Simpson’s 1/3 rule with h = 0.5 applied to ∫₀¹ x² dx gives the exact answer 1/3. The reason is that:

Simpson’s rule is exact for polynomials up to degree three

The step size happens to be small enough

Simpson’s rule is exact for polynomials up to degree three

The integrand is positive on the interval

The number of subintervals is even

The classical fourth-order Runge–Kutta method requires, per step:

One function evaluation and one derivative evaluation

A multi-step method such as Adams–Bashforth cannot be started from the initial condition alone because:

It needs several previous points, which must be supplied by a single-step method

It is unstable for small step sizes

It requires the derivative of f

It only works for linear differential equations

Which of the following statements about Simpson’s 1/3 rule are correct? (More than one option may be correct.)

It requires the number of subintervals to be even

It integrates any cubic polynomial exactly

It requires the number of subintervals to be even

It integrates any cubic polynomial exactly

It requires the derivative of the integrand

Numerical Methods — Roots, Quadrature and Stepping an ODE

Numerical methods is the most predictable topic in Engineering Mathematics, because the questions come in only two shapes. Either you are asked to run one or two iterations by hand, in which case the marks are pure arithmetic care, or you are asked for a property of the method — its order of convergence, the order of its error, how many function evaluations a step costs, whether the number of subintervals must be even. The second kind is where the cheap marks are, and they are cheap because the answers do not depend on the problem at all: Newton–Raphson converges quadratically, bisection linearly, the secant method in between at about 1.618; the trapezoidal rule has error of order h² and is exact for straight lines, Simpson's 1/3 rule has error of order h⁴ and is exact for cubics; Euler's method is first order with one function evaluation per step, and classical Runge–Kutta is fourth order with four. Learn that paragraph and a run of one-markers is settled without touching a calculator. When you do have to iterate, two habits protect the marks. Write the iteration formula down before substituting anything, because the commonest lost mark in Newton–Raphson is a sign — the formula is x − f(x)/f′(x), and a negative f(x) makes the correction positive. And check the answer against what the method is supposed to do: bisection must halve the interval every time, so after two steps the bracket is exactly a quarter of its original width, whatever the function was. One property is worth stating loudly because it produces the most satisfying result in the topic: Simpson's rule integrates any cubic exactly, which is why applying it to ∫₀¹x² dx with just three points returns 1/3 with no error at all, while the trapezoidal rule on the same three points returns 0.375 and is wrong by 1/24.

Root Finding — Three Methods and Their Convergence

What each method costs and what it guarantees
Method	Order of convergence	What it needs, and what it promises
Bisection	Linear (order 1)	Needs a sign change across the bracket. Slow, but it cannot fail — the interval halves every step, so the error is bounded in advance.
Newton–Raphson	Quadratic (order 2)	Needs f′ and a good starting point. Roughly doubles the correct digits each step, but diverges if f′ is near zero or the start is poor.
Secant	About 1.618	Newton with the derivative replaced by a difference quotient. Slower than Newton per step, but each step is cheaper because no f′ is needed.

Newton–Raphson worked, on f(x) = x² − 2 from x₀ = 1. The formula is x₁ = x₀ − f(x₀)/f′(x₀) = 1 − (−1)/2 = 1.5 — note that f(1) = −1 is negative, so the correction adds. One more step: x₂ = 1.5 − (0.25)/3 = 17/12 ≈ 1.41667, against √2 ≈ 1.41421. Two iterations from a crude start have produced three correct digits, which is what quadratic convergence looks like in practice.

🧠 Bisection questions are answered by arithmetic on the interval, not on the function

For f(x) = x³ − x − 1 on [1, 2]: f(1) = −1 and f(2) = 5, so a root is bracketed. The midpoint 1.5 gives f = 0.875 > 0, so the bracket becomes [1, 1.5]; the next midpoint 1.25 gives f = −19/64 ≈ −0.297 < 0, so the bracket becomes [1.25, 1.5]. If the question asks only for the width after two iterations, the answer is (2 − 1)/2² = 0.25 and no function value was ever needed — bisection halves the interval by construction, whatever f does inside it.

Trapezoidal and Simpson — and Why Simpson Wins Here

Both rules approximate an integral by fitting a simple curve through equally spaced points. The trapezoidal rule joins them with straight lines and weights the interior points twice: (h/2)[f₀ + 2f₁ + … + 2f_n−1 + fₙ]. Simpson’s 1/3 rule fits parabolas through points taken three at a time and weights them 1, 4, 2, 4, …, 4, 1: (h/3)[f₀ + 4f₁ + 2f₂ + … + 4f_n−1 + fₙ].

Both rules on ∫₀¹ x² dx with h = 0.5, whose exact value is 1/3
Rule	Computation	Result and error
Trapezoidal	(0.5/2)[0 + 2(0.25) + 1]	0.375, an error of +1/24 ≈ 0.0417. It over-estimates because x² is convex and the chords lie above the curve.
Simpson’s 1/3	(0.5/3)[0 + 4(0.25) + 1]	1/3 exactly, error zero — the rule fits a parabola, and x² is a parabola.

🎯 Simpson is exact for cubics, not just for parabolas

The rule is built by fitting a parabola, so it is obviously exact for a quadratic — but the cubic error term cancels by symmetry over the pair of subintervals, so it is exact for cubics too, and its error is of order h⁴ rather than h³. That is a favourite one-mark question. Two conditions come with it: the number of subintervals must be even (because the points are consumed three at a time, sharing endpoints), and the spacing must be uniform. A question that gives an odd number of subintervals and asks for Simpson’s estimate is testing whether you notice that the rule does not apply.

Single-Step and Multi-Step Methods for an ODE

Euler. y_n+1 = yₙ + h·f(xₙ, yₙ) — one function evaluation, first-order accuracy, and the error accumulates fast. On y′ = y with y(0) = 1 and h = 0.1, one step gives 1 + 0.1 = 1.1 against the exact e^0.1 ≈ 1.10517: an error of about 0.005 after a single step of a very easy problem.
Modified Euler / Heun. Take an Euler step, evaluate the slope there too, and average the two slopes. Two function evaluations, second-order accuracy — the cheapest real improvement available, and the pattern every higher method generalises.
Classical Runge–Kutta (RK4). Four function evaluations per step, fourth-order accuracy. It is the default in practice because halving h cuts the error by a factor of sixteen, so the four evaluations buy far more than four times the accuracy.
Single-step versus multi-step is the distinction the syllabus names. A single-step method — Euler, RK4 — uses only the current point, so it can start from the initial condition alone. A multi-step method such as Adams–Bashforth uses several previous points, which makes each step cheaper but means it cannot start itself: it has to be primed by a single-step method for the first few points.

⚠️ Order of accuracy is not the number of evaluations

For RK4 the two happen to coincide at four, and candidates generalise from that to methods where they do not: Heun uses two evaluations and is second order, but a two-evaluation method need not be, and a fourth-order multi-step method may use only one new evaluation per step by reusing stored values. When a question asks for one of the two, answer the one asked — and read "order" as the exponent of h in the error term, which is the definition the whole topic runs on.

Key takeaways

Convergence orders, free of any problem: bisection 1, secant about 1.618, Newton–Raphson 2. These are one-mark questions with no calculation in them.
Bisection cannot fail once a sign change brackets the root, and it halves the interval every step — so the width after k steps is the original divided by 2^k.
Write x − f(x)/f′(x) down before substituting. A negative f(x) makes the correction positive, and that sign is the commonest lost mark in Newton–Raphson.
Trapezoidal error is order h² and it over-estimates a convex function; Simpson’s 1/3 error is order h⁴ and is exact for cubics.
Simpson’s rule requires an even number of subintervals and uniform spacing. A question with an odd count is testing whether you notice the rule does not apply.
Euler is first order with one evaluation, Heun second with two, RK4 fourth with four. Order means the exponent of h in the error, not the evaluation count.
A single-step method can start from the initial condition alone; a multi-step method has to be primed by one, because it needs several previous points.

Practice questions (9)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

Newton–Raphson is applied to f(x) = x² − 2 with x₀ = 1. The first iterate x₁ is:
1. 1.5
2. 0.5
3. 1.4142
4. 2.0
Show answer
Answer: A — 1.5
x₁ = x₀ − f(x₀)/f′(x₀) with f(1) = 1 − 2 = −1 and f′(1) = 2, so x₁ = 1 − (−1)/2 = 1 + 0.5 = 1.5. Option B is what you get by subtracting instead of adding, that is by dropping the minus sign on f(1) — the single most common error with this formula, and the reason to write the formula down before substituting. Option C is √2 itself, which the method approaches but does not reach in one step: the next iterate is 17/12 ≈ 1.41667, already correct to three decimals, which is what quadratic convergence buys.
The orders of convergence of the bisection, secant and Newton–Raphson methods are respectively:
1. 1, about 1.618, and 2
2. 1, 2 and 3
3. 2, 1.618 and 1
4. All three are linear
Show answer
Answer: A — 1, about 1.618, and 2
Bisection is linear — the error is halved each step, a constant factor, which is exactly what order 1 means. Newton–Raphson is quadratic, order 2, roughly doubling the number of correct digits per step. The secant method sits between them at the golden ratio, about 1.618, because it uses a difference quotient in place of the derivative. Option C reverses the list, which is the answer of a candidate who remembers the three numbers but not which method owns which — and the ordering is easy to reconstruct from cost: the method that needs the most information per step (Newton, which needs f′) converges fastest.
The trapezoidal rule with h = 0.5 applied to ∫₀¹ x² dx gives:
1. 0.3333
2. 0.375
3. 0.25
4. 0.5
Show answer
Answer: B — 0.375
With h = 0.5 the three nodes are 0, 0.5 and 1, where f is 0, 0.25 and 1. The rule gives (h/2)[f₀ + 2f₁ + f₂] = 0.25[0 + 0.5 + 1] = 0.375. The exact value is 1/3 ≈ 0.3333, so the estimate is high by 1/24 ≈ 0.0417 — and it is high rather than low for a structural reason worth knowing: x² is convex, so the chords joining the nodes lie above the curve and the trapezia over-count. Option A is the exact answer, which is what Simpson’s rule returns on the very same three points; a question offering both is checking that you applied the rule it named.
Simpson’s 1/3 rule with h = 0.5 applied to ∫₀¹ x² dx gives the exact answer 1/3. The reason is that:
1. The step size happens to be small enough
2. Simpson’s rule is exact for polynomials up to degree three
3. The integrand is positive on the interval
4. The number of subintervals is even
Show answer
Answer: B — Simpson’s rule is exact for polynomials up to degree three
Simpson’s rule fits a parabola through each group of three nodes, so it is exact for quadratics by construction — and the cubic error term cancels by symmetry across the pair of subintervals, so it is exact for cubics as well. Its error is of order h⁴. Since x² is a quadratic, the rule reproduces the integral exactly with only three points, whatever h is, so option A is wrong about why: a larger h would give the same exact answer. Option D states a real condition — the number of subintervals must be even for the rule to be applicable at all — but that makes the rule usable here, not exact.
Euler’s method with h = 0.1 is applied to y′ = y with y(0) = 1. After one step, y is:
1. 1.10517
2. 1.1
3. 1.01
4. 1.2
Show answer
Answer: B — 1.1
y₁ = y₀ + h·f(x₀, y₀) = 1 + 0.1 × 1 = 1.1. Option A is the exact solution e^0.1 ≈ 1.10517, and the gap of about 0.005 after a single step on the simplest possible ODE is the point of the question: Euler is first order, so its error falls only in proportion to h, and it is offered here beside the exact value to check that you reported the method’s output rather than the truth. Halving h to 0.05 and taking two steps gives 1.1025, closer but still short — that slow approach is exactly what first-order accuracy means.
The classical fourth-order Runge–Kutta method requires, per step:
1. One function evaluation
2. Two function evaluations
3. Four function evaluations
4. One function evaluation and one derivative evaluation
Show answer
Answer: C — Four function evaluations
RK4 computes four slopes k₁ to k₄ per step and combines them in the weighted average (k₁ + 2k₂ + 2k₃ + k₄)/6, so it needs four function evaluations. Option A is Euler and option B is the modified Euler or Heun method, which is second order. Option D describes Newton–Raphson rather than an ODE solver — and note that RK4, like every Runge–Kutta method, deliberately avoids evaluating derivatives of f: extra slopes are cheap, extra derivatives are not, and that trade is the whole design idea of the family.
A multi-step method such as Adams–Bashforth cannot be started from the initial condition alone because:
1. It needs several previous points, which must be supplied by a single-step method
2. It is unstable for small step sizes
3. It requires the derivative of f
4. It only works for linear differential equations
Show answer
Answer: A — It needs several previous points, which must be supplied by a single-step method
A multi-step formula computes y_n+1 from y at several earlier points, so at the very start those earlier points do not exist yet. They have to be generated — primed — by a single-step method such as RK4, after which the multi-step method takes over and is cheaper per step because it reuses slopes it has already computed. Option B inverts the usual stability picture, and option C describes Newton–Raphson. This single-step versus multi-step distinction is exactly what the syllabus phrase "single and multi-step methods" is naming, and it is asked as a concept far more often than as a computation.
The bisection method is applied to f(x) = x³ − x − 1 on the interval [1, 2]. The width of the bracketing interval after two iterations is _____.
Numerical answer — type the value.
Show answer
Answer: 0.25
Bisection halves the interval every iteration by construction, so after two the width is (2 − 1)/2² = 0.25 — and no function value was needed to say so. For the record the brackets are: f(1) = −1 and f(2) = 5 give the sign change; the midpoint 1.5 has f = 0.875 > 0, so the bracket becomes [1, 1.5]; the next midpoint 1.25 has f = −19/64 ≈ −0.297 < 0, so the bracket becomes [1.25, 1.5], of width 0.25 as predicted. This predictability is bisection’s whole selling point: you can state the error bound before running the method, which no faster method lets you do.
Which of the following statements about Simpson’s 1/3 rule are correct? (More than one option may be correct.)
1. It requires the number of subintervals to be even
2. Its error is of order h⁴
3. It integrates any cubic polynomial exactly
4. It requires the derivative of the integrand
Show answer
Answer: A — It requires the number of subintervals to be even; B — Its error is of order h⁴; C — It integrates any cubic polynomial exactly
The first three are correct and they are connected rather than three separate facts. The rule consumes points three at a time with shared endpoints, so the number of subintervals must be even — option A. It fits a parabola through each triple, and the cubic error term cancels by symmetry across the pair, which is why it is exact for cubics — option C — and why its error is of order h⁴ rather than h³ — option B. Only option D is false: quadrature rules use function values at the nodes and never a derivative, which is precisely what makes them usable on tabulated data with no formula behind it. Note that the whole set can be settled from theory with no arithmetic, which is what makes property questions the cheapest marks in this topic.