d}{dp} y(x)\right]$$ Differentiating term-by-term with respect to $p$: $$\mathbf{\left(\frac{dy}{dx}\right)_{x = x_0} = \frac{1}{h}\left[\Delta y_0 - \frac{1}{2}\Delta^2 y_0 + \frac{1}{3}\Delta^3 y_0 - \frac{1}{4}\Delta^4 y_0 + \dots\right]}$$ Differentiating a second time yields the second derivative: $$\mathbf{\left(\frac{d^2y}{dx^2}\right)_{x = x_0} = \frac{1}{h^2}\left[\Delta^2 y_0 - \Delta^3 y_0 + \frac{11}{12}\Delta^4 y_0 - \frac{5}{6}\Delta^5 y_0 + \dots\right]}$$
Derivatives Near the End of Table (Using Newton's Backward Formula):
$$\mathbf{\left(\frac{dy}{dx}\right)_{x = x_n} = \frac{1}{h}\left[\nabla y_n + \frac{1}{2}\nabla^2 y_n + \frac{1}{3}\nabla^3 y_n + \frac{1}{4}\nabla^4 y_n + \dots\right]}$$ $$\mathbf{\left(\frac{d^2y}{dx^2}\right)_{x = x_n} = \frac{1}{h^2}\left[\nabla^2 y_n + \nabla^3 y_n + \frac{11}{12}\nabla^4 y_n + \dots\right]}$$Step-by-Step Solved Numerical: Derivative from Tabulated Data
Problem (CSJMU 2024–25): From the table below, find $\left(\frac{dy}{dx}\right)$ and $\left(\frac{d^2y}{dx^2}\right)$ at $x = 1.2$ ($h = 0.2$):
| $x$ | 1.0 | 1.2 | 1.4 | 1.6 | 1.8 | 2.0 |
|---|---|---|---|---|---|---|
| $y$ | 2.718 | 3.320 | 4.055 | 4.953 | 6.050 | 7.389 |
Constructing the forward differences taking $x_0 = 1.2$ as origin:
$\Delta y_0 = 4.055 - 3.320 = 0.735$
$\Delta^2 y_0 = (4.953 - 4.055) - 0.735 = 0.898 - 0.735 = 0.163$
$\Delta^3 y_0 = (6.050 - 4.953) - 0.898 - 0.163 = 1.097 - 0.898 - 0.163 = 0.036$.
Substitute into the first derivative formula ($h = 0.2$):
$$\left(\frac{dy}{dx}\right)_{x = 1.2} = \frac{1}{0.2}\left[0.735 - \frac{1}{2}(0.163) + \frac{1}{3}(0.036)\right] = 5 \times [0.735 - 0.0815 + 0.0120] = 5 \times 0.6655 = \mathbf{3.3275}$$Substitute into the second derivative formula:
$$\left(\frac{d^2y}{dx^2}\right)_{x = 1.2} = \frac{1}{(0.2)^2}\left[0.163 - 0.036\right] = \frac{1}{0.04}[0.127] = \mathbf{3.175}$$Analytical check: Since $y = e^x$, at $x = 1.2$, $y' = y'' = e^{1.2} \approx 3.3201$. The numerical approximation is remarkably accurate!
Maxima and Minima of a Tabulated Function
Procedure to Find Local Extremum:
1. Write Newton's Forward formula in terms of $p$: $y(p) = y_0 + p \Delta y_0 + \frac{p(p-1)}{2} \Delta^2 y_0 + \dots$
2. Differentiate with respect to $p$ and set $\frac{dy}{dx} = 0 \implies \Delta y_0 + \frac{2p - 1}{2}\Delta^2 y_0 + \frac{3p^2 - 6p + 2}{6}\Delta^3 y_0 = 0$.
3. Solve this polynomial equation for $p$.
4. The location of the extremum is $x = x_0 + ph$, and the extreme value is found by substituting $p$ into $y(p)$!
5. Check the second derivative: if $\frac{d^2y}{dx^2} < 0 \implies$ Maximum; if $> 0 \implies$ Minimum.
1. Write Newton's Forward formula in terms of $p$: $y(p) = y_0 + p \Delta y_0 + \frac{p(p-1)}{2} \Delta^2 y_0 + \dots$
2. Differentiate with respect to $p$ and set $\frac{dy}{dx} = 0 \implies \Delta y_0 + \frac{2p - 1}{2}\Delta^2 y_0 + \frac{3p^2 - 6p + 2}{6}\Delta^3 y_0 = 0$.
3. Solve this polynomial equation for $p$.
4. The location of the extremum is $x = x_0 + ph$, and the extreme value is found by substituting $p$ into $y(p)$!
5. Check the second derivative: if $\frac{d^2y}{dx^2} < 0 \implies$ Maximum; if $> 0 \implies$ Minimum.
The General Quadrature Formula (Master Derivation)
⭐ CSJMU PYQs:
2025–26, 2024–25, 2023–24, 2021, 2019 (Fundamental 10-mark derivation from which all rules stem!)
Let $I = \int_a^b f(x)\,dx$. Divide the interval $[a, b]$ into $n$ equal subintervals of width $h = \frac{b - a}{n}$ with abscissas $x_0 = a, x_1 = x_0 + h, \dots, x_n = x_0 + nh = b$.
Step-by-Step Derivation:
Substitute $x = x_0 + uh \implies dx = h\,du$. When $x = x_0, u = 0$; when $x = x_n, u = n$.
Replacing $f(x)$ with Newton's forward difference interpolating polynomial: $$I = \int_{x_0}^{x_0 + nh} f(x)\,dx = \int_0^n \left[y_0 + u \Delta y_0 + \frac{u(u-1)}{2!} \Delta^2 y_0 + \frac{u(u-1)(u-2)}{3!} \Delta^3 y_0 + \dots\right] h\,du$$ Evaluating the definite integrals with respect to $u$: $$\int_0^n du = n, \quad \int_0^n u\,du = \frac{n^2}{2}, \quad \int_0^n \frac{u^2-u}{2}\,du = \frac{n^3}{6} - \frac{n^2}{4} = \frac{n^2(2n-3)}{12}$$ Factoring out $nh$, we obtain the **General Quadrature Formula**: $$\mathbf{\int_{x_0}^{x_0+nh} f(x)\,dx = nh\left[y_0 + \frac{n}{2}\Delta y_0 + \frac{n(2n-3)}{12}\Delta^2 y_0 + \frac{n(n-2)^2}{24}\Delta^3 y_0 + \dots\right]}$$
Substitute $x = x_0 + uh \implies dx = h\,du$. When $x = x_0, u = 0$; when $x = x_n, u = n$.
Replacing $f(x)$ with Newton's forward difference interpolating polynomial: $$I = \int_{x_0}^{x_0 + nh} f(x)\,dx = \int_0^n \left[y_0 + u \Delta y_0 + \frac{u(u-1)}{2!} \Delta^2 y_0 + \frac{u(u-1)(u-2)}{3!} \Delta^3 y_0 + \dots\right] h\,du$$ Evaluating the definite integrals with respect to $u$: $$\int_0^n du = n, \quad \int_0^n u\,du = \frac{n^2}{2}, \quad \int_0^n \frac{u^2-u}{2}\,du = \frac{n^3}{6} - \frac{n^2}{4} = \frac{n^2(2n-3)}{12}$$ Factoring out $nh$, we obtain the **General Quadrature Formula**: $$\mathbf{\int_{x_0}^{x_0+nh} f(x)\,dx = nh\left[y_0 + \frac{n}{2}\Delta y_0 + \frac{n(2n-3)}{12}\Delta^2 y_0 + \frac{n(n-2)^2}{24}\Delta^3 y_0 + \dots\right]}$$
Deductions from General Quadrature
⭐ CSJMU PYQs:
2025–26, 2024–25, 2023–24, 2021 (State conditions and formulas for Trapezoidal, Simpson 1/3 and 3/8)
1. Trapezoidal Rule (Put $n = 1$)
Putting $n = 1$ in the General Quadrature formula and neglecting 2nd and higher differences ($\Delta^2 y_0 \approx 0$):
$$\int_{x_0}^{x_1} f(x)\,dx = h\left[y_0 + \frac{1}{2}\Delta y_0\right] = h\left[y_0 + \frac{1}{2}(y_1 - y_0)\right] = \frac{h}{2}(y_0 + y_1)$$Summing across all $n$ subintervals gives the **Composite Trapezoidal Rule**:
$$\mathbf{\int_{x_0}^{x_n} f(x)\,dx = \frac{h}{2}\left[(y_0 + y_n) + 2(y_1 + y_2 + y_3 + \dots + y_{n-1})\right]}$$
Rule in Words: $\frac{h}{2} \left[(\text{Sum of First and Last ordinates}) + 2(\text{Sum of all remaining intermediate ordinates})\right]$.
• Constraint: $n$ can be ANY positive integer ($n \ge 1$).
• Geometric shape: Slices area into straight trapezoids.
• Truncation Error: $E_T = -\frac{(b-a)h^2}{12} f''(\xi)$. Degree of precision = 1.
• Constraint: $n$ can be ANY positive integer ($n \ge 1$).
• Geometric shape: Slices area into straight trapezoids.
• Truncation Error: $E_T = -\frac{(b-a)h^2}{12} f''(\xi)$. Degree of precision = 1.
2. Simpson's 1/3rd Rule (Put $n = 2$)
Putting $n = 2$ in the General Quadrature formula and neglecting 3rd and higher differences ($\Delta^3 y_0 \approx 0$):
$$\int_{x_0}^{x_2} f(x)\,dx = 2h\left[y_0 + \frac{2}{2}\Delta y_0 + \frac{2(4-3)}{12}\Delta^2 y_0\right] = 2h\left[y_0 + (y_1 - y_0) + \frac{1}{6}(y_2 - 2y_1 + y_0)\right] = \frac{h}{3}(y_0 + 4y_1 + y_2)$$Summing across all pairs of subintervals gives the **Composite Simpson's 1/3rd Rule**:
$$\mathbf{\int_{x_0}^{x_n} f(x)\,dx = \frac{h}{3}\left[(y_0 + y_n) + 4(y_1 + y_3 + y_5 + \dots) + 2(y_2 + y_4 + y_6 + \dots)\right]}$$
Rule in Words: $\frac{h}{3} \left[(\text{First} + \text{Last}) + 4(\text{Sum of ODD ordinates}) + 2(\text{Sum of EVEN ordinates})\right]$.
• MANDATORY CONSTRAINT: The number of subintervals $n$ MUST BE AN **EVEN NUMBER** ($n = 2, 4, 6, 8, \dots$).
• Geometric shape: Replaces the curve with second-degree parabolas through sets of 3 points.
• Truncation Error: $E_{S1/3} = -\frac{(b-a)h^4}{180} f^{(4)}(\xi)$. Degree of precision = 3! (Exact for polynomials up to degree 3).
• MANDATORY CONSTRAINT: The number of subintervals $n$ MUST BE AN **EVEN NUMBER** ($n = 2, 4, 6, 8, \dots$).
• Geometric shape: Replaces the curve with second-degree parabolas through sets of 3 points.
• Truncation Error: $E_{S1/3} = -\frac{(b-a)h^4}{180} f^{(4)}(\xi)$. Degree of precision = 3! (Exact for polynomials up to degree 3).
3. Simpson's 3/8th Rule (Put $n = 3$)
Putting $n = 3$ in the General Quadrature formula and neglecting 4th and higher differences:
$$\mathbf{\int_{x_0}^{x_n} f(x)\,dx = \frac{3h}{8}\left[(y_0 + y_n) + 3(y_1 + y_2 + y_4 + y_5 + \dots) + 2(y_3 + y_6 + y_9 + \dots)\right]}$$
Rule in Words: $\frac{3h}{8} \left[(\text{First} + \text{Last}) + 3(\text{Sum of non-multiples of 3}) + 2(\text{Sum of multiples of 3})\right]$.
• MANDATORY CONSTRAINT: The number of subintervals $n$ MUST BE A **MULTIPLE OF 3** ($n = 3, 6, 9, 12, \dots$).
• Geometric shape: Fits third-degree cubic polynomials through sets of 4 points.
• Truncation Error: $E_{S3/8} = -\frac{(b-a)h^4}{80} f^{(4)}(\xi)$.
• MANDATORY CONSTRAINT: The number of subintervals $n$ MUST BE A **MULTIPLE OF 3** ($n = 3, 6, 9, 12, \dots$).
• Geometric shape: Fits third-degree cubic polynomials through sets of 4 points.
• Truncation Error: $E_{S3/8} = -\frac{(b-a)h^4}{80} f^{(4)}(\xi)$.
4. Weddle's Rule (Put $n = 6$) — Extremely High CSJMU PYQ Frequency
Putting $n = 6$ in the General Quadrature formula and neglecting 7th and higher differences ($\Delta^7 y_0 \approx 0$):
$$\mathbf{\int_{x_0}^{x_0+6h} f(x)\,dx = \frac{3h}{10}\left[y_0 + 5y_1 + y_2 + 6y_3 + y_4 + 5y_5 + y_6\right]}$$
Mnemonic for Weddle's Coefficients: `1 - 5 - 1 - 6 - 1 - 5 - 1`.
• Constraint: Number of subintervals $n$ must be a **multiple of 6** ($n = 6, 12, 18, \dots$).
• Truncation Error: $E_W = -\frac{(b-a)h^6}{140} f^{(6)}(\xi)$. Degree of precision = 5! Offers extraordinary accuracy for 6 intervals.
• Constraint: Number of subintervals $n$ must be a **multiple of 6** ($n = 6, 12, 18, \dots$).
• Truncation Error: $E_W = -\frac{(b-a)h^6}{140} f^{(6)}(\xi)$. Degree of precision = 5! Offers extraordinary accuracy for 6 intervals.
5. Boole's Rule (Put $n = 4$) & Romberg's Integration
Boole's Rule ($n = 4$):
$$\mathbf{\int_{x_0}^{x_0+4h} f(x)\,dx = \frac{2h}{45}\left[7y_0 + 32y_1 + 12y_2 + 32y_3 + 7y_4\right]}$$ Coefficients pattern: `7 - 32 - 12 - 32 - 7`. Constraint: $n$ must be a multiple of 4.
Romberg's Integration (Richardson Extrapolation):
If $I_1$ is the Trapezoidal estimate with step $h$, and $I_2$ is with step $h/2$, the higher-order Romberg estimate is: $$\mathbf{I = I_2 + \frac{I_2 - I_1}{3} = \frac{4I_2 - I_1}{3}}$$
$$\mathbf{\int_{x_0}^{x_0+4h} f(x)\,dx = \frac{2h}{45}\left[7y_0 + 32y_1 + 12y_2 + 32y_3 + 7y_4\right]}$$ Coefficients pattern: `7 - 32 - 12 - 32 - 7`. Constraint: $n$ must be a multiple of 4.
Romberg's Integration (Richardson Extrapolation):
If $I_1$ is the Trapezoidal estimate with step $h$, and $I_2$ is with step $h/2$, the higher-order Romberg estimate is: $$\mathbf{I = I_2 + \frac{I_2 - I_1}{3} = \frac{4I_2 - I_1}{3}}$$
Master Solved University Numerical Problem
⭐ CSJMU PYQs:
2025–26, 2024–25, 2023–24, 2021 (The classic $\int_0^6 \frac{dx}{1+x^2}$ problem!)
Problem: Evaluate $I = \int_0^6 \frac{dx}{1 + x^2}$ by dividing the range $[0, 6]$ into $n = 6$ equal subintervals using:
(i) Trapezoidal Rule
(ii) Simpson's 1/3rd Rule
(iii) Simpson's 3/8th Rule
(iv) Weddle's Rule
Compare your answers with the exact analytical value.
Step 1: Compute Step Size $h$ and Table of Ordinates:
$$h = \frac{b - a}{n} = \frac{6 - 0}{6} = \mathbf{1}$$Compute $y = f(x) = \frac{1}{1 + x^2}$ for $x = 0, 1, 2, 3, 4, 5, 6$:
| Ordinate | $x$ | $y = f(x) = \frac{1}{1 + x^2}$ | Decimal Value (to 4 places) |
|---|---|---|---|
| $y_0$ (First) | $0$ | $\frac{1}{1 + 0} = 1.0000$ | $1.0000$ |
| $y_1$ (Odd) | $1$ | $\frac{1}{1 + 1} = 0.5000$ | $0.5000$ |
| $y_2$ (Even) | $2$ | $\frac{1}{1 + 4} = 0.2000$ | $0.2000$ |
| $y_3$ (Odd/Mult of 3) | $3$ | $\frac{1}{1 + 9} = 0.1000$ | $0.1000$ |
| $y_4$ (Even) | $4$ | $\frac{1}{1 + 16} = 0.0588$ | $0.0588$ |
| $y_5$ (Odd) | $5$ | $\frac{1}{1 + 25} = 0.0385$ | $0.0385$ |
| $y_6$ (Last) | $6$ | $\frac{1}{1 + 36} = 0.0270$ | $0.0270$ |
Part (i): By Trapezoidal Rule ($h = 1$)
$$I_{Trap} = \frac{h}{2}\left[(y_0 + y_6) + 2(y_1 + y_2 + y_3 + y_4 + y_5)\right]$$ $$I_{Trap} = \frac{1}{2}\left[(1.0000 + 0.0270) + 2(0.5000 + 0.2000 + 0.1000 + 0.0588 + 0.0385)\right]$$ $$I_{Trap} = \frac{1}{2}\left[1.0270 + 2(0.8973)\right] = \frac{1}{2}[1.0270 + 1.7946] = \frac{2.8216}{2} = \mathbf{1.4108}$$Part (ii): By Simpson's 1/3rd Rule ($n = 6$ is even!)
$$I_{S1/3} = \frac{h}{3}\left[(y_0 + y_6) + 4(y_1 + y_3 + y_5) + 2(y_2 + y_4)\right]$$ $$I_{S1/3} = \frac{1}{3}\left[(1.0000 + 0.0270) + 4(0.5000 + 0.1000 + 0.0385) + 2(0.2000 + 0.0588)\right]$$ $$I_{S1/3} = \frac{1}{3}\left[1.0270 + 4(0.6385) + 2(0.2588)\right] = \frac{1}{3}[1.0270 + 2.5540 + 0.5176] = \frac{4.0986}{3} = \mathbf{1.3662}$$Part (iii): By Simpson's 3/8th Rule ($n = 6$ is a multiple of 3!)
$$I_{S3/8} = \frac{3h}{8}\left[(y_0 + y_6) + 3(y_1 + y_2 + y_4 + y_5) + 2(y_3)\right]$$ $$I_{S3/8} = \frac{3(1)}{8}\left[(1.0000 + 0.0270) + 3(0.5000 + 0.2000 + 0.0588 + 0.0385) + 2(0.1000)\right]$$ $$I_{S3/8} = \frac{3}{8}\left[1.0270 + 3(0.7973) + 0.2000\right] = \frac{3}{8}[1.0270 + 2.3919 + 0.2000] = \frac{3(3.6189)}{8} = \mathbf{1.3571}$$Part (iv): By Weddle's Rule ($n = 6$)
$$I_{Weddle} = \frac{3h}{10}\left[y_0 + 5y_1 + y_2 + 6y_3 + y_4 + 5y_5 + y_6\right]$$ $$I_{Weddle} = \frac{3(1)}{10}\left[1.0000 + 5(0.5000) + 0.2000 + 6(0.1000) + 0.0588 + 5(0.0385) + 0.0270\right]$$ $$I_{Weddle} = \frac{3}{10}\left[1.0000 + 2.5000 + 0.2000 + 0.6000 + 0.0588 + 0.1925 + 0.0270\right] = \frac{3}{10}[4.5783] = \mathbf{1.3735}$$Exact Analytical Comparison:
$$\int_0^6 \frac{dx}{1 + x^2} = [\arctan(x)]_0^6 = \arctan(6) - \arctan(0) \approx \mathbf{1.4056}$$Weddle's rule and Simpson's rules give exceptional accuracy without needing continuous calculus!
Exam Answer Blueprint: 10-Mark Model Structure
How to Score Full 10 Marks on "Derive Trapezoidal & Simpson's 1/3":
- General Quadrature (4 Marks): Integrate Newton's forward difference series $I = \int_0^n [\dots] h\,du$ to derive the master formula $nh[y_0 + \frac{n}{2}\Delta y_0 + \dots]$.
- Deduce Trapezoidal (2 Marks): Substitute $n = 1$, neglect $\Delta^2$, show single strip $\frac{h}{2}(y_0+y_1)$, then write composite formula.
- Deduce Simpson's 1/3 (3 Marks): Substitute $n = 2$, keep up to $\Delta^2$, show double strip $\frac{h}{3}(y_0+4y_1+y_2)$, then write composite formula. Emphasize that $n$ must be EVEN!
- Error Comparison (1 Mark): State that Trapezoidal error is $O(h^2)$ while Simpson's 1/3 is $O(h^4)$.
⚠️ Common Exam Pitfalls:
• Violating Parity Constraints: Applying Simpson's 1/3rd rule when $n$ is ODD (e.g. $n = 5$) is mathematically invalid and receives 0 marks! Simpson's 1/3rd strictly requires $n$ to be EVEN.
• Multipliers in Simpson's 1/3: Remember: 4 multiplies ODD ordinates ($y_1, y_3, y_5$) and 2 multiplies EVEN ordinates ($y_2, y_4, y_6$). Don't invert them!
• Violating Parity Constraints: Applying Simpson's 1/3rd rule when $n$ is ODD (e.g. $n = 5$) is mathematically invalid and receives 0 marks! Simpson's 1/3rd strictly requires $n$ to be EVEN.
• Multipliers in Simpson's 1/3: Remember: 4 multiplies ODD ordinates ($y_1, y_3, y_5$) and 2 multiplies EVEN ordinates ($y_2, y_4, y_6$). Don't invert them!
Unit III Examination Mastery MCQ Quiz
Verify your conceptual grasp across numerical differentiation, general quadrature, Trapezoidal, Simpson 1/3, and Simpson 3/8 with this 10-question quiz.