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UPSC 2026 Maths Optional Paper 2 Q7b — Step-by-Step Solution

15 marks · Section B

Simpson's 1/3 and 3/8 rules · Numerical Analysis · asked 6× in 14 yrs · Read the full method →

Question

Evaluate the integral $\displaystyle\int_{0}^{0\cdot8} f(x)\,dx$, where

by using single application of Simpson’s $\dfrac{3}{8}$ rule.

(The raised dots are the Indian decimal point: $0\cdot8 = 0.8$, $0\cdot2 = 0.2$.)

Technique

“Single application” is the operative phrase: it means the basic (un-composited) Newton–Cotes rule on four equally spaced nodes, so the number of strips is fixed at $n = 3$ and $h = (b-a)/3$ — not a free choice. Compute the four ordinates and apply $\frac{3h}{8}\bigl[f_0+3f_1+3f_2+f_3\bigr]$. Then, because the integrand is an explicit polynomial, integrate it exactly and report the absolute and relative error: the $3/8$ rule is exact only up to cubics and this integrand is quintic, so there is a genuine error, and quantifying it against the error term $-\frac{3h^5}{80}f^{(4)}(\xi)$ is what converts a rote answer into a full-marks one.

Solution

Step 1 — The rule, and what “single application” fixes

Simpson’s $3/8$ rule replaces $f$ on $[x_0,x_3]$ by the cubic through four equally spaced points $x_0,x_1,x_2,x_3$ (Newton–Cotes, $n=3$) and integrates that cubic exactly:

with error term

A single application therefore uses exactly $n=3$ sub-intervals and $4$ ordinates over the whole range $[0,\,0.8]$. Hence

A useful arithmetic shortcut, worth taking in the hall: the multiplier is

Step 2 — The four ordinates

The nodes are $x_i = ih$, i.e. $0,\ \tfrac{4}{15},\ \tfrac{8}{15},\ \tfrac{4}{5}$. Evaluating $f(x) = 0.2 + 25x - 200x^2 + 675x^3 - 900x^4 + 400x^5$ (all values carried to $7$ decimals; the exact rationals are given because they make the check reproducible):

Sample working for $f_1$ (the one candidates get wrong), using $x_1 = \tfrac4{15}$:

And for the endpoint,

Step 3 — Apply the rule

Step 4 — The exact value (the integrand is a polynomial, so this is free)

exactly $\dfrac{3076}{1875} = 1.640533\overline{3}$.

Step 5 — Error, absolute and relative

Why the error is there, and why it is this size. The $3/8$ rule integrates the interpolating cubic exactly, so it is exact for every polynomial of degree $\le 3$. Our integrand is quintic; the terms $-900x^4$ and $+400x^5$ are precisely what the rule cannot see, and they carry the whole error.

Quantitatively, from (1) with

the crude bound uses $\max_{[0,0.8]}|f^{(4)}| = \max\{21600,\,16800\} = 21600$:

which is satisfied but slack by exactly the factor $21600/2400 = 9$. The sharp statement is available here because $f^{(4)}$ is affine and the $3/8$ rule is symmetric about the midpoint of $[a,b]$: its degree-$4$ Peano kernel $K$ is symmetric, so $\int t\,K(t)\,dt = \bar{x}\int K(t)\,dt$ with $\bar x = \tfrac{a+b}{2}$, and for an affine $f^{(4)}$ the mean-value point is exactly the midpoint,

in exact agreement with the directly computed error. $\qquad\blacksquare$

Interpretation for the answer script (one line, worth writing): $h = 0.2667$ is a coarse step for an integrand that swings from $0.2$ up to $3.49$ and back down to $0.232$ on this range; a single application of any four-point rule cannot track that, and $\approx 7.4\%$ error is the honest price. Composite application (two $3/8$ panels, $h=0.1333$) would cut the error by the expected factor $\approx 2^5 = 32$.

Answer