Numerical Integration Approximation
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Direct answer
Numerical integration estimates a definite integral when the antiderivative is unavailable or awkward, by sampling the integrand at equally spaced points. Split [a, b] into n strips of width h = (b − a)/n with ordinates y0, y1, ..., yn. The trapezoidal rule joins sample points by straight lines: T = h/2[(y0 + yn) + 2(sum of the middle ordinates)]. Simpson's one-third rule fits a parabola through each consecutive triple: S = h/3[(y0 + yn) + 4(odd-indexed ordinates) + 2(even-indexed ordinates)], and it needs n even. Simpson's error is O(h^4) against the trapezoid's O(h^2), so it converges far faster on smooth integrands.
What you must remember
- Trapezoidal rule: T = h/2[(y0 + yn) + 2(sum of middle ordinates)] with h = (b − a)/n; endpoints weigh once, every interior ordinate twice.
- Simpson's one-third rule: S = h/3[(y0 + yn) + 4(odd-indexed ordinates) + 2(even-indexed ordinates)]; it applies to even intervals only — with odd n, run Simpson over the first n − 1 strips and the trapezoid over the last.
- Error orders: trapezoid error is O(h^2), involving f''; Simpson error is O(h^4), involving the fourth derivative — halving h cuts trapezoid error 4-fold and Simpson error 16-fold.
- Exactness: Simpson is exact for polynomials up to degree 3, because its fourth-derivative error term vanishes on cubics.
- Over or under: the trapezoid overestimates a convex integrand (f'' > 0) and underestimates a concave one; if f'' changes sign on the interval, compute rather than guess.
- Signature example: ∫₀¹ dx/(1 + x²) = π/4 ≈ 0.7854 — five points with n = 4 give Simpson ≈ 0.78539, near-perfect agreement.
- Small-change companion: dy = f'(x)dx handles one-off estimates such as √26.3 ≈ 5 + 1.3/(2×5) = 5.13.
Working an estimate by hand
Estimate ∫₀¹ dx/(1 + x²) with n = 4 — the standard JEE-style setup, chosen because the true value is π/4 ≈ 0.7854 and the answer checks itself. Here h = (1 − 0)/4 = 0.25, the nodes sit at 0, 0.25, 0.5, 0.75, 1, and y = 1/(1 + x²) gives y0 = 1, y1 = 1/1.0625 ≈ 0.9412, y2 = 0.8, y3 = 1/1.5625 = 0.64, y4 = 0.5. Trapezoidal: T = (0.25/2)[(1 + 0.5) + 2(0.9412 + 0.8 + 0.64)] = 0.125 × 6.2624 ≈ 0.7828, short of π/4 by about 0.0026. Simpson: S = (0.25/3)[(1 + 0.5) + 4(0.9412 + 0.64) + 2(0.8)] = (1/12) × 9.4247 ≈ 0.78539, matching π/4 to four decimal places from five sample points. A subtlety most solutions skip: y'' = (6x² − 2)/(1 + x²)³ changes sign at x = 1/√3, so on this interval the trapezoid's underestimate cannot be predicted by inspection — it is confirmed only after computing. The discipline generalises: fix h first, list every ordinate in order, apply the weights last, and treat a Simpson result drifting from π/4 as an alarm that a 4 and a 2 got swapped.
How JEE actually tests it
Numerical integration occupies an odd slot in JEE: it has no dedicated chapter in most syllabus listings and appears rarely on JEE Main — and even then usually disguised as a differential-approximation item such as (1.02)^4 ≈ 1.0824 — while JEE Advanced has used it experimentally, supplying a small ordinate table and asking for the rule value directly. The examiner's leverage is procedural. A question asking for |T − S| rewards the student who subtracts the two rule values instead of recomputing the integral; an "even number of subintervals" phrase decides whether Simpson is legal; unequal-looking x-labels tempt a wrong h. Prepare it as arithmetic discipline — correct h, correct weight pattern, values carried to four decimals — and the rare Main question is nearly free marks, while the Advanced variant tests precisely that discipline under time pressure.
Frequently asked questions
Can Simpson's rule be used with an odd number of strips?
No — the one-third rule needs an even count. Apply Simpson to the first n − 1 strips and finish the last strip with the trapezoidal rule.
Why is Simpson's rule more accurate than the trapezoidal rule?
Its error term involves the fourth derivative and is O(h^4), against the trapezoid's O(h^2) second-derivative term. Doubling n cuts Simpson's error 16-fold but the trapezoid's only 4-fold.
For which integrands is Simpson's rule exact?
For every polynomial of degree up to 3, since the fourth derivative vanishes. It also performs very well on smooth integrands with small fourth derivatives, such as 1/(1 + x²) on [0, 1].
Does the trapezoidal rule overestimate or underestimate?
It overestimates a convex integrand (f'' > 0), where chords lie above the curve, and underestimates a concave one. When f'' changes sign across the interval, the sign of the error cannot be read off by inspection.
Is numerical integration actually asked in JEE?
Not as a named chapter, and JEE Main asks it rarely — usually as a differential approximation. JEE Advanced has explored it experimentally with supplied ordinate tables, so both rules are worth knowing cold.