Numerical Integration | NEB Class 11 Mathematics Notes

Chapter Overview

This chapter is a highly practical branch of calculus — bridging the gap between theoretical integration and real-world computation. While standard calculus relies on finding anti-derivatives (which is impossible for functions like $e^{-x^2}$ or $\frac{\sin x}{x}$), numerical integration allows us to approximate definite integrals with remarkable accuracy using discrete points. In the NEB board exams, numerical integration is a scoring area, appearing as 1-mark MCQs, 2-mark conceptual questions, and 4-to-5-mark long calculation tasks. Master these structured algorithms, and you are guaranteed full marks in this section.

What this chapter covers

Concept of Definite Integrals as AreaLimitations of Analytical IntegrationSub-intervals, Step Size, and OrdinatesMid-point Approximation RuleSimple and Composite Trapezoidal RulesSimpson's 1/3 Rule (Parabolic Approximation)Key Differences and Applicability ConditionsError Estimation and Absolute BoundsTabulating Function Values AccuratelyRadian Mode in Trigonometric Integrals

Learning Objectives

1

Explain the geometric principles behind the Mid-point, Trapezoidal, and Simpson's 1/3 rules

2

Calculate step size ($h$) correctly and construct error-free tables of ordinates

3

Apply the Composite Trapezoidal Rule for any integer $n \ge 1$

4

Apply Simpson's 1/3 Rule correctly, ensuring the sub-interval count ($n$) is strictly even

5

Compare numerical approximations with exact analytical values to compute absolute, relative, and percentage errors

6

Select the most appropriate numerical integration method based on given boundary conditions or tabular data

The Core Philosophy of Numerical Integration

Key Insight

Every numerical integration technique works by replacing a complex, non-integrable curve $y = f(x)$ with a simple, easily-integrable approximating function over discrete sub-intervals.

The Mid-point Rule uses horizontal line segments (rectangles).
The Trapezoidal Rule connects points with straight, sloping lines (first-degree polynomials) to form trapeziums.
Simpson's 1/3 Rule fits quadratic parabolas (second-degree polynomials) through groups of three successive points, capturing curves far better.

As a rule of thumb, Simpson's rule yields significantly higher accuracy than the Trapezoidal rule for the same number of sub-intervals.

In NEB exams: always keep your calculator in RADIAN mode when working with trigonometric functions like $\sin x$ or $\cos x$. Using degree mode is the most common reason students lose all marks on a question.

Key Definitions

Numerical Integration

A mathematical process of approximating the numerical value of a definite integral $\int_{a}^{b} f(x) dx$ when the analytical antiderivative of $f(x)$ is difficult or impossible to find.

Sub-interval (n)

The number of equal parts into which the integration interval $[a, b]$ is divided. For Simpson's 1/3 rule, $n$ must always be an even integer.

Step Size (h)

The uniform width of each sub-interval, representing the horizontal step along the x-axis.

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

Ordinates (y_i)

The calculated values of the function $f(x)$ at the boundary points of each sub-interval. For $n$ sub-intervals, there are always $n+1$ ordinates.

$$y_i = f(x_i) = f(a + i \cdot h)$$

Mid-point Rule

An approximation method where the height of each rectangle is taken as the value of the function at the midpoint of each sub-interval.

$$I \approx (b - a) \cdot f\left(\frac{a + b}{2}\right)$$

Composite Trapezoidal Rule

The integration method that estimates the area under $f(x)$ by summing the areas of $n$ distinct trapeziums formed under the curve.

$$T_n = \frac{h}{2} \left[ y_0 + y_n + 2(y_1 + y_2 + \dots + y_{n-1}) \right]$$

Simpson's 1/3 Rule

A highly accurate numerical method that uses parabolic arcs to approximate the boundary of the curve, requiring $n$ to be even.

$$S_n = \frac{h}{3} \left[ y_0 + y_n + 4(y_1 + y_3 + \dots + y_{n-1}) + 2(y_2 + y_4 + \dots + y_{n-2}) \right]$$

Absolute Error

The magnitude of the difference between the exact analytical value of a definite integral ($I_{exact}$) and its numerically approximated value ($I_{approx}$).

$$E_{abs} = |I_{exact} - I_{approx}|$$

Trapezoidal Rule vs Simpson's 1/3 Rule

The two main numerical integration techniques compared. Memorize their conditions and formulas.

Trapezoidal Rule

$$T_n = \frac{h}{2} [y_0 + y_n + 2(\sum y_{rem})]$$

Approximates the curve using first-degree linear polynomials (straight line segments)

Works for any number of sub-intervals $n \ge 1$ (both even and odd)

Geometrically constructs $n$ trapeziums under the curve

Less accurate than Simpson's rule for curved boundaries

Error is proportional to $h^2$ (second-order accuracy)

VS

Simpson's 1/3 Rule

$$S_n = \frac{h}{3} [y_0 + y_n + 4(\sum y_{odd}) + 2(\sum y_{even})]$$

Approximates the curve using second-degree quadratic parabolas

Strictly requires the number of sub-intervals $n$ to be even

Geometrically fits parabolas across successive groups of three points

Highly accurate, capturing curved pathways effectively

Error is proportional to $h^4$ (fourth-order accuracy)

Quick check: Is $n$ odd? You can only use the Trapezoidal Rule. Is $n$ even? Use Simpson's 1/3 Rule for superior accuracy.

Complete Formula Sheet with Derivations

Interval Width / Step Size

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

Always calculate this step first. Here, $a$ is the lower limit, $b$ is the upper limit, and $n$ is the number of intervals.

1We wish to partition the closed interval $[a, b]$ into $n$ equal pieces.

2The total length of the span is $b - a$.

3Dividing this total horizontal distance by $n$ equal segments gives the uniform step size: $$h = \frac{b-a}{n}$$

Mid-point Rule

\int_{a}^{b} f(x) dx \approx (b - a) f\left(\frac{a + b}{2}\right)

Estimates the entire integral using a single rectangle evaluated at the center.

1Approximate the function $f(x)$ over $[a, b]$ by a constant function $y = f(x_m)$, where $x_m = \frac{a+b}{2}$ is the midpoint.

2Integrating this constant function gives: $$\int_{a}^{b} f(x) dx \approx \int_{a}^{b} f(x_m) dx = f(x_m) \cdot [x]_{a}^{b} = (b - a)f(x_m)$$

Composite Trapezoidal Rule

T_n = \frac{h}{2} \left[ y_0 + y_n + 2\sum_{i=1}^{n-1} y_i \right]

End-point ordinates are taken once; all internal ordinates are doubled.

1Divide $[a, b]$ into $n$ sub-intervals: $[x_0, x_1], [x_1, x_2], \dots, [x_{n-1}, x_n]$.

2The area of the first trapezium over $[x_0, x_1]$ is $\frac{h}{2}(y_0 + y_1)$. The second is $\frac{h}{2}(y_1 + y_2)$, and so on.

3Summing these areas together: $$T_n = \frac{h}{2}(y_0 + y_1) + \frac{h}{2}(y_1 + y_2) + \dots + \frac{h}{2}(y_{n-1} + y_n)$$

4Factor out $\frac{h}{2}$ to get: $$T_n = \frac{h}{2} \left[ y_0 + y_n + 2(y_1 + y_2 + \dots + y_{n-1}) \right]$$

Simpson's 1/3 Rule

S_n = \frac{h}{3} \left[ y_0 + y_n + 4\sum_{\text{odd } i} y_i + 2\sum_{\text{even } i} y_i \right]

Remember: Ends are single, odds are multiplied by 4, evens (excluding ends) are multiplied by 2.

1Approximate $f(x)$ using a quadratic parabola $y = Ax^2 + Bx + C$ over each double sub-interval $[x_0, x_2]$.

2The area under this parabola from $-h$ to $h$ (shifting the origin to $x_1$ for algebraic ease) is: $$A_1 = \int_{-h}^{h} (Ax^2 + Bx + C) dx = \left[ \frac{Ax^3}{3} + \frac{Bx^2}{2} + Cx \right]_{-h}^{h} = \frac{2Ah^3}{3} + 2Ch$$

3Since the curve passes through $(-h, y_0)$, $(0, y_1)$, and $(h, y_2)$, we solve for $A$ and $C$ to get: $y_0 = Ah^2 - Bh + C$, $y_1 = C$, $y_2 = Ah^2 + Bh + C$. Therefore, $y_0 + 4y_1 + y_2 = 2Ah^2 + 6C$.

4Substituting this back into the area equation gives: $$A_1 = \frac{h}{3}(y_0 + 4y_1 + y_2)$$

5Summing this over all pairs of intervals yields: $$S_n = \frac{h}{3} \left[ y_0 + y_n + 4(y_1 + y_3 + \dots) + 2(y_2 + y_4 + \dots) \right]$$

Error Bound for Trapezoidal Rule

|E_T| \le \frac{M(b - a)^3}{12n^2}

M is the maximum value of the absolute second derivative |f''(x)| on [a, b].

1The local error of a single trapezium over interval size $h$ is of the order $-\frac{h^3}{12}f''(\xi)$.

2Summing this local error over $n$ intervals gives the global error: $$E_T = -\frac{n h^3}{12} f''(\eta) = -\frac{(b-a)h^2}{12} f''(\eta)$$ since $nh = b-a$.

3Substituting $h = \frac{b-a}{n}$ and taking absolute values, we establish the maximum error bound: $$|E_T| \le \frac{M(b - a)^3}{12n^2}$$ where $M = \max_{x \in [a,b]} |f''(x)|$.

Error Bound for Simpson's Rule

|E_S| \le \frac{M(b - a)^5}{180n^4}

M is the maximum value of the absolute fourth derivative |f^(4)(x)| on [a, b].

1The error of Simpson's rule over a double interval is $-\frac{h^5}{90}f^{(4)}(\xi)$.

2Over $n/2$ pairs of intervals, the accumulated global error is: $$E_S = -\frac{n}{2} \frac{h^5}{90} f^{(4)}(\eta) = -\frac{(b-a)h^4}{180} f^{(4)}(\eta)$$

3Substituting $h = \frac{b-a}{n}$ and taking absolute values, we get: $$|E_S| \le \frac{M(b - a)^5}{180n^4}$$ where $M = \max_{x \in [a,b]} |f^{(4)}(x)|$.

Which Integration Method Should I Choose?

Use this structured decision tree to select the correct integration rule for any board problem.

Is the function $f(x)$ non-integrable but you only have a single interval with no sub-intervals?

Yes

Use the basic Mid-point Rule: Area ≈ (b - a) * f((a+b)/2)

No

Continue ↓

Are you asked specifically to approximate with an ODD number of sub-intervals (e.g., n = 3, n = 5)?

Yes

Use the Composite Trapezoidal Rule (Simpson's 1/3 cannot be applied here).

No

Continue ↓

Is the number of sub-intervals (n) EVEN (e.g., n = 2, 4, 6) or are you free to choose?

Yes

Use Simpson's 1/3 Rule (it yields significantly lower error margins).

No

Check coordinates carefully. If intervals are unequal, standard composite formulas do not apply.

The Coordinate Table and Radian Warning

The foundation of any numerical integration problem is constructing a clean, accurate coordinate table mapping inputs $x_i$ to functional outputs $y_i$.

$$x_i = a + i \cdot h \quad \text{and} \quad y_i = f(x_i)$$

Why it works (Derivation):

1Constructing the table of coordinates is the first step of any calculation.

2For $n$ sub-intervals, you must list $n+1$ values of $x$ starting from $x_0 = a$ up to $x_n = b$.

3At each $x_i$, compute $y_i = f(x_i)$. Keep at least 4 decimal places during intermediate steps to minimize rounding errors.

4Critical Trigonometric Rule: When evaluating integrals like $\int_0^{\pi/2} \sin x \, dx$, the limits and steps are in radians. Your scientific calculator MUST be in Radian Mode (R), not Degree Mode (D).

Example: For $\int_0^{\pi/2} \sin x \, dx$ with $n=2$, step size $h = \frac{\pi/2 - 0}{2} = \frac{\pi}{4}$. Points are $x_0 = 0$, $x_1 = \frac{\pi}{4}$, $x_2 = \frac{\pi}{2}$. In radian mode: $y_0 = \sin(0) = 0$, $y_1 = \sin(\pi/4) = 0.7071$, $y_2 = \sin(\pi/2) = 1.0$.

When to use: Construct the table systematically before writing down or substituting values into any integration formula.

Memory Tricks — Never Forget These

Trick: Three quick memory hooks that cover the most common formula confusions in this chapter.

Steps to Remember:

The Trapezoidal Rhythm — 'Ends are Single, Middle is Double': In the Trapezoidal formula, the extreme values ($y_0$ and $y_n$) have a coefficient of 1, while all intermediate values are multiplied by 2.
The Simpson Rhythm — '1, 4, 2, 4, 2... 1': In Simpson's 1/3 rule, coefficients start at 1, alternate between 4 (for odd steps) and 2 (for even steps), and end with 1.
The H-Denominator Trick: The fraction in front of the bracket matches the rule name. For Trapezoidal (2 parallel sides), it is $\frac{h}{2}$. For Simpson's 1/3 rule, it is $\frac{h}{3}$.

Mnemonic: Ends are single, mid-doubled for Trap | Ends + 4(Odds) + 2(Evens) for Simpson | Radian mode for trigs

Common Mistakes

These mistakes cause the most mark losses in NEB exams for this chapter.

Degree Mode Pitfall: Evaluating trigonometric functions ($\sin x, \cos x$) with the calculator in Degree mode. This yields completely incorrect coordinate values.
Applying Simpson's with Odd n: Trying to use Simpson's 1/3 rule on a dataset or problem where $n$ is odd (like $n = 3$ or $n = 5$). Simpson's rule strictly requires an even number of intervals.
Ordinate vs Interval Count: Confusing the number of intervals ($n$) with the number of ordinates ($n+1$). If a question says 'use 5 ordinates', then $n = 4$.
Early Rounding Errors: Rounding intermediate values in your table to 1 or 2 decimal places. This propagates errors. Always maintain at least 4 decimal places.
Formula Mismatch: Swapping the multipliers for odd and even indices in Simpson's rule. Remember, odd steps ($y_1, y_3$) get a multiplier of 4, while even steps ($y_2, y_4$) get a multiplier of 2.

NEB Exam Traps — These Appear Every Year

NEB Exam Warning

Board examiners specifically test these edge cases. Knowing them can save 2–5 marks.

'Use n ordinates' trick: The question asks to integrate with '5 ordinates'. Many students set $n = 5$. In reality, $n = 5 - 1 = 4$ intervals. Always check the wording carefully.
Comparing with Exact Value: Many 5-mark long questions ask you to compare your numerical result with the exact analytical value. Do not skip integrating the function using standard calculus rules first.
Logarithmic Integrals: For $\int_1^2 \frac{dx}{x}$, the analytical value is $\ln(2)$. Remember that $\ln(2)$ is the natural logarithm (base $e$), evaluated using the 'ln' key on your calculator, not 'log' (base 10).
Constant or Linear Functions: If $f(x) = kx + c$, both Trapezoidal and Simpson's rules give the exact value. If $f(x) = Ax^2 + Bx + C$, Simpson's rule is exact, while Trapezoidal is an approximation.

Question Recognition Guide

Keyword / Phrase in Question	Use This
Question asks for 'Trapezoidal Rule' with any given $n$	Calculate $h = \frac{b-a}{n}$ and use $T_n = \frac{h}{2} [y_0 + y_n + 2\sum y_{\text{rem}}]$
Question asks for 'Simpson's Rule' with even $n$	Calculate $h = \frac{b-a}{n}$ and use $S_n = \frac{h}{3} [y_0 + y_n + 4\sum y_{\text{odds}} + 2\sum y_{\text{evens}}]$
Question specifies 'using 5 ordinates'	Set $n = 4$ intervals, construct 5 coordinate points, then apply the rule
Question says 'compare with exact value' or 'find absolute error'	Find the definite integral analytically first, then calculate $\|I_{\text{exact}} - I_{\text{approx}}\|$
Integrand contains $\sin(x)$, $\cos(x)$, or $e^x$ limits	Construct table of values with calculator set strictly to Radian Mode
Dataset table is given directly with uneven spacing	Calculate individual trapeziums manually; standard composite formulas cannot be used

Solved Examples (Easy → Hard)

1

Evaluate $\int_1^2 x^2 \, dx$ with $n = 2$ using the Trapezoidal Rule.

Easy

Solution

Step 1 — Identify parameters:
$a = 1$, $b = 2$, $n = 2$, $f(x) = x^2$.

Step 2 — Calculate step size:
$$h = \frac{b - a}{n} = \frac{2 - 1}{2} = 0.5$$

Step 3 — Construct the coordinate table:
• $x_0 = 1.0 \implies y_0 = f(1.0) = 1.0^2 = 1.0$
• $x_1 = 1.5 \implies y_1 = f(1.5) = 1.5^2 = 2.25$
• $x_2 = 2.0 \implies y_2 = f(2.0) = 2.0^2 = 4.0$

Step 4 — Apply the Trapezoidal formula:
$$T_2 = \frac{h}{2} [y_0 + y_2 + 2(y_1)]$$
$$T_2 = \frac{0.5}{2} [1.0 + 4.0 + 2(2.25)]$$
$$T_2 = 0.25 [5.0 + 4.5] = 0.25 [9.5] = 2.375$$

∴ Trapezoidal approximation = 2.375

2

Evaluate $\int_{0}^{1} \frac{dx}{1+x}$ with $n = 4$ using Simpson's 1/3 Rule.

Medium

Solution

Step 1 — Identify parameters:
$a = 0$, $b = 1$, $n = 4$ (even, so Simpson's rule is valid), $f(x) = \frac{1}{1+x}$.

Step 2 — Calculate step size:
$$h = \frac{1 - 0}{4} = 0.25$$

Step 3 — Construct the coordinate table:
• $x_0 = 0.00 \implies y_0 = \frac{1}{1+0} = 1.0000$
• $x_1 = 0.25 \implies y_1 = \frac{1}{1+0.25} = 0.8000$
• $x_2 = 0.50 \implies y_2 = \frac{1}{1+0.5} = 0.6667$
• $x_3 = 0.75 \implies y_3 = \frac{1}{1+0.75} = 0.5714$
• $x_4 = 1.00 \implies y_4 = \frac{1}{1+1} = 0.5000$

Step 4 — Apply Simpson's 1/3 Rule formula:
$$S_4 = \frac{h}{3} [y_0 + y_4 + 4(y_1 + y_3) + 2(y_2)]$$
$$S_4 = \frac{0.25}{3} [1.0000 + 0.5000 + 4(0.8000 + 0.5714) + 2(0.6667)]$$
$$S_4 = \frac{0.25}{3} [1.5000 + 4(1.3714) + 1.3334]$$
$$S_4 = \frac{0.25}{3} [1.5000 + 5.4856 + 1.3334] = \frac{0.25}{3} [8.3190] \approx 0.69325$$

∴ Simpson's approximation = 0.6933

3

Find the error bound for the approximation of $\int_{1}^{2} \frac{1}{x} \, dx$ using the Trapezoidal Rule with $n = 4$.

Hard

Solution

Step 1 — Find the derivatives of $f(x)$:
$f(x) = x^{-1}$
$f'(x) = -x^{-2}$
$f''(x) = 2x^{-3} = \frac{2}{x^3}$

Step 2 — Find the maximum value $M$ of $|f''(x)|$ on $[1, 2]$:
Since $|f''(x)| = \frac{2}{x^3}$ is a decreasing function on $[1, 2]$, its maximum value occurs at the lower limit $x = 1$.
$$M = |f''(1)| = \frac{2}{1^3} = 2$$

Step 3 — Apply the error bound formula:
$$|E_T| \le \frac{M(b-a)^3}{12n^2}$$
Substitute $M=2$, $a=1$, $b=2$, and $n=4$:
$$|E_T| \le \frac{2(2-1)^3}{12(4)^2} = \frac{2(1)}{12(16)} = \frac{2}{192} \approx 0.01042$$
Thus, the absolute error cannot exceed $0.01042$.

∴ Error Bound ≤ 0.0104

MCQ Practice

Mixed Level — MCQ

1

Numerical integration is primarily used to compute which mathematical quantity?

Correct!

Incorrect. Correct: The approximate value of a definite integral

Solution

Numerical integration computes approximate definite integrals when analytic antiderivatives are difficult or unavailable.

∴ The approximate value of a definite integral

2

Which of the following conditions must be met to apply Simpson's 1/3 Rule?

Correct!

Incorrect. Correct: The number of sub-intervals (n) must be even

Solution

Simpson's 1/3 Rule approximates the integrand with parabolas across successive pairs of intervals, which strictly requires an even value of $n$.

∴ The number of sub-intervals (n) must be even

3

If a definite integral is partitioned into 8 sub-intervals (n = 8), how many ordinates (y-values) must be calculated?

Correct!

Incorrect. Correct: 9

Solution

The coordinates range from $x_0$ to $x_n$, giving a total count of $n+1$ points. Here, $8 + 1 = 9$.

∴ 9

4

What is the uniform step size (h) when evaluating an integral from a = 1 to b = 4 with n = 6?

Correct!

Incorrect. Correct: 0.50

Solution

Use the step size formula: $h = \frac{b-a}{n} = \frac{4 - 1}{6} = \frac{3}{6} = 0.5$.

∴ 0.50

5

Simpson's 1/3 Rule approximates the boundary curve of an integrand using which type of polynomial?

Correct!

Incorrect. Correct: Parabolas (second-degree)

Solution

Simpson's 1/3 Rule fits a quadratic polynomial (parabola) $Ax^2 + Bx + C$ over every successive pair of intervals.

∴ Parabolas (second-degree)

6

In the Composite Trapezoidal Rule, what coefficient is applied to the intermediate ordinates (y₁, y₂, ..., yₙ₋₁)?

Correct!

Incorrect. Correct: 2

Solution

The Composite Trapezoidal Rule is structured as: $T_n = \frac{h}{2} [y_0 + y_n + 2(y_1 + y_2 + \dots + y_{n-1})]$.

∴ 2

7

For which of the following functions will the Trapezoidal Rule yield the exact value of the definite integral?

Correct!

Incorrect. Correct: f(x) = 3x + 5

Solution

The Trapezoidal rule approximates curves with straight line segments. For a linear function (first-degree polynomial), the approximation is exact.

∴ f(x) = 3x + 5

8

In Simpson's 1/3 Rule, what coefficient is applied to the odd-indexed intermediate ordinates (y₁, y₃, y₅)?

Correct!

Incorrect. Correct: 4

Solution

The coefficient scheme for Simpson's 1/3 Rule is: Ends + 4(Odds) + 2(Evens). Thus, odd ordinates get a coefficient of 4.

∴ 4

9

The global truncation error in the Composite Trapezoidal Rule is proportional to:

Correct!

Incorrect. Correct: h²

Solution

The global error bound formula is $|E_T| \le \frac{M(b-a)h^2}{12}$, demonstrating that error scales with $h^2$.

∴ h²

10

When solving trigonometric definite integrals numerically, why is Radian Mode mandatory on calculators?

Correct!

Incorrect. Correct: Degree mode values are conceptually incompatible with linear limits

Solution

Limits of integration are real numbers. Since $\pi$ radians equals $\pi \approx 3.1416$, calculating trigonometric functions like $\sin(x)$ requires the input angle to be treated in radians.

∴ Degree mode values are conceptually incompatible with linear limits

Short Questions (2 Marks Each)

1

Define numerical integration and state when it is used.

[2 marks]

Solution

Numerical integration is a technique used to compute the approximate value of a definite integral $\int_{a}^{b} f(x) \, dx$. It is used when:
1. The integrand $f(x)$ has no analytical antiderivative (e.g., $e^{-x^2}$).
2. The integrand is known only as a discrete set of data points (e.g., experimental measurements).

∴ Numerical integration approximates definite integrals when analytic antiderivatives are unavailable.

2

Write down the Composite Trapezoidal Rule formula for $n = 4$.

[2 marks]

Solution

For $n = 4$, the partition points are $x_0, x_1, x_2, x_3, x_4$. The corresponding ordinates are $y_0, y_1, y_2, y_3, y_4$.
The formula is written as:
$$T_4 = \frac{h}{2} [y_0 + y_4 + 2(y_1 + y_2 + y_3)]$$
where $h = \frac{b-a}{4}$.

∴ T₄ = (h/2) [y₀ + y₄ + 2(y₁ + y₂ + y₃)]

3

Explain why Simpson's 1/3 Rule cannot be applied when the number of sub-intervals is $n = 5$.

[2 marks]

Solution

Simpson's 1/3 Rule fits a quadratic parabola across groups of three consecutive points ($x_{2k}, x_{2k+1}, x_{2k+2}$), which requires pairs of sub-intervals.
Therefore, the total number of sub-intervals ($n$) must be an even integer. Since $n = 5$ is odd, the rule cannot be applied.

∴ Simpson's rule requires an even number of sub-intervals; n = 5 is odd.

4

Calculate the step size ($h$) and list the partition points ($x_i$) for $\int_{1}^{3} f(x) \, dx$ with $n = 4$.

[2 marks]

Solution

Here, $a = 1$, $b = 3$, $n = 4$.
Step size:
$$h = \frac{b - a}{n} = \frac{3 - 1}{4} = 0.5$$
The partition points are:
• $x_0 = a = 1.0$
• $x_1 = 1.0 + 0.5 = 1.5$
• $x_2 = 1.5 + 0.5 = 2.0$
• $x_3 = 2.0 + 0.5 = 2.5$
• $x_4 = b = 3.0$

∴ h = 0.5; points are {1.0, 1.5, 2.0, 2.5, 3.0}

5

Approximate $\int_{1}^{3} x^3 \, dx$ using the Mid-point Rule with a single interval ($n = 1$).

[2 marks]

Solution

Here, $a = 1$, $b = 3$, $f(x) = x^3$.
The midpoint of the interval is:
$$x_m = \frac{a + b}{2} = \frac{1 + 3}{2} = 2$$
Evaluate the function at the midpoint:
$$f(x_m) = f(2) = 2^3 = 8$$
Applying the Mid-point Rule:
$$I \approx (b - a) f(x_m) = (3 - 1) \cdot 8 = 2 \cdot 8 = 16$$

∴ Approximation = 16

6

What is the geometric interpretation of the Trapezoidal Rule?

[2 marks]

Solution

Geometrically, the Trapezoidal Rule approximates the area under the curve $y = f(x)$ by dividing it into a series of vertical strip trapeziums.
The curved boundary of the function over each sub-interval is replaced with a straight line segment connecting the points $(x_i, y_i)$ and $(x_{i+1}, y_{i+1})$.

∴ Approximates the area under a curve using a sum of straight-topped trapeziums.

7

Find the ordinates $y_0, y_1, y_2$ for $\int_{0}^{1} \frac{dx}{1+x^2}$ with $n = 2$.

[2 marks]

Solution

Here, $a = 0$, $b = 1$, $n = 2$.
Step size:
$$h = \frac{1 - 0}{2} = 0.5$$
The points are $x_0 = 0$, $x_1 = 0.5$, $x_2 = 1.0$.
The corresponding ordinates are:
• $y_0 = \frac{1}{1 + 0^2} = 1.0000$
• $y_1 = \frac{1}{1 + (0.5)^2} = \frac{1}{1.25} = 0.8000$
• $y_2 = \frac{1}{1 + 1^2} = 0.5000$

∴ y₀ = 1.0, y₁ = 0.8, y₂ = 0.5

8

State the global error bound formula for the Composite Trapezoidal Rule.

[2 marks]

Solution

The absolute error $|E_T|$ for the Composite Trapezoidal Rule is bounded by:
$$|E_T| \le \frac{M(b - a)^3}{12n^2}$$
where:
• $a$ and $b$ are the integration limits.
• $n$ is the number of sub-intervals.
• $M$ is the maximum value of $|f''(x)|$ on the interval $[a, b]$.

∴ |E_T| ≤ M(b-a)³ / (12n²)

9

If $f(x) = e^x$, find the exact analytical value of $\int_{0}^{1} e^x \, dx$.

[2 marks]

Solution

Using standard integration rules:
$$\int_{0}^{1} e^x \, dx = \left[ e^x \right]_{0}^{1} = e^1 - e^0 = e - 1$$
Using $e \approx 2.71828$:
$$\text{Exact Value} \approx 2.71828 - 1 = 1.71828$$

∴ Exact Value = e − 1 ≈ 1.7183

10

State the coefficients of the ordinates in Simpson's 1/3 Rule when $n = 4$.

[2 marks]

Solution

The formula for $n = 4$ is:
$$S_4 = \frac{h}{3} [y_0 + 4y_1 + 2y_2 + 4y_3 + y_4]$$
The coefficients applied to the ordinates $[y_0, y_1, y_2, y_3, y_4]$ inside the brackets are:
1, 4, 2, 4, 1.

∴ Coefficients are 1, 4, 2, 4, 1

Long Questions (4–5 Marks Each)

1

Evaluate $\int_{0}^{1} \frac{dx}{1+x^2}$ using the Trapezoidal Rule with $n = 4$. Compare your approximation with the exact analytical value to find the absolute error.

[4 marks]

Solution

Step 1 — Calculate step size:
$a = 0$, $b = 1$, $n = 4 \implies h = \frac{1-0}{4} = 0.25$.

Step 2 — Construct the coordinate table:
• $x_0 = 0.00 \implies y_0 = \frac{1}{1 + 0^2} = 1.0000$
• $x_1 = 0.25 \implies y_1 = \frac{1}{1 + 0.25^2} = 0.94118$
• $x_2 = 0.50 \implies y_2 = \frac{1}{1 + 0.50^2} = 0.80000$
• $x_3 = 0.75 \implies y_3 = \frac{1}{1 + 0.75^2} = 0.64000$
• $x_4 = 1.00 \implies y_4 = \frac{1}{1 + 1.00^2} = 0.50000$

Step 3 — Apply the Composite Trapezoidal Rule:
$$T_4 = \frac{h}{2} [y_0 + y_4 + 2(y_1 + y_2 + y_3)]$$
$$T_4 = \frac{0.25}{2} [1.0000 + 0.5000 + 2(0.94118 + 0.80000 + 0.64000)]$$
$$T_4 = 0.125 [1.5000 + 2(2.38118)] = 0.125 [1.5000 + 4.76236]$$
$$T_4 = 0.125 [6.26236] = 0.782795 \approx 0.7828$$

Step 4 — Calculate the exact value:
$$I_{\text{exact}} = \int_{0}^{1} \frac{dx}{1+x^2} = [\tan^{-1}x]_0^1 = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} \approx 0.785398$$

Step 5 — Find the absolute error:
$$\text{Absolute Error} = |I_{\text{exact}} - T_4| = |0.785398 - 0.782795| = 0.002603 \approx 0.0026$$

∴ Approximation = 0.7828; Absolute Error = 0.0026

2

Approximate the value of $\int_{1}^{3} \frac{1}{x} \, dx$ using Simpson's 1/3 Rule with $n = 4$ sub-intervals. Verify your result using standard integration.

[4 marks]

Solution

Step 1 — Calculate step size:
$a = 1$, $b = 3$, $n = 4$ (even, valid) $\implies h = \frac{3-1}{4} = 0.5$.

Step 2 — Construct the coordinate table:
• $x_0 = 1.0 \implies y_0 = \frac{1}{1.0} = 1.0000$
• $x_1 = 1.5 \implies y_1 = \frac{1}{1.5} \approx 0.6667$
• $x_2 = 2.0 \implies y_2 = \frac{1}{2.0} = 0.5000$
• $x_3 = 2.5 \implies y_3 = \frac{1}{2.5} = 0.4000$
• $x_4 = 3.0 \implies y_4 = \frac{1}{3.0} \approx 0.3333$

Step 3 — Apply Simpson's 1/3 Rule formula:
$$S_4 = \frac{h}{3} [y_0 + y_4 + 4(y_1 + y_3) + 2(y_2)]$$
$$S_4 = \frac{0.5}{3} [1.0000 + 0.3333 + 4(0.6667 + 0.4000) + 2(0.5000)]$$
$$S_4 = \frac{0.5}{3} [1.3333 + 4(1.0667) + 1.0000] = \frac{0.5}{3} [1.3333 + 4.2668 + 1.0000]$$
$$S_4 = \frac{0.5}{3} [6.6001] \approx 1.1000$$

Step 4 — Calculate exact analytical value:
$$I_{\text{exact}} = \int_{1}^{3} \frac{dx}{x} = [\ln x]_1^3 = \ln(3) - \ln(1) = \ln(3) \approx 1.09861$$

Step 5 — Compare values:
$$\text{Absolute Error} = |1.09861 - 1.1000| = 0.00139$$

∴ Simpson's approximation = 1.1000; Exact value = 1.0986

3

Evaluate $\int_{0}^{\pi} \sin x \, dx$ using Simpson's 1/3 Rule with $n = 6$. Keep calculations in Radian Mode.

[5 marks]

Solution

Step 1 — Calculate step size:
$a = 0$, $b = \pi$, $n = 6 \implies h = \frac{\pi - 0}{6} = \frac{\pi}{6}$.

Step 2 — Construct coordinate table (using Radian Mode):
• $x_0 = 0 \implies y_0 = \sin(0) = 0$
• $x_1 = \frac{\pi}{6} \implies y_1 = \sin(\pi/6) = 0.5000$
• $x_2 = \frac{2\pi}{6} = \frac{\pi}{3} \implies y_2 = \sin(\pi/3) \approx 0.8660$
• $x_3 = \frac{3\pi}{6} = \frac{\pi}{2} \implies y_3 = \sin(\pi/2) = 1.0000$
• $x_4 = \frac{4\pi}{6} = \frac{2\pi}{3} \implies y_4 = \sin(2\pi/3) \approx 0.8660$
• $x_5 = \frac{5\pi}{6} \implies y_5 = \sin(5\pi/6) = 0.5000$
• $x_6 = \pi \implies y_6 = \sin(\pi) = 0$

Step 3 — Apply Simpson's 1/3 Rule formula:
$$S_6 = \frac{h}{3} [y_0 + y_6 + 4(y_1 + y_3 + y_5) + 2(y_2 + y_4)]$$
$$S_6 = \frac{\pi/6}{3} [0 + 0 + 4(0.5000 + 1.0000 + 0.5000) + 2(0.8660 + 0.8660)]$$
$$S_6 = \frac{\pi}{18} [4(2.0000) + 2(1.7320)] = \frac{\pi}{18} [8.0000 + 3.4640]$$
$$S_6 = \frac{\pi}{18} [11.4640] \approx 2.0008$$

Step 4 — Verification:
The exact value of $\int_0^\pi \sin x \, dx$ is $[-\cos x]_0^\pi = -\cos(\pi) - (-\cos 0) = -(-1) - (-1) = 2.0$.
Simpson's approximation (2.0008) matches the exact value closely.

∴ Simpson's approximation = 2.0008 (Exact = 2.0000)

4

Use the Trapezoidal Rule to approximate the value of $\int_{1}^{2} e^{-x^2} \, dx$ using $n = 4$ sub-intervals.

[4 marks]

Solution

Step 1 — Calculate step size:
$a = 1$, $b = 2$, $n = 4 \implies h = \frac{2-1}{4} = 0.25$.

Step 2 — Construct the coordinate table:
• $x_0 = 1.00 \implies y_0 = e^{-1.0^2} = e^{-1} \approx 0.36788$
• $x_1 = 1.25 \implies y_1 = e^{-1.25^2} = e^{-1.5625} \approx 0.20961$
• $x_2 = 1.50 \implies y_2 = e^{-1.5^2} = e^{-2.25} \approx 0.10540$
• $x_3 = 1.75 \implies y_3 = e^{-1.75^2} = e^{-3.0625} \approx 0.04677$
• $x_4 = 2.00 \implies y_4 = e^{-2.0^2} = e^{-4} \approx 0.01832$

Step 3 — Apply the Composite Trapezoidal formula:
$$T_4 = \frac{h}{2} [y_0 + y_4 + 2(y_1 + y_2 + y_3)]$$
$$T_4 = \frac{0.25}{2} [0.36788 + 0.01832 + 2(0.20961 + 0.10540 + 0.04677)]$$
$$T_4 = 0.125 [0.38620 + 2(0.36178)] = 0.125 [0.38620 + 0.72356]$$
$$T_4 = 0.125 [1.10976] \approx 0.13872$$

∴ Trapezoidal approximation = 0.1387

5

Evaluate $\int_{0}^{2} \sqrt{1+x^3} \, dx$ using Simpson's 1/3 Rule with $n = 4$.

[4 marks]

Solution

Step 1 — Calculate step size:
$a = 0$, $b = 2$, $n = 4 \implies h = \frac{2-0}{4} = 0.5$.

Step 2 — Construct the coordinate table:
• $x_0 = 0.0 \implies y_0 = \sqrt{1 + 0} = 1.0000$
• $x_1 = 0.5 \implies y_1 = \sqrt{1 + 0.5^3} = \sqrt{1.125} \approx 1.0607$
• $x_2 = 1.0 \implies y_2 = \sqrt{1 + 1^3} = \sqrt{2} \approx 1.4142$
• $x_3 = 1.5 \implies y_3 = \sqrt{1 + 1.5^3} = \sqrt{4.375} \approx 2.0917$
• $x_4 = 2.0 \implies y_4 = \sqrt{1 + 2^3} = \sqrt{9} = 3.0000$

Step 3 — Apply Simpson's 1/3 Rule formula:
$$S_4 = \frac{h}{3} [y_0 + y_4 + 4(y_1 + y_3) + 2(y_2)]$$
$$S_4 = \frac{0.5}{3} [1.0000 + 3.0000 + 4(1.0607 + 2.0917) + 2(1.4142)]$$
$$S_4 = \frac{0.5}{3} [4.0000 + 4(3.1524) + 2.8284] = \frac{0.5}{3} [4.0000 + 12.6096 + 2.8284]$$
$$S_4 = \frac{0.5}{3} [19.4380] \approx 3.23967$$

∴ Simpson's approximation = 3.2397

6

Approximate the value of $\int_2^4 (x^3 - 2x) \, dx$ using:
(i) Trapezoidal Rule with $n = 4$
(ii) Simpson's 1/3 Rule with $n = 4$
Compare both with the exact analytical value.

[5 marks]

Solution

Step 1 — Calculate step size:
$a = 2$, $b = 4$, $n = 4 \implies h = \frac{4-2}{4} = 0.5$.

Step 2 — Construct the coordinate table:
• $x_0 = 2.0 \implies y_0 = 2^3 - 2(2) = 8 - 4 = 4.0$
• $x_1 = 2.5 \implies y_1 = 2.5^3 - 2(2.5) = 15.625 - 5 = 10.625$
• $x_2 = 3.0 \implies y_2 = 3^3 - 2(3) = 27 - 6 = 21.0$
• $x_3 = 3.5 \implies y_3 = 3.5^3 - 2(3.5) = 42.875 - 7 = 35.875$
• $x_4 = 4.0 \implies y_4 = 4^3 - 2(4) = 64 - 8 = 56.0$

Step 3 — (i) Apply the Trapezoidal Rule:
$$T_4 = \frac{h}{2} [y_0 + y_4 + 2(y_1 + y_2 + y_3)]$$
$$T_4 = \frac{0.5}{2} [4.0 + 56.0 + 2(10.625 + 21.0 + 35.875)]$$
$$T_4 = 0.25 [60.0 + 2(67.5)] = 0.25 [60.0 + 135.0] = 48.75$$

Step 4 — (ii) Apply Simpson's 1/3 Rule:
$$S_4 = \frac{h}{3} [y_0 + y_4 + 4(y_1 + y_3) + 2(y_2)]$$
$$S_4 = \frac{0.5}{3} [4.0 + 56.0 + 4(10.625 + 35.875) + 2(21.0)]$$
$$S_4 = \frac{0.5}{3} [60.0 + 4(46.5) + 42.0] = \frac{0.5}{3} [60.0 + 186.0 + 42.0] = \frac{0.5}{3} [288.0] = 48.00$$

Step 5 — Find exact value:
$$\int_2^4 (x^3-2x)\,dx = \left[ \frac{x^4}{4} - x^2 \right]_2^4 = \left(\frac{256}{4} - 16\right) - \left(\frac{16}{4} - 4\right) = (64-16) - (4-4) = 48.00$$

Conclusion: Simpson's 1/3 rule gives the exact answer ($48.00$) because the integrand is a cubic function, for which Simpson's rule is theoretically exact. The Trapezoidal rule yields a slight overestimation ($48.75$).

∴ Trapezoidal = 48.75, Simpson's = 48.00, Exact = 48.00

7

A river is 80 meters wide. The depth $d$ (in meters) at distance $x$ (in meters) from one bank is given by the following table:

x	0	10	20	30	40	50	60	70	80
d	0	4	7	9	12	15	14	8	3

Find the approximate cross-sectional area of the river using Simpson's 1/3 Rule.

[5 marks]

Solution

Step 1 — Identify the parameters from the dataset:
The width intervals are uniform: $h = 10$ meters.
The boundary coordinates represent indices $0$ to $8$ ($n = 8$ intervals, which is even and thus valid for Simpson's rule).
The ordinates are:
$y_0 = 0$, $y_1 = 4$, $y_2 = 7$, $y_3 = 9$, $y_4 = 12$, $y_5 = 15$, $y_6 = 14$, $y_7 = 8$, $y_8 = 3$.

Step 2 — Group the terms for Simpson's Rule:
• Extreme points: $y_0 + y_8 = 0 + 3 = 3$
• Odd-indexed points: $y_1 + y_3 + y_5 + y_7 = 4 + 9 + 15 + 8 = 36$
• Even-indexed points (excluding extremes): $y_2 + y_4 + y_6 = 7 + 12 + 14 = 33$

Step 3 — Apply Simpson's 1/3 Rule:
$$\text{Area} = \frac{h}{3} [y_0 + y_8 + 4(\sum y_{\text{odds}}) + 2(\sum y_{\text{evens}})]$$
$$\text{Area} = \frac{10}{3} [3 + 4(36) + 2(33)]$$
$$\text{Area} = \frac{10}{3} [3 + 144 + 66] = \frac{10}{3} [213]$$
$$\text{Area} = 10 \cdot 71 = 710 \text{ square meters}$$

∴ Cross-sectional Area = 710 m²

8

Evaluate $\dots \int_1^5 \ln x \, dx$ using the Composite Trapezoidal Rule with $n = 4$. Compare with the exact value to find the percentage error.

[5 marks]

Solution

Step 1 — Calculate step size:
$a = 1$, $b = 5$, $n = 4 \implies h = \frac{5-1}{4} = 1.0$.

Step 2 — Construct coordinate table:
• $x_0 = 1.0 \implies y_0 = \ln(1) = 0.0000$
• $x_1 = 2.0 \implies y_1 = \ln(2) \approx 0.6931$
• $x_2 = 3.0 \implies y_2 = \ln(3) \approx 1.0986$
• $x_3 = 4.0 \implies y_3 = \ln(4) \approx 1.3863$
• $x_4 = 5.0 \implies y_4 = \ln(5) \approx 1.6094$

Step 3 — Apply the Trapezoidal formula:
$$T_4 = \frac{h}{2} [y_0 + y_4 + 2(y_1 + y_2 + y_3)]$$
$$T_4 = \frac{1.0}{2} [0.0000 + 1.6094 + 2(0.6931 + 1.0986 + 1.3863)]$$
$$T_4 = 0.5 [1.6094 + 2(3.1780)] = 0.5 [1.6094 + 6.3560] = 0.5 [7.9654] \approx 3.9827$$

Step 4 — Calculate exact analytical value:
Use integration by parts: $\int \ln x \, dx = x \ln x - x$.
$$I_{\text{exact}} = [x \ln x - x]_1^5 = (5 \ln 5 - 5) - (1 \ln 1 - 1) = 5(1.6094) - 5 + 1 = 8.0472 - 4 = 4.0472$$

Step 5 — Find percentage error:
$$\text{Absolute Error} = |4.0472 - 3.9827| = 0.0645$$
$$\text{Percentage Error} = \left( \frac{0.0645}{4.0472} \right) \times 100\% \approx 1.59\%$$

∴ Approximation = 3.9827; Percentage Error = 1.59%

9

For the integral $\int_1^2 \frac{dx}{x}$, find the maximum absolute error in its approximation using the Trapezoidal Rule with $n = 4$. Compare it with the actual error.

[5 marks]

Solution

Step 1 — Calculate actual error:
From Example 2 in this section, we know $T_4$ approximation of $\int_1^2 \frac{dx}{x}$ is:
$h = 0.25$. Points: $1.0, 1.25, 1.5, 1.75, 2.0$.
Ordinates: $1.0000, 0.8000, 0.6667, 0.5714, 0.5000$.
$$T_4 = \frac{0.25}{2} [1.0 + 0.5 + 2(0.8 + 0.6667 + 0.5714)] = 0.125 [1.5 + 4.0762] \approx 0.6970$$
Exact value: $I_{\text{exact}} = \ln(2) \approx 0.6931$.
$$\text{Actual Error} = |0.6931 - 0.6970| = 0.0039$$

Step 2 — Find the theoretical error bound:
For $f(x) = \frac{1}{x}$, the second derivative is $f''(x) = \frac{2}{x^3}$.
The maximum value of $|f''(x)|$ on $[1, 2]$ is at $x=1$, so $M = 2$.
$$|E_T| \le \frac{M(b-a)^3}{12n^2} = \frac{2(1)^3}{12(4^2)} = \frac{2}{192} \approx 0.0104$$

Step 3 — Compare:
The actual error ($0.0039$) is less than the maximum absolute error bound ($0.0104$), validating the formula.

∴ Actual Error = 0.0039; Error Bound = 0.0104

10

Evaluate $\int_0^{1.2} e^x \, dx$ using Simpson's 1/3 Rule with $n = 6$. Compare your result with the exact analytical value to demonstrate accuracy.

[5 marks]

Solution

Step 1 — Calculate step size:
$a = 0$, $b = 1.2$, $n = 6 \implies h = \frac{1.2 - 0}{6} = 0.2$.

Step 2 — Construct the coordinate table:
• $x_0 = 0.0 \implies y_0 = e^{0.0} = 1.0000$
• $x_1 = 0.2 \implies y_1 = e^{0.2} \approx 1.2214$
• $x_2 = 0.4 \implies y_2 = e^{0.4} \approx 1.4918$
• $x_3 = 0.6 \implies y_3 = e^{0.6} \approx 1.8221$
• $x_4 = 0.8 \implies y_4 = e^{0.8} \approx 2.2255$
• $x_5 = 1.0 \implies y_5 = e^{1.0} \approx 2.7183$
• $x_6 = 1.2 \implies y_6 = e^{1.2} \approx 3.3201$

Step 3 — Apply Simpson's 1/3 Rule:
$$S_6 = \frac{h}{3} [y_0 + y_6 + 4(y_1 + y_3 + y_5) + 2(y_2 + y_4)]$$
$$S_6 = \frac{0.2}{3} [1.0000 + 3.3201 + 4(1.2214 + 1.8221 + 2.7183) + 2(1.4918 + 2.2255)]$$
$$S_6 = \frac{0.2}{3} [4.3201 + 4(5.7618) + 2(3.7173)] = \frac{0.2}{3} [4.3201 + 23.0472 + 7.4346]$$
$$S_6 = \frac{0.2}{3} [34.8019] \approx 2.320127$$

Step 4 — Calculate exact analytical value:
$$I_{\text{exact}} = \int_0^{1.2} e^x \, dx = [e^x]_0^{1.2} = e^{1.2} - e^0 = 3.320117 - 1.0 = 2.320117$$

Step 5 — Find error:
$$\text{Absolute Error} = |2.320117 - 2.320127| = 0.000010$$
This demonstrates the high accuracy of Simpson's rule.

∴ Simpson's approximation = 2.32013; Exact = 2.32012

Past NEB Board Questions

1

Evaluate $\int_1^2 \frac{dx}{x}$ by using the Trapezoidal Rule for $n = 4$.

NEB 2079 [4 marks]

Solution

Step 1 — Identify parameters:
$a = 1$, $b = 2$, $n = 4 \implies h = \frac{2-1}{4} = 0.25$.

Step 2 — Coordinate Table:
$x_0 = 1.0 \implies y_0 = 1.0$
$x_1 = 1.25 \implies y_1 = 0.8$
$x_2 = 1.5 \implies y_2 = 0.6667$
$x_3 = 1.75 \implies y_3 = 0.5714$
$x_4 = 2.0 \implies y_4 = 0.5$

Step 3 — Formula and Calculation:
$$T_4 = \frac{h}{2} [y_0 + y_4 + 2(y_1 + y_2 + y_3)]$$
$$T_4 = \frac{0.25}{2} [1.0 + 0.5 + 2(0.8 + 0.6667 + 0.5714)]$$
$$T_4 = 0.125 [1.5 + 2(2.0381)] = 0.125 [1.5 + 4.0762] \approx 0.6970$$

∴ Trapezoidal approximation = 0.6970

2

Use Simpson's 1/3 Rule to evaluate $\int_{0}^{1} \frac{dx}{1+x}$ with $n = 4$.

NEB 2078 [4 marks]

Solution

Step 1 — Step size:
$h = \frac{1-0}{4} = 0.25$.

Step 2 — Coordinates:
• $y_0 = f(0) = 1.0$
• $y_1 = f(0.25) = 0.8$
• $y_2 = f(0.5) = 0.6667$
• $y_3 = f(0.75) = 0.5714$
• $y_4 = f(1.0) = 0.5$

Step 3 — Formula:
$$S_4 = \frac{h}{3} [y_0 + y_4 + 4(y_1 + y_3) + 2(y_2)]$$
$$S_4 = \frac{0.25}{3} [1.0 + 0.5 + 4(0.8 + 0.5714) + 2(0.6667)] = 0.6933$$

∴ Simpson's approximation = 0.6933

3

Estimate the area under the curve $y = \sqrt{x}$ from $x = 1$ to $x = 4$ using the Trapezoidal Rule with $n = 3$.

NEB 2077 [4 marks]

Solution

Step 1 — Step size:
$a = 1$, $b = 4$, $n = 3 \implies h = \frac{4-1}{3} = 1.0$.

Step 2 — Coordinate Table:
• $x_0 = 1 \implies y_0 = \sqrt{1} = 1.0000$
• $x_1 = 2 \implies y_1 = \sqrt{2} \approx 1.4142$
• $x_2 = 3 \implies y_2 = \sqrt{3} \approx 1.7321$
• $x_3 = 4 \implies y_3 = \sqrt{4} = 2.0000$

Step 3 — Formula and Calculation:
$$T_3 = \frac{h}{2} [y_0 + y_3 + 2(y_1 + y_2)]$$
$$T_3 = \frac{1.0}{2} [1.0000 + 2.0000 + 2(1.4142 + 1.7321)]$$
$$T_3 = 0.5 [3.0000 + 2(3.1463)] = 0.5 [3.0000 + 6.2926] = 4.6463$$

∴ Estimated Area = 4.6463

4

Calculate the approximate value of $\int_0^{\pi/2} \cos x \, dx$ with $n = 4$ using Simpson's 1/3 Rule.

NEB 2076 [4 marks]

Solution

Step 1 — Step size (Radian Mode):
$a = 0$, $b = \pi/2$, $n = 4 \implies h = \frac{\pi/2 - 0}{4} = \frac{\pi}{8}$.

Step 2 — Coordinate Table:
• $x_0 = 0 \implies y_0 = \cos(0) = 1.0000$
• $x_1 = \pi/8 \implies y_1 = \cos(\pi/8) \approx 0.9239$
• $x_2 = \pi/4 \implies y_2 = \cos(\pi/4) \approx 0.7071$
• $x_3 = 3\pi/8 \implies y_3 = \cos(3\pi/8) \approx 0.3827$
• $x_4 = \pi/2 \implies y_4 = \cos(\pi/2) = 0.0000$

Step 3 — Simpson's Formula:
$$S_4 = \frac{h}{3} [y_0 + y_4 + 4(y_1 + y_3) + 2(y_2)]$$
$$S_4 = \frac{\pi/8}{3} [1.0000 + 0 + 4(0.9239 + 0.3827) + 2(0.7071)]$$
$$S_4 = \frac{\pi}{24} [1.0000 + 4(1.3066) + 1.4142] = \frac{\pi}{24} [7.6406] \approx 1.0003$$

∴ Simpson's approximation = 1.0003

Chapter Test — Full Mixed Paper

Attempt this test in 45 minutes without consulting your notes. This matches standard NEB grading guidelines.

1 Section A — MCQs (1 Mark Each)

Mixed Level — MCQ

1

Which of the following numerical methods uses straight lines to approximate boundaries?

Correct!

Incorrect. Correct: Trapezoidal Rule

Solution

The Trapezoidal rule approximates curves by linear interpolation over each sub-interval.

∴ Trapezoidal Rule

2

If $n = 6$ for an integration interval $[2, 5]$, the step size $h$ is:

Correct!

Incorrect. Correct: 0.5

Solution

Use the formula $h = \frac{b-a}{n} = \frac{5-2}{6} = 0.5$.

∴ 0.5

3

Why is $n = 3$ invalid for Simpson's 1/3 Rule?

Correct!

Incorrect. Correct: Because Simpson's rule requires $n$ to be even

Solution

Simpson's 1/3 rule requires an even number of sub-intervals ($n = 2, 4, 6, \dots$).

∴ Because Simpson's rule requires $n$ to be even

2 Section B — Short Questions (2 Marks Each)

1

Define step size and state how it changes with $n$.

[2 marks]

Solution

Step size ($h$) is the horizontal width of each sub-interval: $h = \frac{b-a}{n}$. As the number of sub-intervals ($n$) increases, the step size ($h$) decreases, leading to more accurate approximations.

∴ h = (b-a)/n; it decreases as n increases

2

Write the Composite Trapezoidal formula for $n = 3$.

[2 marks]

Solution

For $n = 3$, the partition points are $x_0, x_1, x_2, x_3$. The formula is:
$$T_3 = \frac{h}{2} [y_0 + y_3 + 2(y_1 + y_2)]$$
where $h = \frac{b-a}{3}$.

∴ T₃ = (h/2) [y₀ + y₃ + 2(y₁ + y₂)]

3 Section C — Long Questions (4–5 Marks Each)

1

Evaluate $\int_1^2 \frac{dx}{x^2}$ using Simpson's 1/3 Rule with $n = 4$. Compare it with the exact analytical value.

[5 marks]

Solution

Step 1 — Calculate step size:
$h = \frac{2-1}{4} = 0.25$.

Step 2 — Coordinate Table:
• $x_0 = 1.00 \implies y_0 = 1/1^2 = 1.0000$
• $x_1 = 1.25 \implies y_1 = 1/1.25^2 = 0.6400$
• $x_2 = 1.50 \implies y_2 = 1/1.5^2 \approx 0.4444$
• $x_3 = 1.75 \implies y_3 = 1/1.75^2 \approx 0.3265$
• $x_4 = 2.00 \implies y_4 = 1/2^2 = 0.2500$

Step 3 — Apply Simpson's Rule:
$$S_4 = \frac{0.25}{3} [1.0000 + 0.2500 + 4(0.6400 + 0.3265) + 2(0.4444)]$$
$$S_4 = \frac{0.25}{3} [1.2500 + 3.8660 + 0.8888] = \frac{0.25}{3} [6.0048] \approx 0.5004$$

Step 4 — Find exact value:
$$\int_1^2 x^{-2} \, dx = [-x^{-1}]_1^2 = -\frac{1}{2} - (-1) = 0.5000$$

Step 5 — Compare:
The error is $|0.5004 - 0.5000| = 0.0004$, showing high accuracy.

∴ Simpson's = 0.5004; Exact = 0.5000

2

Evaluate $\int_0^1 e^x \, dx$ using the Trapezoidal Rule with $n = 4$.

[4 marks]

Solution

Step 1 — Step size:
$h = \frac{1-0}{4} = 0.25$.

Step 2 — Coordinate Table:
• $y_0 = e^{0} = 1.0000$
• $y_1 = e^{0.25} \approx 1.2840$
• $y_2 = e^{0.50} \approx 1.6487$
• $y_3 = e^{0.75} \approx 2.1170$
• $y_4 = e^{1.00} \approx 2.7183$

Step 3 — Apply Trapezoidal Rule:
$$T_4 = \frac{0.25}{2} [1.0000 + 2.7183 + 2(1.2840 + 1.6487 + 2.1170)]$$
$$T_4 = 0.125 [3.7183 + 2(5.0497)] = 0.125 [13.8177] \approx 1.7272$$

∴ Trapezoidal approximation = 1.7272

One-Page Revision Cheat Sheet

1

Step size (h): $h = \frac{b-a}{n}$ — always calculate this first.

2

Table Setup: Write down all $n+1$ points $x_i$ and evaluate $y_i = f(x_i)$.

3

Mid-point Rule: Area $\approx (b-a) f\left(\frac{a+b}{2}\right)$ — uses a single central rectangle.

4

Trapezoidal Rule: $T_n = \frac{h}{2} [y_0 + y_n + 2(y_1 + y_2 + \dots + y_{n-1})]$. Connects points with straight lines.

5

Simpson's 1/3 Rule: $S_n = \frac{h}{3} [y_0 + y_n + 4(\sum y_{\text{odds}}) + 2(\sum y_{\text{evens}})]$. Connects points with parabolic arcs.

6

Simpson's Constraint: The interval count $n$ must be even.

7

Error bound (Trap): $|E_T| \le \frac{M(b-a)^3}{12n^2}$ where $M = \max|f''(x)|$.

8

Error bound (Simpson): $|E_S| \le \frac{M(b-a)^5}{180n^4}$ where $M = \max|f^{(4)}(x)|$.

9

Trig Warning: Set your calculator strictly to Radian Mode for trigonometric limits.

10

Exactness: Trapezoidal is exact for linear functions. Simpson's is exact for polynomials up to degree 3 (cubics).

11

Precision: Keep intermediate table values rounded to at least 4 decimal places to prevent rounding error accumulation.