← 2025 Paper 2

UPSC Maths 2025 Paper 2 Q8b — Solution

15 marks · Section B

Question

Find the constant $p$ and error term for the quadrature formula $\displaystyle\int_{x_0}^{x_1} f(x)\,dx = \dfrac{h}{2}(f_0 + f_1) + ph^2(f_0' - f_1')$ where $x_0 + h = x_1$, $f_0 = f(x_0)$, $f_1 = f(x_1)$ and prime ($'$) represents derivative with respect to $x$. Hence deduce the composite rule for integrating $\displaystyle\int_a^b f(x)\,dx$, $a = x_0 < x_1 < \cdots < x_N = b$.

Technique

Use the method of undetermined coefficients: expand both sides in powers of $h$ about $x_0$ and choose $p$ to maximise the order of accuracy; the first non-vanishing residual gives the error term. This is the corrected trapezoidal (Euler–Maclaurin) rule.

Solution

Taylor expansion of both sides

Expand $f$ about $x_0$. Write $f^{(k)} \equiv f^{(k)}(x_0)$. With $x_1 = x_0 + h$: $\int_{x_0}^{x_0+h} f\,dx = f h + \frac{f' h^2}{2} + \frac{f'' h^3}{6} + \frac{f''' h^4}{24} + \frac{f^{(4)} h^5}{120} + \cdots$

Right-hand side pieces: $f_0 = f,\qquad f_1 = f + f'h + \tfrac{f''}{2}h^2 + \tfrac{f'''}{6}h^3 + \tfrac{f^{(4)}}{24}h^4 + \cdots,$ $f_0' = f',\qquad f_1' = f' + f''h + \tfrac{f'''}{2}h^2 + \tfrac{f^{(4)}}{6}h^3 + \cdots$

Then $\frac{h}{2}(f_0 + f_1) = fh + \frac{f'}{2}h^2 + \frac{f''}{4}h^3 + \frac{f'''}{12}h^4 + \frac{f^{(4)}}{48}h^5 + \cdots,$ $ph^2(f_0' - f_1') = ph^2\Big(-f''h - \tfrac{f'''}{2}h^2 - \tfrac{f^{(4)}}{6}h^3 - \cdots\Big) = -pf''h^3 - \tfrac{p}{2}f'''h^4 - \tfrac{p}{6}f^{(4)}h^5 - \cdots$

Subtract to get the error

$E = \int_{x_0}^{x_1}f\,dx - \text{RHS}.$ Term-by-term (the $h$ and $h^2$ terms cancel exactly): $E = \Big(\tfrac{1}{6} - \tfrac14 + p\Big)f''h^3 + \Big(\tfrac{1}{24} - \tfrac{1}{12} + \tfrac{p}{2}\Big)f'''h^4 + \Big(\tfrac{1}{120} - \tfrac{1}{48} + \tfrac{p}{6}\Big)f^{(4)}h^5 + \cdots$

Choose $p$ to kill the $h^3$ term: $\frac16 - \frac14 + p = 0 \implies p = \frac14 - \frac16 = \frac{1}{12}.$

$\boxed{p = \frac{1}{12}.}$

With $p = \tfrac{1}{12}$, the $h^4$ coefficient becomes $\tfrac{1}{24} - \tfrac{1}{12} + \tfrac{1}{24} = 0$ — it also vanishes. The first surviving term is the $h^5$ term: $\frac{1}{120} - \frac{1}{48} + \frac{1}{72} = \frac{6 - 15 + 10}{720} = \frac{1}{720}.$

So the leading error is $E = \frac{h^5}{720}f^{(4)}(\xi),\qquad \xi \in (x_0,x_1),$ via the mean-value form of the remainder.

Single-interval formula

$\int_{x_0}^{x_1} f\,dx = \frac{h}{2}(f_0 + f_1) + \frac{h^2}{12}(f_0' - f_1') + \frac{h^5}{720}f^{(4)}(\xi).$

Composite rule on $[a,b]$

Partition $[a,b]$ into $N$ equal subintervals of width $h = (b-a)/N$, nodes $x_i = a + ih$. Sum the single-interval rule: $\int_a^b f\,dx = \sum_{i=0}^{N-1}\left[\frac{h}{2}(f_i + f_{i+1}) + \frac{h^2}{12}(f_i' - f_{i+1}')\right] + \sum_{i=0}^{N-1}\frac{h^5}{720}f^{(4)}(\xi_i).$

The trapezoidal sum telescopes, and the derivative correction telescopes ($f_i' - f_{i+1}'$ sums to $f_0' - f_N'$): $\boxed{\int_a^b f\,dx = h\Big[\tfrac12 f_0 + f_1 + f_2 + \cdots + f_{N-1} + \tfrac12 f_N\Big] + \frac{h^2}{12}\big(f'(a) - f'(b)\big) + E_N,}$ with total error $E_N = \sum_{i=0}^{N-1}\frac{h^5}{720}f^{(4)}(\xi_i) = \frac{(b-a)h^4}{720}f^{(4)}(\eta),\quad \eta\in(a,b)$ (using $Nh = b-a$ and the intermediate-value theorem).

This is the corrected (end-corrected) trapezoidal rule — the leading Euler–Maclaurin term; it raises the trapezoidal accuracy from $O(h^2)$ to $O(h^4)$ globally.

Answer

$p = \dfrac{1}{12}$. Single-panel error $\dfrac{h^5}{720}f^{(4)}(\xi)$. Composite rule: $\int_a^b f\,dx = h\Big[\tfrac12 f_0 + \textstyle\sum_{i=1}^{N-1} f_i + \tfrac12 f_N\Big] + \frac{h^2}{12}\big(f'(a)-f'(b)\big) + \frac{(b-a)h^4}{720}f^{(4)}(\eta).$