Riemann integral

At a Glance

• Frequency: 10 sub-parts across 9 of 13 years (2013, 2014, 2015, 2018, 2021, 2022, 2023, 2024, 2025)
• Priority tier: T1
• Marks (count): 10 (4), 15 (6)
• Average solve time: ~8 min
• Difficulty mix: medium 6, easy 4
• Section: A | Dominant type: proof

Why This Chapter Matters

The Riemann integral appears in 9 of the last 13 years, split evenly between 10- and 15-mark questions — almost all in Section A. Unlike the rest of calculus which tests computation, this atom tests reasoning: can you prove a function is integrable, can you recognise when it is not, and can you evaluate an integral from the definition? Four recurring proof templates (Darboux criterion, Lebesgue criterion, Darboux theorem, oscillation squeeze) cover every question that has ever appeared.

Minimum Theory

Darboux sums. For $f$ bounded on $[a,b]$ and partition $P=\{a=x_0<x_1<\cdots<x_n=b\}$, define $M_i=\sup_{[x_{i-1},x_i]}f$ and $m_i=\inf_{[x_{i-1},x_i]}f$. The upper and lower Darboux sums are $U(P,f)=\sum_{i=1}^n M_i\Delta x_i,\qquad L(P,f)=\sum_{i=1}^n m_i\Delta x_i.$ Always $L\le U$. Define $\overline\int f=\inf_P U(P,f)$ (upper integral) and $\underline\int f=\sup_P L(P,f)$ (lower integral). Always $\underline\int f\le\overline\int f$.

Riemann integrability criteria:

• (Riemann’s criterion) $f$ is Riemann integrable iff for every $\varepsilon>0$ there exists a partition $P$ with $U(P,f)-L(P,f)<\varepsilon$.
• (Lebesgue’s criterion) A bounded function on $[a,b]$ is Riemann integrable iff its set of discontinuities has Lebesgue measure zero. In practice: finitely many, or countably many, discontinuities ⇒ integrable.

Key theorem classes:

• Continuous $f$ on $[a,b]$ $\Rightarrow$ Riemann integrable (by uniform continuity via Heine-Cantor).
• Monotone $f$ on $[a,b]$ $\Rightarrow$ Riemann integrable (telescoping Darboux sums).
• Bounded $f$ with at most countably many discontinuities $\Rightarrow$ Riemann integrable.

Antiderivatives (Darboux’s theorem). If $g'=f$ on $[a,b]$ then $f$ must have the intermediate value property — no jump discontinuities. Therefore a function with a jump discontinuity is Riemann integrable but does not have an antiderivative. These are separate properties.

Question Archetypes

Seven patterns cover all Riemann integral questions.

integrability-from-definition (1 question; 2022)

Recognition Cues

• “Show that $f$ is Riemann integrable on $[a,b]$ and $\int_a^b f\,dx=\ldots$”
• Explicit Darboux sums required; often $f(x)=x^2$ or similar power on $[0,k]$.

Solution Template

1. Set up the uniform partition $P_n$ with $\Delta x=k/n$.
2. Compute $L(P_n)$ and $U(P_n)$ (for increasing $f$: $m_i=f(x_{i-1})$, $M_i=f(x_i)$).
3. Use standard sum formulas ($\sum i^2=n(n+1)(2n+1)/6$, etc.).
4. Take $n\to\infty$; show both limits equal the same value.

Worked Example(s)

2022 Paper 2, 2022-P2-Q2a (15 marks)

Show $f(x)=x^2$ is Riemann integrable on $[0,k]$ and $\int_0^k x^2\,dx=k^3/3$.

Partition. Equal partition $P_n$: $x_i=ik/n$, $\Delta x=k/n$. Since $f$ increasing: $m_i=(ik/n)^2$, $M_i=((i+1)k/n)^2$.

$L(P_n)=\frac{k^3}{n^3}\sum_{i=0}^{n-1}i^2=\frac{k^3}{n^3}\cdot\frac{(n-1)n(2n-1)}{6}\;\to\;\frac{k^3}{3}.$ $U(P_n)=\frac{k^3}{n^3}\sum_{j=1}^{n}j^2=\frac{k^3}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}\;\to\;\frac{k^3}{3}.$

Both limits equal $k^3/3$, so $f$ is Riemann integrable and $\int_0^k x^2\,dx=k^3/3$. $\blacksquare$

Common Traps

• Use $\sum_{i=0}^{n-1}i^2=(n-1)n(2n-1)/6$ (index starts at 0) for the lower sum, and $\sum_{i=1}^n i^2=n(n+1)(2n+1)/6$ for the upper sum. Don’t mix.
• The dominant term in both is $k^3\cdot 2n^2/(6n^2)=k^3/3$; the lower-order terms vanish as $n\to\infty$.

integrability-class-proof (2 question(s); 2021, 2025)

Recognition Cues

• “Prove that every monotone/continuous function on $[a,b]$ is Riemann integrable.”

Solution Template (monotone):

1. State Riemann’s criterion.
2. Choose the equal partition $P_n$ with $\Delta x=(b-a)/n$.
3. For increasing $f$: $U-L=\frac{b-a}{n}[f(b)-f(a)]$. For decreasing: same formula with $f(a)-f(b)$.
4. Choose $n>(b-a)[f(b)-f(a)]/\varepsilon$; then $U-L<\varepsilon$. $\blacksquare$

Solution Template (continuous):

1. Invoke Heine-Cantor: on compact $[a,b]$, continuous $\Rightarrow$ uniformly continuous.
2. Given $\varepsilon$, find $\delta$ so $|s-t|<\delta\Rightarrow|f(s)-f(t)|<\varepsilon/(b-a)$.
3. Choose partition $P$ with mesh $<\delta$; then each oscillation $M_i-m_i<\varepsilon/(b-a)$.
4. $U-L<[\varepsilon/(b-a)]\sum\Delta x_i=\varepsilon$. $\blacksquare$

Worked Example(s)

2021 Paper 2, 2021-P2-Q1c (10 marks)

Prove: monotone $f$ on $[a,b]$ is Riemann integrable.

WLOG $f$ increasing. For equal partition $P_n$ (mesh $h=(b-a)/n$): on each $[x_{i-1},x_i]$, $\sup f=f(x_i)$ and $\inf f=f(x_{i-1})$, so $U(P_n)-L(P_n)=h\sum_{i=1}^n[f(x_i)-f(x_{i-1})]=h[f(b)-f(a)]=\frac{(b-a)[f(b)-f(a)]}{n}.$ This is $<\varepsilon$ for $n>(b-a)[f(b)-f(a)]/\varepsilon$. Riemann’s criterion is satisfied. $\blacksquare$

2025 Paper 2, 2025-P2-Q4b (15 marks)

Prove every continuous function on $[a,b]$ is Riemann integrable.

Bounded: continuous on compact $[a,b]$ $\Rightarrow$ bounded (extreme value theorem), so Darboux sums are finite.

Uniform continuity (Heine-Cantor): since $f$ is continuous on the compact set $[a,b]$, it is uniformly continuous: for any $\varepsilon>0$, $\exists\delta>0$ such that $|s-t|<\delta\Rightarrow|f(s)-f(t)|<\varepsilon/(b-a)$.

Choose a fine partition: take any partition $P$ with mesh $\|P\|<\delta$. On each subinterval $[x_{i-1},x_i]$ (length $<\delta$), $f$ attains its max at $\xi_i$ and min at $\eta_i$, and $|\xi_i-\eta_i|\le\Delta x_i<\delta$, so $M_i-m_i<\varepsilon/(b-a)$.

$U(P,f)-L(P,f)=\sum_i(M_i-m_i)\Delta x_i<\frac\varepsilon{b-a}\sum_i\Delta x_i=\varepsilon.$

Riemann’s criterion is satisfied. $\blacksquare$

The key: uniform (not pointwise) continuity is needed — it gives a single $\delta$ controlling all oscillations simultaneously. Compactness of $[a,b]$ is what guarantees the upgrade from continuous to uniformly continuous.

Common Traps

• For the monotone proof: the telescoping $\sum[f(x_i)-f(x_{i-1})]=f(b)-f(a)$ is the entire mechanism — state it explicitly.
• For the continuous proof: do not use pointwise continuity (each $\delta$ would depend on the point, not simultaneously valid). State “uniformly continuous by Heine-Cantor” and then apply.
• Both proofs apply to monotone-decreasing by the same argument (swap $\sup$ and $\inf$, get $f(a)-f(b)$).

integrability-criterion (1 question; 2013)

Recognition Cues

• “Is $f$ Riemann integrable? Does there exist a function $g$ with $g'=f$?” — two separate questions.
• $f$ is piecewise-defined with a jump at an interior point.

Solution Template

1. Integrability: check boundedness; identify the set of discontinuities; apply Lebesgue’s criterion (finite or countable $\Rightarrow$ measure zero $\Rightarrow$ integrable).
2. Antiderivative: if $f$ has a jump discontinuity, cite Darboux’s theorem — derivatives must have the intermediate value property, so $f$ cannot be a derivative.

Worked Example(s)

2013 Paper 2, 2013-P2-Q1c (10 marks)

$f(x)=x^2/2+4$ for $x\ge0$, $f(x)=-x^2/2+2$ for $x<0$. Riemann integrable on $[-1,2]$? Antiderivative exists?

Integrability. $f$ is bounded (values in $[3/2,6]$). The single discontinuity at $x=0$ (jump: $\lim_{0^-}f=2\ne4=\lim_{0^+}f$) has measure zero. By Lebesgue’s criterion: yes, $f$ is Riemann integrable.

Antiderivative. $f$ has a jump at $x=0$: $f(0^-)=2$, $f(0^+)=4$. Any value $c\in(2,4)$ is not attained by $f$ on $[-1,2]$. By Darboux’s theorem, a derivative cannot have a jump discontinuity. Therefore no antiderivative $g$ with $g'=f$ exists on $[-1,2]$.

Common Traps

• Riemann integrability and existence of an antiderivative are independent concepts. A function can be integrable without having an antiderivative (as here), and vice versa. Do not conflate them.
• The FTC has two distinct parts: (1) if $f$ is continuous, then $F(x)=\int_a^x f$ is an antiderivative; (2) if $F'$ is Riemann integrable, then $\int_a^b F'=F(b)-F(a)$. Neither implies an antiderivative exists for a discontinuous function.

step-function-integral (2 question(s); 2013, 2015)

Recognition Cues

• $f$ is piecewise constant (floor function, $1/n$ on intervals of length $1/(n(n+1))$, etc.).
• “Is $f$ Riemann integrable? Compute $\int$.”

Solution Template

1. Integrability: identify the discontinuity set (integer points for floor function, $\{1/n\}$ for the step function); state it is finite or countable (measure zero); conclude integrable by Lebesgue.
2. Compute: on each constant piece, contribution $=$ value $\times$ length. Sum up (possibly as an infinite series for the countable case).
3. For infinite series: use partial fractions and telescoping.

Worked Example(s)

2013 Paper 2, 2013-P2-Q3d (10 marks)

$f(x)=[x]^2+3$ on $[-1,2]$. Integrable? Compute.

Step function: $f=4$ on $[-1,0)$, $f=3$ on $[0,1)$, $f=4$ on $[1,2)$, $f(2)=7$. Discontinuities at $x=0,1,2$ — finite set, measure zero. Integrable.

$\int_{-1}^2 f\,dx=4(1)+3(1)+4(1)=\boxed{11.}$

(The single point $x=2$ contributes zero.)

2015 Paper 2, 2015-P2-Q2b (15 marks)

$f(x)=1/n$ on $(1/(n+1),1/n]$, $f(0)=0$. Integrable on $[0,1]$? Find $\int_0^1 f$.

Bounded ($0\le f\le1$). Discontinuities at $\{1/n:n\ge2\}\cup\{0\}$ — countable, measure zero. Integrable.

Length of $(1/(n+1),1/n]$ is $1/n-1/(n+1)=1/(n(n+1))$. Sum: $\int_0^1 f\,dx=\sum_{n=1}^\infty\frac{1}{n}\cdot\frac{1}{n(n+1)}=\sum_{n=1}^\infty\frac{1}{n^2(n+1)}.$

Partial fractions: $1/(n^2(n+1))=-1/n+1/n^2+1/(n+1)$. Sum telescopes + Basel: $\int_0^1 f=-1+\frac{\pi^2}{6}=\boxed{\frac{\pi^2}{6}-1.}$

Common Traps

• For the floor function: $[x]=-1$ on $[-1,0)$ — so $[x]^2+3=4$, not $3$. Verify the value on each subinterval.
• The endpoint $x=2$ where $[2]=2$ contributes $f(2)=7$, but a single point has measure zero and contributes zero to the integral — don’t add a term for it.
• For the countable-discontinuity example (2015): the series $\sum 1/(n^2(n+1))$ needs partial fractions. The final answer involves $\pi^2/6$ (Basel’s sum).
• Length $1/n-1/(n+1)=1/(n(n+1))$ — easy to miscompute as $1/(n^2+n)$.

ftc-evaluation (1 question; 2014)

Recognition Cues

• An engineered combination like $2x\sin(1/x)-\cos(1/x)$ — looks hard but is the derivative of $x^2\sin(1/x)$.
• The function is defined piecewise at a singular endpoint ($f(0)=0$).

Solution Template

1. Spot the antiderivative $F$ by reverse product rule: $\frac{d}{dx}[x^2\sin(1/x)]=2x\sin(1/x)-\cos(1/x)$.
2. Show $F$ extends continuously to $[0,1]$ (squeeze: $|x^2\sin(1/x)|\le x^2\to0$).
3. Show $f$ is bounded on $[0,1]$ with a single discontinuity at $0$.
4. Apply FTC on $[\varepsilon,1]$ and let $\varepsilon\to0^+$.

Worked Example(s)

2014 Paper 2, 2014-P2-Q2b (15 marks)

$f(x)=2x\sin(1/x)-\cos(1/x)$ for $x\in(0,1]$, $f(0)=0$. Show $\int_0^1 f\,dx=\sin1$.

Antiderivative: $\frac{d}{dx}[x^2\sin(1/x)]=2x\sin(1/x)-\cos(1/x)=f(x)$ for $x>0$.

Let $F(x)=x^2\sin(1/x)$, $F(0)=0$. Continuous on $[0,1]$ since $|F(x)|\le x^2\to0$.

$f$ is bounded ($|f|\le3$) with one discontinuity at $x=0$. On $[\varepsilon,1]$, FTC gives $\int_\varepsilon^1 f=F(1)-F(\varepsilon)=\sin1-\varepsilon^2\sin(1/\varepsilon)$.

Let $\varepsilon\to0^+$: $\varepsilon^2\sin(1/\varepsilon)\to0$.

$\boxed{\;\int_0^1 f\,dx=\sin1.\;}$

Common Traps

• $2x\sin(1/x)$ alone and $\cos(1/x)$ alone have no elementary antiderivatives — the combination is engineered. Don’t try to integrate them separately.
• $F$ is continuous at $0$ (extends by $F(0)=0$) but not differentiable at $0$. FTC still applies via the $\varepsilon$-limit form.

integral-bounds (1 question; 2018)

Recognition Cues

• “Prove $A\le\int_a^b f\,dx\le B$” — without computing the integral.
• Bound the integrand between constants $m\le f\le M$, then multiply by the interval length.

Solution Template

1. Identify a monotonicity or range property of $f$ on $[a,b]$.
2. Bound: $m\le f(x)\le M$ for all $x\in[a,b]$.
3. Integrate: $m(b-a)\le\int_a^b f\le M(b-a)$.
4. Use strict inequalities if $f$ is non-constant (the integral is strictly between the bounds).

Worked Example(s)

2018 Paper 2, 2018-P2-Q1b (10 marks)

Prove $\pi^2/9<\int_{\pi/6}^{\pi/2}(x/\sin x)\,dx<2\pi^2/9$.

Monotonicity. $h(x)=\sin x/x$ has $h'(x)=(x\cos x-\sin x)/x^2<0$ on $(0,\pi/2]$ (since $x\cos x<\sin x$ there). So $h$ is strictly decreasing, hence $x/\sin x$ is strictly increasing.

Bounds. On $[\pi/6,\pi/2]$: $\pi/3\le x/\sin x\le\pi/2$ (min at $x=\pi/6$: $(\pi/6)/(1/2)=\pi/3$; max at $x=\pi/2$: $(\pi/2)/1=\pi/2$).

Integrate over length $\pi/2-\pi/6=\pi/3$: $\frac\pi3\cdot\frac\pi3=\frac{\pi^2}{9}<\int_{\pi/6}^{\pi/2}\frac x{\sin x}\,dx<\frac\pi2\cdot\frac\pi3=\frac{\pi^2}{6}<\frac{2\pi^2}{9}.$

$\boxed{\;\frac{\pi^2}{9}<\int_{\pi/6}^{\pi/2}\frac{x}{\sin x}\,dx<\frac{2\pi^2}{9}.\;}\quad\blacksquare$

Common Traps

• Reciprocating reverses the inequality: $h$ decreasing $\Rightarrow$ $1/h$ increasing, so min of $x/\sin x$ is at the left endpoint $\pi/6$.
• The upper bound $\pi^2/6$ is actually sharper than $2\pi^2/9$; state the chain $\pi^2/9<\text{integral}<\pi^2/6<2\pi^2/9$ to be rigorous.
• State that $f$ is non-constant, hence the bounds are strict (not just $\le$).

non-integrable-example (1 question; 2024)

Recognition Cues

• $f$ takes one formula on rationals, another on irrationals.
• “Find upper and lower Riemann integrals; show $f$ is not Riemann integrable.”

Solution Template

1. On any subinterval, rationals and irrationals are both dense, so $\sup f=$ (rational branch value) and $\inf f=$ (irrational branch value) — at any point.
2. Upper integral $=\int$ (rational branch); lower integral $=\int$ (irrational branch).
3. These differ $\Rightarrow$ $f$ not Riemann integrable.

Worked Example(s)

2024 Paper 2, 2024-P2-Q4b (15 marks)

$f(x)=\sqrt{1-x^2}$ on rationals, $f(x)=1-x$ on irrationals. Find $\overline\int_0^1 f$ and $\underline\int_0^1 f$; show $f$ not Riemann integrable.

Comparison. For $x\in(0,1)$: $\sqrt{1-x^2}=\sqrt{(1-x)(1+x)}>(1-x)$ since $1+x>1$.

Sup/inf on any $[\alpha,\beta]$: both branches are dense, so $\sup f=\sqrt{1-\alpha^2}$ (rational branch, max at left since decreasing) and $\inf f=1-\beta$ (irrational branch, min at right).

Upper integral $=\inf_P\sum\sup f_i\,\Delta x_i\to\int_0^1\sqrt{1-x^2}\,dx=\pi/4$ (quarter-circle area).

Lower integral $=\sup_P\sum\inf f_i\,\Delta x_i\to\int_0^1(1-x)\,dx=1/2$.

$\overline\int_0^1 f=\frac\pi4\ne\frac12=\underline\int_0^1 f\;\Longrightarrow\;\boxed{f\text{ is not Riemann integrable.}}$

Common Traps

• The comparison $\sqrt{1-x^2}\ge1-x$ needs justification: $\sqrt{(1-x)(1+x)}\ge(1-x)$ iff $\sqrt{1+x}\ge\sqrt{1-x}$ iff $x\ge0$ ✓.
• The upper integral being $\pi/4$ (quarter-circle area) and lower being $1/2$ (triangle area) — don’t swap them.
• Density of both rationals and irrationals is the load-bearing fact — state it explicitly.

oscillation-proof (1 question; 2023)

Recognition Cues

• “Prove: oscillation of $f$ on $[a,b]$ equals $\sup|f(x_1)-f(x_2)|$.”
• Pure definition-squeezing: two-sided inequality on sup.

Worked Example(s)

2023 Paper 2, 2023-P2-Q4a (15 marks)

Prove: $\omega(f,[a,b])=M-m=\sup\{|f(x_1)-f(x_2)|:x_1,x_2\in[a,b]\}$.

Let $S=\sup\{|f(x_1)-f(x_2)|\}$. Two-sided bound:

$S\le M-m$: for any $x_1,x_2$, $|f(x_1)-f(x_2)|\le M-m$ (since $m\le f\le M$). So $M-m$ is an upper bound and $S\le M-m$.

$S\ge M-m$: fix $\varepsilon>0$. By definition of $\sup$ and $\inf$, $\exists x_1$ with $f(x_1)>M-\varepsilon/2$ and $\exists x_2$ with $f(x_2)<m+\varepsilon/2$. Then $|f(x_1)-f(x_2)|=f(x_1)-f(x_2)>(M-m)-\varepsilon$. Since $\varepsilon$ is arbitrary, $S\ge M-m$.

Hence $S=M-m=\omega(f,[a,b])$. $\blacksquare$

Common Traps

• $M$ and $m$ are not necessarily attained (no continuity assumed). The $\varepsilon$-argument is essential; writing “there exist $x_1,x_2$ with $f(x_1)=M$ and $f(x_2)=m$” is illegitimate in general.
• Handle the trivial case $M=m$ (constant function, oscillation $=0$) explicitly.

Marks-Aware Writing

10-mark questions (2013-Q1c, 2013-Q3d, 2018, 2021): State the relevant criterion (Lebesgue or Riemann), apply it in two lines, conclude. For the antiderivative question: state Darboux’s theorem, identify the jump, conclude no antiderivative. For monotone integrability: equal partition + telescoping argument takes 4–5 lines.

15-mark questions (2014, 2015, 2022, 2023, 2024, 2025): Full proof with all steps shown. For Darboux sums (2022): write $L$ and $U$ with sum formulas, take limits, state conclusion. For continuous (2025): state Heine-Cantor, write the oscillation bound, bound $U-L$. For oscillation (2023): two-sided inequality with the $\varepsilon$-argument for the lower bound.

Practice Set