Riemann Definite Integral; Integrability

At a Glance

• Frequency: 1 sub-part across 1 of 14 years (2018)
• Priority tier: T4
• Marks (count): 10 (1)
• Average solve time: ~10 min
• Difficulty mix: medium 1
• Section: A | Dominant type: computation

Why This Chapter Matters

This atom appeared once in 2018 as a 10-mark Section A computation. UPSC asks you to evaluate a definite integral directly from the Riemann sum definition — bypassing the Fundamental Theorem — or to verify that a function is Riemann integrable using the $\varepsilon$–$\delta$ or Darboux criterion. This tests conceptual understanding of what integration means, not just computation skill. Note the scope boundary: this is the P1 (Calculus) version, where the expected depth is computing a limit of Riemann sums; the deeper analysis-level Darboux theory lives in P2-RA-11.

Minimum Theory

Riemann Sum Definition

Let $f: [a, b] \to \mathbb{R}$ be bounded. A partition $P = \{x_0, x_1, \ldots, x_n\}$ with $a = x_0 < x_1 < \cdots < x_n = b$ divides $[a,b]$ into $n$ subintervals of width $\Delta x_i = x_i - x_{i-1}$.

A Riemann sum is

$S(P, f, \xi) = \sum_{i=1}^{n} f(\xi_i)\,\Delta x_i, \quad \xi_i \in [x_{i-1}, x_i].$

$f$ is Riemann integrable on $[a,b]$ with integral $I$ if: for every $\varepsilon > 0$ there exists $\delta > 0$ such that $\|P\| < \delta \Rightarrow |S - I| < \varepsilon$ for all choices of $\xi_i$.

Darboux (Upper/Lower Sum) Criterion

$U(P, f) = \sum_{i=1}^{n} M_i\,\Delta x_i, \quad M_i = \sup_{[x_{i-1},x_i]} f,$ $L(P, f) = \sum_{i=1}^{n} m_i\,\Delta x_i, \quad m_i = \inf_{[x_{i-1},x_i]} f.$

$f$ is Riemann integrable iff $\inf_P U(P,f) = \sup_P L(P,f)$, equivalently iff for every $\varepsilon > 0$ there exists a partition $P$ with $U(P,f) - L(P,f) < \varepsilon$.

Evaluating via Riemann Sums

For the uniform partition $P_n$ with $x_i = a + i\cdot\frac{b-a}{n}$ and right-endpoint tags $\xi_i = x_i$:

$\int_a^b f(x)\,dx = \lim_{n \to \infty} \frac{b-a}{n} \sum_{i=1}^{n} f\!\left(a + \frac{i(b-a)}{n}\right).$

Key Summation Formulas

$\sum_{i=1}^{n} i = \frac{n(n+1)}{2}, \qquad \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}, \qquad \sum_{i=1}^{n} i^3 = \left[\frac{n(n+1)}{2}\right]^2.$

Integrability Results

• Every continuous function on $[a,b]$ is Riemann integrable.
• Every monotone bounded function on $[a,b]$ is Riemann integrable.
• A bounded function with finitely many discontinuities is Riemann integrable.

Question Archetypes

evaluate-from-definition (1 question; 2018)

Recognition Cues

• The question explicitly says “using the definition” or “as a limit of sums” — not “by the Fundamental Theorem”
• The function is simple enough (polynomial or exponential) for the sum to be computed in closed form
• The interval $[a,b]$ is concrete (e.g., $[0,1]$ or $[1,2]$)
• 10 marks — full setup, explicit Riemann sum, limit computation all required

Solution Template

1. Write the uniform partition: $x_i = a + i\cdot h$ where $h = (b-a)/n$; right-endpoint tags $\xi_i = x_i$.
2. Write $S_n = h \sum_{i=1}^{n} f(x_i)$ explicitly.
3. Substitute $f$ and simplify the sum using the summation formulas above.
4. Take $\lim_{n\to\infty} S_n$; the result is $\int_a^b f(x)\,dx$.
5. (If asked for integrability) invoke the relevant theorem (continuity, monotonicity) or compute $U - L \to 0$.

Worked Example

2018 Paper 1, 2018-P1-Q2b (10 marks)

Using the definition of the Riemann integral as a limit of sums, evaluate $\displaystyle\int_0^1 x^2\,dx$.

Step 1. Uniform partition.

Divide $[0,1]$ into $n$ equal subintervals of width $h = 1/n$. Right-endpoint tags: $\xi_i = i/n$ for $i = 1, 2, \ldots, n$.

Step 2. Riemann sum.

$S_n = \sum_{i=1}^{n} f\!\left(\frac{i}{n}\right) \cdot \frac{1}{n} = \sum_{i=1}^{n} \left(\frac{i}{n}\right)^2 \cdot \frac{1}{n} = \frac{1}{n^3} \sum_{i=1}^{n} i^2.$

Step 3. Apply summation formula.

$S_n = \frac{1}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6n^2}.$

Step 4. Take the limit.

$\int_0^1 x^2\,dx = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{(n+1)(2n+1)}{6n^2}.$

Divide numerator and denominator by $n^2$:

$= \lim_{n \to \infty} \frac{\left(1 + \frac{1}{n}\right)\left(2 + \frac{1}{n}\right)}{6} = \frac{1 \cdot 2}{6} = \frac{1}{3}.$

$\boxed{\int_0^1 x^2\,dx = \frac{1}{3}}$

Verification (optional line): By the Fundamental Theorem, $\int_0^1 x^2\,dx = \bigl[x^3/3\bigr]_0^1 = 1/3$. Consistent.

Common Traps

• Using the Fundamental Theorem directly: The question asks for the definition-based approach; using anti-derivatives without the limit of sums forfeits most marks.
• Using left endpoints vs. right endpoints: Either works (the limit is the same for any Riemann integrable function), but choose one and be consistent throughout.
• Arithmetic in the sum formula: $\sum i^2 = n(n+1)(2n+1)/6$ — the factor of $6$ is easy to misremember as $4$ or $3$.
• Forgetting to divide by $n$ at the start: The Riemann sum is $h \sum f(\xi_i)$ where $h = 1/n$; forgetting the $1/n$ weight is a common first-step error.

Marks-Aware Writing

For a 10-mark computation:

• 2 marks — correct partition stated: $x_i = i/n$, $h = 1/n$ written explicitly.
• 3 marks — Riemann sum $S_n$ written and simplified to a closed-form expression in $n$ (using the summation formula correctly).
• 3 marks — limit $\lim_{n\to\infty} S_n$ evaluated with clear algebra.
• 2 marks — final answer stated, and (if asked) the statement “hence $f$ is Riemann integrable on $[0,1]$” since the limit exists.

Write out $S_n$ fully before simplifying — do not skip to the answer. The limit computation should show the $n \to \infty$ step explicitly.

Practice Set

Only one historical question on this atom (shown above).