03 — Integral Calculus

Comprehensive study guide for Bocconi Math Module 2 (cod. 30063), General Exam. Source materials: Alice Sicconi's Proofs - Integral Calculus.pdf, Integral Calculus.pdf, lect{1..9}_calc_int*.pdf, TA41_calc_int{1,2}.pdf, General_24524_ENG_SOL.pdf, Final exercises solution.pdf.

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§1. Overview & Exam Relevance

Integral Calculus is the third block of the first partial and accounts for approximately 18% of the general exam (typically 1 MCQ worth 5 pts and one open-ended question worth up to 20 pts — about 25 pts of the 150-pt exam). It is the topic where a correctly stated First Fundamental Theorem of Calculus (Theorem 1894) is often the single highest-yield 20-pt question on the whole paper.

Topic scope. The exam tests:

• Riemann integration: partitions $\Pi$ of $[a,b]$, lower and upper Riemann sums $I(f,\Pi), S(f,\Pi)$, lower and upper integrals, the integrability criterion (Theorem 1869), integrability of continuous functions (Theorem 1875);
• Elementary properties of the definite integral: linearity, monotonicity, boundedness (Prop. 1884), absolute value inequality (Prop. 1883), and the Integral Mean Value Theorem (Theorem 1885);
• The Fundamental Theorems of Calculus: Newton–Leibniz (Theorem 1894 — "state and prove" on May 2024 Q9), and the fact that $F(x) = \int_a^x f(t)\,dt$ is a primitive of a continuous $f$ (Theorem 1899);
• Antiderivatives / primitives: the indefinite integral $\int f(x)\,dx + C$, linearity of primitives (Theorem 1903), integration by parts (Prop. 1905), integration by substitution (Prop. 1908);
• Standard antiderivatives table: $x^a, 1/x, e^x, \sin x, \cos x, 1/(1+x^2), 1/\sqrt{1-x^2}$ and composition shortcuts;
• Improper integrals of the first kind (infinite bounds) and second kind (unbounded integrand); convergence via the comparison criterion (Theorem 1932), asymptotic comparison (Theorem 1933 / 1928 monotonicity), and absolute convergence (Theorem 1937);
• Cauchy principal value $\operatorname{PV}\int_{-\infty}^{+\infty} f(x)\,dx$ — a symmetric limit that can converge even when the integral diverges;
• Barrow–Torricelli's Theorem (2030): the bijection $T: C^1_0([a,b]) \leftrightarrow C([a,b])$ via differentiation / integration.

Typical MCQ pattern (from May 2024 General exam).

• MCQ3 (Mode A) / MCQ2 (Mode B): "The improper integral $\int_1^{+\infty} \frac{x + \ln x}{x^3}\,dx$ is …" — asymptotically $\frac{x + \ln x}{x^3} \sim \frac{1}{x^2}$ at $+\infty$ so the integral is finite; moreover $\frac{x + \ln x}{x^3} \ge \frac{1}{x^2}$ on $[1, +\infty)$ so by monotonicity of the improper integral (Theorem 1928), $\int_1^{+\infty} \frac{x + \ln x}{x^3}\,dx \ge \int_1^{+\infty} \frac{1}{x^2}\,dx$.

Typical open-ended pattern (May 2024 Q9 Mode A / Q8 Mode B).

• Part (a) — State and prove the First Fundamental Theorem of Calculus (Theorem 1894). This is the flagship proof of Integral Calculus and must be reproduced verbatim.
• Part (b1) — Compute an antiderivative of $f(x) = x^2 \ln x$: integrate by parts with $f'_1 = x^2, g_1 = \ln x$ to get $\int x^2 \ln x\,dx = \frac{x^3}{3}\ln x - \frac{x^3}{9} + C$.
• Part (b2) — Compute $\int_1^e x^2 \ln x\,dx = \left[\frac{x^3}{3}\ln x - \frac{x^3}{9}\right]_1^e = \frac{2}{9}e^3 + \frac{1}{9}$, invoking Theorem 1894.

Why this topic is high-leverage for the final partial.

• The indicator $\int f'(s)\,ds$ shows up again in Mathematical Finance (continuous-compounding capital $M(t) = 1 + \int_0^t f'(s)\,ds$, Final exercises Q5).
• The Riemann integral is the rigorous scaffolding for the expected value of a continuous random variable in Probability (since $E[X] = \int_{\mathbb{R}} x f_X(x)\,dx$).
• Improper integrals of the form $\int_0^{+\infty} \lambda e^{-\lambda x}\,dx$ underpin the exponential distribution.

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§2. Definitions

2.1 Partition (subdivision) $\Pi$ of $[a,b]$

A partition (or subdivision) $\Pi$ of $[a, b]$ is a finite set of points $\Pi = \{x_0, x_1, \ldots, x_n\}, \qquad a = x_0 < x_1 < x_2 < \cdots < x_n = b.$ It breaks $[a,b]$ into $n$ sub-intervals $[x_{i-1}, x_i]$ for $i = 1, \ldots, n$, each of length $\Delta x_i = x_i - x_{i-1}$.

The mesh (or norm) of $\Pi$ is $|\Pi| = \max_{i} \Delta x_i$ — the length of the largest sub-interval. We write $\Pi \in \pi$ for the collection of all possible partitions of $[a,b]$.

A partition $\Pi'$ is a refinement of $\Pi$ if every point of $\Pi$ is also a point of $\Pi'$ (i.e., $\Pi \subseteq \Pi'$). The union $\Pi_1 \cup \Pi_2$ of two partitions is always a refinement of both.

2.2 Tagged partition $(\xi, \Pi)$

A tagged partition is a pair $(\xi, \Pi)$ where $\Pi$ is a partition and $\xi = (\xi_1, \ldots, \xi_n)$ is a choice of tags with $\xi_i \in [x_{i-1}, x_i]$. The tags will be evaluation points for the integrand.

Remark. In Sicconi's notation (and Marinacci's textbook), the lower and upper sums are defined via the infimum and supremum of $f$ on each sub-interval — independent of any tag choice — so the tag drops out of the final Riemann integral. We still introduce the tagged notation because it is the standard setup for the Riemann sum $\sum_i f(\xi_i)\Delta x_i$ that appears in the proof of Theorem 1894 (FTC I).

2.3 Lower and upper Riemann sums $I(f, \Pi)$, $S(f, \Pi)$

Let $f : [a, b] \to \mathbb{R}$ be bounded. For a partition $\Pi$, define on each sub-interval $m_i = \inf_{x \in [x_{i-1}, x_i]} f(x), \qquad M_i = \sup_{x \in [x_{i-1}, x_i]} f(x).$

The lower (Riemann) sum and upper (Riemann) sum are $I(f, \Pi) = \sum_{i=1}^n m_i \cdot \Delta x_i, \qquad S(f, \Pi) = \sum_{i=1}^n M_i \cdot \Delta x_i.$

Geometrically, for a positive $f$, $I(f, \Pi)$ is the total area of the largest inscribed rectangles (heights $m_i$) and $S(f, \Pi)$ is the total area of the smallest circumscribed rectangles (heights $M_i$). One always has $I(f, \Pi) \le S(f, \Pi), \qquad I(f, \Pi) \le I(f, \Pi') \le S(f, \Pi') \le S(f, \Pi) \qquad \text{whenever $\Pi'$ refines $\Pi$.}$ Refining a partition always gives a better approximation of the area.

2.4 Lower and upper integrals

Let $f : [a, b] \to \mathbb{R}$ be bounded. The lower integral and upper integral of $f$ on $[a, b]$ are $\underline{\int_a^b} f(x)\,dx \;=\; \sup_{\Pi \in \pi} I(f, \Pi), \qquad \overline{\int_a^b} f(x)\,dx \;=\; \inf_{\Pi \in \pi} S(f, \Pi).$

One always has $\underline{\int_a^b} f \le \overline{\int_a^b} f$ (Theorem 1857 below).

2.5 Riemann integrability and the Riemann integral

A bounded function $f : [a, b] \to \mathbb{R}$ is Riemann-integrable on $[a, b]$ if $\underline{\int_a^b} f(x)\,dx \;=\; \overline{\int_a^b} f(x)\,dx.$ In that case the common value is the Riemann integral of $f$ on $[a, b]$: $\int_a^b f(x)\,dx \;=\; \underline{\int_a^b} f(x)\,dx \;=\; \overline{\int_a^b} f(x)\,dx.$

We denote by $\mathcal{R}[a,b]$ the set of Riemann-integrable functions on $[a,b]$. It is a linear subspace of the space $F([a,b])$ of all real-valued functions on $[a,b]$, and $\int_a^b \cdot\,dx$ is a linear functional on it (the linearity property).

Counterexample (Dirichlet). The function $f(x) = 1$ if $x \in \mathbb{Q} \cap [a,b]$ and $f(x) = 0$ otherwise is bounded but not Riemann-integrable: on every partition $m_i = 0$ and $M_i = 1$, giving $I(f, \Pi) = 0$ and $S(f, \Pi) = b - a$. So $\underline{\int_a^b} f = 0 \neq b - a = \overline{\int_a^b} f$.

2.6 Oscillation on a sub-interval

The oscillation of $f$ on $[x_{i-1}, x_i]$ is $M_i - m_i = \sup f - \inf f \ge 0$. Theorem 1869 recasts integrability as "oscillations can be made arbitrarily small in total length". For a continuous $f$, the Heine–Cantor theorem (uniform continuity on a compact set) controls the oscillation uniformly — that is the mechanism behind Theorem 1875.

2.7 Primitive (antiderivative) $P$

Let $I$ be an interval (not necessarily closed). A function $P : I \to \mathbb{R}$ is a primitive (or antiderivative) of $f : I \to \mathbb{R}$ if

1. $P$ is differentiable on $I$;
2. $P'(x) = f(x)$ for every $x \in I$.

Theorem (uniqueness up to a constant). If $P_1, P_2$ are primitives of the same $f$ on an interval $I$, then there exists $K \in \mathbb{R}$ with $P_2(x) = P_1(x) + K$ for every $x \in I$.

Existence. If $f$ is continuous on $I$, then $f$ has primitives (this is essentially Theorem 1899). If $f$ has a jump or eliminable discontinuity in $I$, $f$ does not have primitives on $I$; essential discontinuities may or may not block primitivability.

2.8 Indefinite integral $\int f(x)\,dx + C$

The indefinite integral of $f$ is the family of all primitives: $\int f(x)\,dx \;=\; P(x) + C, \qquad C \in \mathbb{R},$ where $P$ is any single primitive of $f$. In the textbook the symbols $\int f(x)\,dx$ and "antiderivative" are used interchangeably, with the understanding that a constant of integration is always present.

2.9 Definite integral $\int_a^b f(x)\,dx$

The definite integral of a Riemann-integrable $f$ on $[a, b]$ is the number $\int_a^b f(x)\,dx$ defined in §2.5. By convention: $\int_b^a f(x)\,dx \;=\; -\int_a^b f(x)\,dx, \qquad \int_a^a f(x)\,dx \;=\; 0.$ These conventions make the additivity with respect to the interval valid for any ordering: $\int_a^c f + \int_c^b f \;=\; \int_a^b f \qquad \forall\, a, b, c \in I \text{ where $f$ is integrable}.$

Crucial distinction. The definite integral is a number (depends on $a$, $b$, and $f$); the indefinite integral is a family of functions (depends only on $f$).

2.10 Positive / negative parts and signed integration

Every function decomposes as $f(x) = f^+(x) - f^-(x)$, where $f^+(x) = \max\{f(x), 0\}$ (positive part) and $f^-(x) = -\min\{f(x), 0\} = \max\{-f(x), 0\}$ (negative part). Both $f^+, f^-$ are non-negative. A bounded $f$ is Riemann-integrable iff both $f^+, f^-$ are, and then $\int_a^b f(x)\,dx \;=\; \int_a^b f^+(x)\,dx \;-\; \int_a^b f^-(x)\,dx.$ Geometrically this is the signed area — area above the $x$-axis minus area below.

2.11 Improper integral of the first kind (infinite bounds)

Let $f : [a, +\infty) \to \mathbb{R}$ be integrable on every $[a, t] \subseteq [a, +\infty)$. The improper integral of the first kind is $\int_a^{+\infty} f(x)\,dx \;=\; \lim_{t \to +\infty} \int_a^t f(x)\,dx$ provided the limit exists in $\mathbb{R} \cup \{\pm \infty\}$. Symmetrically define $\int_{-\infty}^b$. If both bounds are infinite, split at any $c$: $\int_{-\infty}^{+\infty} f(x)\,dx \;=\; \lim_{s \to -\infty} \int_s^c f(x)\,dx \;+\; \lim_{t \to +\infty} \int_c^t f(x)\,dx,$ with the two limits taken independently — both must exist and be finite (or at least not an indeterminate form $+\infty - \infty$) for the integral to make sense.

Convergence / divergence / oscillation.

• Converges if the limit is a finite number $L \in \mathbb{R}$.
• Diverges to $+\infty$ if the limit equals $+\infty$; similarly to $-\infty$.
• Does not exist (oscillates, or yields an indeterminate form) otherwise — e.g., $\int_{-\infty}^{+\infty} (2x - 3)\,dx$ gives $+\infty + (-\infty)$, which does not exist.

2.12 Improper integral of the second kind (unbounded integrand)

Let $f : [a, b) \to \mathbb{R}$ be integrable on every $[a, t] \subseteq [a, b)$ but possibly unbounded as $x \to b^-$. The improper integral of the second kind is $\int_a^b f(x)\,dx \;=\; \lim_{t \to b^-} \int_a^t f(x)\,dx.$ Similarly for a singularity at the left endpoint, or at an interior point (split and take two separate limits).

Typical example. $\int_0^1 \frac{1}{\sqrt{x}}\,dx = \lim_{t \to 0^+} \int_t^1 x^{-1/2}\,dx = \lim_{t \to 0^+} [2\sqrt{x}]_t^1 = 2$, converges; $\int_0^1 \frac{1}{x}\,dx = \lim_{t \to 0^+} (-\ln t) = +\infty$, diverges.

Harmonic rule for finite-singularity improper integrals. $\int_0^1 \frac{1}{x^p}\,dx$ converges iff $p < 1$ (and equals $\frac{1}{1-p}$); diverges iff $p \ge 1$.

2.13 Cauchy principal value (PV)

Let $f : \mathbb{R} \to \mathbb{R}$ be integrable on every $[-K, K]$. The Cauchy principal value of $\int_{-\infty}^{+\infty} f(x)\,dx$ is $\operatorname{PV}\int_{-\infty}^{+\infty} f(x)\,dx \;=\; \lim_{K \to +\infty} \int_{-K}^{K} f(x)\,dx$ provided this single symmetric limit exists. The definition uses one parameter $K$ for both bounds.

Key fact. If the classical improper integral $\int_{-\infty}^{+\infty} f(x)\,dx$ converges, then its PV equals the classical value. However, the PV may exist even when the classical improper integral diverges — because the symmetric cancellation can kill a divergence. See Example 4.6 below ($\operatorname{PV}\int_{-\infty}^{+\infty}(2x - 3)\,dx$).

A similar PV definition applies at a finite singularity: $\operatorname{PV}\int_a^b f = \lim_{\varepsilon \to 0^+}\bigl(\int_a^{c - \varepsilon} + \int_{c + \varepsilon}^b\bigr)$ with the same $\varepsilon$ on both sides of an interior singularity $c$.

2.14 Integral function $F(x) = \int_a^x f(t)\,dt$

Let $f : [a, b] \to \mathbb{R}$ be integrable. The integral function of $f$ with base point $a$ is $F : [a, b] \to \mathbb{R}, \qquad F(x) \;=\; \int_a^x f(t)\,dt.$ (Here $t$ is a dummy integration variable and $x$ is the upper bound.) By convention $F(a) = 0$.

Properties.

• $F$ is always Lipschitz continuous on $[a,b]$ (and hence uniformly continuous): $|F(x) - F(y)| \le M|x - y|$ where $M = \sup|f|$. Proof sketch. $|F(x) - F(y)| = |\int_y^x f\,dt| \le \int_y^x |f|\,dt \le M|x-y|$ using the absolute-value property 1883.
• If moreover $f$ is continuous, then $F$ is differentiable and $F'(x) = f(x)$ (Theorem 1899, Second FTC) — so $F$ is a primitive of $f$.

Geometric interpretation. $F(x)$ is the signed area between the graph of $f$ and the $x$-axis from $a$ to $x$. Increasing $x$ adds more area on the right.

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§3. Theorems, Propositions & Proofs

3.1 Theorem 1857 — Lower and upper integrals are finite

Statement. Let $f : [a, b] \to \mathbb{R}$ be positive and bounded on $[a, b]$. Then (1) the lower and upper integrals $\underline{\int_a^b} f(x)\,dx$ and $\overline{\int_a^b} f(x)\,dx$ are both finite; (2) $\underline{\int_a^b} f(x)\,dx \le \overline{\int_a^b} f(x)\,dx$.

Source: Proofs - Integral Calculus.pdf p.1 (handwritten theorem 1857).

Proof.

$f$ is positive and bounded, so there exists $M > 0$ with $0 \le f(x) \le M$ for every $x \in [a, b]$. For any subdivision $\Pi = \{x_0, \ldots, x_n\}$ and every $i$, $0 \le m_i \le M_i \le M$. Therefore $0 \;=\; \sum_{i=1}^n 0 \cdot \Delta x_i \;\le\; \sum_{i=1}^n m_i \cdot \Delta x_i \;\le\; \sum_{i=1}^n M_i \cdot \Delta x_i \;\le\; \sum_{i=1}^n M \cdot \Delta x_i \;=\; M \sum_{i=1}^n \Delta x_i \;=\; M(b - a).$ That is, 0 \;\le\; I(f, \Pi) \;\le\; S(f, \Pi) \;\le\; M(b - a) \qquad \forall\, \Pi \in \pi. \tag{$\ast$} Taking supremum over $\Pi$ in the first inequality of $(\ast)$: $0 \;\le\; \underline{\int_a^b} f(x)\,dx \;=\; \sup_{\Pi \in \pi} I(f, \Pi) \;\le\; M(b - a) \;<\; +\infty \qquad \text{(finite).}$ Taking infimum over $\Pi$ in the last inequality of $(\ast)$: $0 \;\le\; \overline{\int_a^b} f(x)\,dx \;=\; \inf_{\Pi \in \pi} S(f, \Pi) \;\le\; M(b - a) \;<\; +\infty \qquad \text{(finite).}$ This proves (1).

For (2), suppose by contradiction that $\underline{\int_a^b} f > \overline{\int_a^b} f$ and set $\delta \;:=\; \underline{\int_a^b} f(x)\,dx \;-\; \overline{\int_a^b} f(x)\,dx \;>\; 0.$ Using the properties of infimums and supremums, for any $\varepsilon > 0$ (take $\varepsilon = \delta / 2$) there exist partitions $\Pi', \Pi''$ such that $I(f, \Pi') \;>\; \underline{\int_a^b} f(x)\,dx \;-\; \frac{\delta}{2}, \qquad S(f, \Pi'') \;<\; \overline{\int_a^b} f(x)\,dx \;+\; \frac{\delta}{2}.$ Subtracting the two inequalities: $I(f, \Pi') - S(f, \Pi'') \;>\; \Bigl(\underline{\int_a^b} f - \frac{\delta}{2}\Bigr) - \Bigl(\overline{\int_a^b} f + \frac{\delta}{2}\Bigr) \;=\; \underline{\int_a^b} f - \overline{\int_a^b} f - \delta \;=\; \delta - \delta \;=\; 0.$

Now consider the common refinement $\Pi = \Pi' \cup \Pi''$. Refinement monotonicity gives $I(f, \Pi) \ge I(f, \Pi')$ and $S(f, \Pi) \le S(f, \Pi'')$. Combining: $I(f, \Pi) - S(f, \Pi) \;\ge\; I(f, \Pi') - S(f, \Pi'') \;>\; 0,$ i.e., $I(f, \Pi) > S(f, \Pi)$ — which is absurd since the lower sum never exceeds the upper sum. Therefore $\underline{\int_a^b} f \le \overline{\int_a^b} f$. $\blacksquare$

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3.2 Theorem 1869 — Riemann integrability criterion ($S - I < \varepsilon$)

Statement. Let $f : [a, b] \to \mathbb{R}$ be bounded. Then $f$ is Riemann-integrable on $[a, b]$ if and only if for every $\varepsilon > 0$ there exists a subdivision $\Pi$ of $[a, b]$ such that $S(f, \Pi) \;-\; I(f, \Pi) \;<\; \varepsilon.$

Source: Proofs - Integral Calculus.pdf p.2 (handwritten theorem 1869).

Proof.

($\Rightarrow$) Riemann integrability $\Longrightarrow$ existence of a partition with $S - I < \varepsilon$.

Assume $f$ is Riemann-integrable, so $\underline{\int_a^b} f = \overline{\int_a^b} f$. Fix $\varepsilon > 0$ and set $\delta = \varepsilon / 2 > 0$. Using the properties of infimums and supremums, there exist partitions $\Pi', \Pi''$ with $I(f, \Pi') \;>\; \underline{\int_a^b} f \;-\; \delta, \qquad S(f, \Pi'') \;<\; \overline{\int_a^b} f \;+\; \delta.$ Consider the common refinement $\Pi = \Pi' \cup \Pi''$. By refinement monotonicity, $I(f, \Pi) \ge I(f, \Pi')$ and $S(f, \Pi) \le S(f, \Pi'')$. Hence $S(f, \Pi) - I(f, \Pi) \;\le\; S(f, \Pi'') - I(f, \Pi') \;<\; \Bigl(\overline{\int_a^b} f + \delta\Bigr) - \Bigl(\underline{\int_a^b} f - \delta\Bigr) \;=\; \underbrace{\overline{\int_a^b} f - \underline{\int_a^b} f}_{= 0} \;+\; 2\delta \;=\; 2\delta.$ The underbraced term is zero by integrability. With $\delta = \varepsilon / 2$, this gives $S(f, \Pi) - I(f, \Pi) < \varepsilon$. $\checkmark$

($\Leftarrow$) Existence of such partitions for every $\varepsilon > 0$ $\Longrightarrow$ Riemann integrability.

Assume: $\forall\,\varepsilon > 0$, $\exists$ a partition $\Pi$ of $[a,b]$ with $S(f, \Pi) - I(f, \Pi) < \varepsilon$. For every $\Pi$: $\underline{\int_a^b} f \;\ge\; I(f, \Pi), \qquad \overline{\int_a^b} f \;\le\; S(f, \Pi)$ (by the sup / inf definitions). Therefore $\overline{\int_a^b} f \;-\; \underline{\int_a^b} f \;\le\; S(f, \Pi) - I(f, \Pi) \;<\; \varepsilon.$ Since this holds for every $\varepsilon > 0$ and the difference $\overline{\int_a^b} f - \underline{\int_a^b} f \ge 0$ by Theorem 1857, we conclude $\overline{\int_a^b} f \;=\; \underline{\int_a^b} f,$ i.e., $f$ is Riemann-integrable. $\blacksquare$

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3.3 Theorem 1875 — Every continuous function on $[a,b]$ is Riemann-integrable

Statement. Every continuous function $f : [a, b] \to \mathbb{R}$ is Riemann-integrable on $[a, b]$.

Source: Proofs - Integral Calculus.pdf p.3 (handwritten theorem 1875).

Proof.

Because $f$ is continuous on the compact interval $[a, b]$:

• By the Heine–Cantor theorem, $f$ is uniformly continuous on $[a, b]$.
• By the Weierstrass theorem, $f$ attains a minimum and maximum on $[a, b]$, hence $f$ is bounded.

Fix $\varepsilon > 0$. By uniform continuity there exists $\delta = \delta(\varepsilon) > 0$ such that $\forall\,x, y \in [a, b], \quad |x - y| < \delta \;\Longrightarrow\; |f(x) - f(y)| < \varepsilon.$

Choose a subdivision $\Pi = \{x_0, \ldots, x_n\}$ of $[a, b]$ with mesh $|\Pi| < \delta$. Then for every $i$ and every $x, y \in [x_{i-1}, x_i]$ we have $|x - y| \le \Delta x_i \le |\Pi| < \delta$, so $|f(x) - f(y)| < \varepsilon$.

Applying Weierstrass on each sub-interval $[x_{i-1}, x_i]$ (a compact), $f$ attains its min and max there: $M_i - m_i \;=\; \max_{x \in [x_{i-1}, x_i]} f(x) \;-\; \min_{x \in [x_{i-1}, x_i]} f(x) \;<\; \varepsilon.$

Therefore $S(f, \Pi) - I(f, \Pi) \;=\; \sum_{i=1}^n M_i \Delta x_i - \sum_{i=1}^n m_i \Delta x_i \;=\; \sum_{i=1}^n (M_i - m_i)\,\Delta x_i \;<\; \sum_{i=1}^n \varepsilon \cdot \Delta x_i \;=\; \varepsilon \sum_{i=1}^n \Delta x_i \;=\; \varepsilon (b - a).$

We have shown: for every $\varepsilon > 0$, there exists a partition $\Pi$ with $S(f, \Pi) - I(f, \Pi) < \varepsilon(b - a)$. To match the hypothesis of Theorem 1869 exactly, rescale: given any target $\varepsilon^\ast > 0$, apply the above with $\varepsilon' = \varepsilon^\ast / (b - a)$; we obtain a partition $\Pi'$ with $S(f, \Pi') - I(f, \Pi') < \varepsilon'(b - a) = \varepsilon^\ast$. By Theorem 1869, $f$ is Riemann-integrable on $[a, b]$. $\blacksquare$

Remark. Other useful integrability criteria in the course (not proved here, but good to know):

• If $f$ is bounded on $[a,b]$ and has a finite or countable number of discontinuities, then $f$ is integrable.
• If $f$ is monotonic on $[a,b]$, then $f$ is integrable (monotonic functions have at most countably many discontinuities).
• $f$ unbounded on $[a,b]$ $\Rightarrow$ $f$ not Riemann-integrable (requires the improper framework).

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3.4 Property 1883 — Absolute value property for definite integrals

Statement. Let $f : [a, b] \to \mathbb{R}$ be integrable (so $|f|$ is also integrable). Then $\Bigl|\int_a^b f(x)\,dx\Bigr| \;\le\; \int_a^b |f(x)|\,dx.$

Source: Proofs - Integral Calculus.pdf p.7 (handwritten theorem 1883).

Proof.

Since $f(x) \le |f(x)|$ and $-f(x) \le |f(x)|$ pointwise on $[a,b]$, the monotonicity property (Prop. 1882) gives $\int_a^b f(x)\,dx \;\le\; \int_a^b |f(x)|\,dx, \qquad \int_a^b (-f)(x)\,dx \;\le\; \int_a^b |f(x)|\,dx.$ By linearity (homogeneity with $\alpha = -1$), $\int_a^b (-f) = -\int_a^b f$, so the second inequality is $-\int_a^b f(x)\,dx \;\le\; \int_a^b |f(x)|\,dx.$ Together: $-\int_a^b |f|\,dx \;\le\; \int_a^b f\,dx \;\le\; \int_a^b |f|\,dx \qquad \Longleftrightarrow \qquad \Bigl|\int_a^b f\,dx\Bigr| \;\le\; \int_a^b |f|\,dx. \qquad \blacksquare$

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3.5 Property 1884 — Boundedness of the definite integral

Statement. Let $f : [a, b] \to \mathbb{R}$ be integrable. If $m = \inf_{[a,b]} f(x)$ and $M = \sup_{[a,b]} f(x)$, then $m(b - a) \;\le\; \int_a^b f(x)\,dx \;\le\; M(b - a).$

Source: Proofs - Integral Calculus.pdf p.8 (handwritten theorem 1884).

Proof.

Since $m \le f(x) \le M$ for every $x \in [a, b]$, by the monotonicity property $\int_a^b m\,dx \;\le\; \int_a^b f(x)\,dx \;\le\; \int_a^b M\,dx.$ But $\int_a^b m\,dx = m \int_a^b 1\,dx = m \cdot [x]_a^b = m(b - a)$, and similarly $\int_a^b M\,dx = M(b - a)$. Therefore $m(b - a) \;\le\; \int_a^b f(x)\,dx \;\le\; M(b - a). \qquad \blacksquare$

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3.6 Theorem 1885 — Integral Mean Value Theorem

Statement. Let $f : [a, b] \to \mathbb{R}$ be bounded and integrable, with $m = \inf f$ and $M = \sup f$. Then: (1) there exists $\lambda \in [m, M]$ such that $\int_a^b f(x)\,dx = \lambda(b - a)$; (2) if moreover $f$ is continuous on $[a, b]$, then there exists $c \in [a, b]$ with $f(c) = \lambda$, i.e., $\int_a^b f(x)\,dx \;=\; f(c)(b - a) \qquad \text{for some } c \in [a, b].$

Source: Proofs - Integral Calculus.pdf p.9 (handwritten theorem 1885).

Proof of (1). By the boundedness property 1884, $m(b - a) \le \int_a^b f \le M(b - a)$. Divide by $b - a > 0$: $m \;\le\; \frac{\int_a^b f(x)\,dx}{b - a} \;\le\; M.$ Set $\lambda := \dfrac{\int_a^b f(x)\,dx}{b - a}$. Then $\lambda \in [m, M]$ and $\int_a^b f(x)\,dx = \lambda(b - a)$ by construction.

Proof of (2). If $f$ is continuous on $[a, b]$, by the Intermediate Value Theorem $f$ takes every value in $[\min f, \max f] = [m, M]$. Since $\lambda \in [m, M]$, there is $c \in [a, b]$ with $f(c) = \lambda$. Substituting into (1): $\int_a^b f(x)\,dx \;=\; f(c)(b - a). \qquad \blacksquare$

Why this matters. This is the engine of the proof of Theorem 1899 (Second FTC). The mean value $\lambda$ is the "average height" of $f$; $f(c) \cdot (b-a)$ says there exists a horizontal line at height $f(c)$ that encloses the same area under it as $f$ does.

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3.7 Theorem 1894 — First Fundamental Theorem of Calculus (Newton–Leibniz) ⭐

Statement. Let $f : [a, b] \to \mathbb{R}$. If (1) $f$ is integrable on $[a, b]$, and (2) $f$ has primitives on $[a, b]$ and $P : [a, b] \to \mathbb{R}$ is one of them (so $P'(x) = f(x)$ on $[a,b]$),

then $\int_a^b f(x)\,dx \;=\; \bigl[P(x)\bigr]_a^b \;=\; P(b) - P(a).$

Source: Proofs - Integral Calculus.pdf p.11 (handwritten theorem 1894).

⭐ This is the proof most likely to be requested on the exam (May 2024 Q9).

Proof.

Consider an arbitrary subdivision $\Pi = \{x_0, x_1, \ldots, x_n\}$ of $[a, b]$ with $a = x_0 < x_1 < \cdots < x_n = b$.

Step 1 — Telescoping sum. Write $P(b) - P(a)$ as a telescoping sum by inserting and cancelling all intermediate $P(x_i)$: $P(b) - P(a) \;=\; P(x_n) - P(x_{n-1}) + P(x_{n-1}) - \cdots - P(x_1) + P(x_1) - P(x_0) \;=\; \sum_{i=1}^n \bigl[P(x_i) - P(x_{i-1})\bigr].$ All intermediate terms cancel in pairs: $+P(x_{n-1}) - P(x_{n-1}) = 0$, etc.

Step 2 — Apply Lagrange's Mean Value Theorem on each sub-interval. $P$ is a primitive of $f$, so $P$ is differentiable on $(x_{i-1}, x_i)$ and continuous on $[x_{i-1}, x_i]$ for every $i$. By Lagrange's Mean Value Theorem, for each $i \in \{1, \ldots, n\}$ there exists $\xi_i \in [x_{i-1}, x_i]$ such that $P'(\xi_i) \;=\; \frac{P(x_i) - P(x_{i-1})}{x_i - x_{i-1}}.$ Since $P$ is a primitive of $f$, we have $P'(\xi_i) = f(\xi_i)$. Rearranging: $P(x_i) - P(x_{i-1}) \;=\; f(\xi_i)(x_i - x_{i-1}) \;=\; f(\xi_i) \Delta x_i.$

Step 3 — Recognise the Riemann sum. Substituting into the telescoping sum: $P(b) - P(a) \;=\; \sum_{i=1}^n \bigl[P(x_i) - P(x_{i-1})\bigr] \;=\; \sum_{i=1}^n f(\xi_i) \Delta x_i.$ This is a Riemann sum for $f$: it is an approximation of the definite integral $\int_a^b f(x)\,dx$, sandwiched between the lower and upper sums. Indeed, for each $i$: $m_i \;\le\; f(\xi_i) \;\le\; M_i \qquad \Longrightarrow \qquad m_i \Delta x_i \;\le\; f(\xi_i)\Delta x_i \;\le\; M_i \Delta x_i,$ and summing, $I(f, \Pi) \;\le\; \sum_{i=1}^n f(\xi_i)\Delta x_i \;\le\; S(f, \Pi).$ Combined with Step 2: I(f, \Pi) \;\le\; P(b) - P(a) \;\le\; S(f, \Pi) \qquad \forall\, \Pi \in \pi. \tag{$\star$}

Step 4 — Pass to the sup / inf. The value $P(b) - P(a)$ does not depend on $\Pi$. Taking supremum over $\Pi$ in the left inequality and infimum in the right inequality of $(\star)$: $\sup_{\Pi \in \pi} I(f, \Pi) \;\le\; P(b) - P(a) \;\le\; \inf_{\Pi \in \pi} S(f, \Pi),$ i.e., $\underline{\int_a^b} f(x)\,dx \;\le\; P(b) - P(a) \;\le\; \overline{\int_a^b} f(x)\,dx.$

Step 5 — Conclude using integrability. By hypothesis $f$ is integrable, so $\underline{\int_a^b} f = \overline{\int_a^b} f = \int_a^b f$. Squeezed between two equal quantities: $\int_a^b f(x)\,dx \;=\; P(b) - P(a) \;=\; \bigl[P(x)\bigr]_a^b. \qquad \blacksquare$

Exam tip. When reproducing this proof, make sure you explicitly flag the three ingredients: (i) integrability of $f$ so that upper = lower, (ii) primitivability so that $P$ exists and is differentiable, (iii) Lagrange's MVT on each sub-interval to turn $P$-differences into $f$-values.

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3.8 Theorem 1899 — Second Fundamental Theorem of Calculus

Statement. Let $f : [a, b] \to \mathbb{R}$ be continuous. Then its integral function $F(x) \;=\; \int_a^x f(t)\,dt$ is a primitive of $f$ on $[a, b]$: $F$ is differentiable on $[a, b]$ with $F'(x) \;=\; f(x) \qquad \forall\, x \in [a, b].$

Source: Proofs - Integral Calculus.pdf p.13 (handwritten theorem 1899). See also p.12 for the preliminary result that $F$ is Lipschitz-continuous for any integrable $f$.

Proof.

Let $x_0 \in [a, b]$. We show $F'(x_0) = f(x_0)$ by computing the right and left derivatives separately and showing they agree.

Right derivative ($h > 0$). For $x_0, x_0 + h \in [a, b]$ with $h > 0$: $F(x_0 + h) - F(x_0) \;=\; \int_a^{x_0 + h} f(t)\,dt - \int_a^{x_0} f(t)\,dt \;=\; \int_{x_0}^{x_0 + h} f(t)\,dt.$ The last equality uses additivity with respect to the interval.

Since $f$ is continuous on $[x_0, x_0 + h]$ (a compact sub-interval of $[a,b]$), by the Integral Mean Value Theorem (Theorem 1885, part 2), there exists $c \in [x_0, x_0 + h]$ with $\int_{x_0}^{x_0 + h} f(t)\,dt \;=\; f(c) \cdot h.$ Hence $\frac{F(x_0 + h) - F(x_0)}{h} \;=\; \frac{f(c) \cdot h}{h} \;=\; f(c).$

Take $h \to 0^+$: $c \in [x_0, x_0 + h]$ forces $c \to x_0^+$, and since $f$ is continuous at $x_0$, $f(c) \to f(x_0)$. Therefore $F'_+(x_0) \;=\; \lim_{h \to 0^+} \frac{F(x_0 + h) - F(x_0)}{h} \;=\; \lim_{c \to x_0^+} f(c) \;=\; f(x_0).$

Left derivative ($h < 0$). Repeat the procedure with $h < 0$: using $F(x_0 + h) - F(x_0) \;=\; -\bigl[F(x_0) - F(x_0 + h)\bigr] \;=\; -\int_{x_0 + h}^{x_0} f(t)\,dt,$ and applying the Integral Mean Value Theorem on $[x_0 + h, x_0]$, there exists $c \in [x_0 + h, x_0]$ with $\int_{x_0 + h}^{x_0} f(t)\,dt \;=\; f(c) \cdot (-h).$ Hence $\frac{F(x_0 + h) - F(x_0)}{h} \;=\; \frac{-f(c) \cdot (-h)}{h} \;=\; f(c),$ and $c \to x_0^-$ as $h \to 0^-$, so $F'_-(x_0) = f(x_0)$ by continuity.

Conclusion. $F'_+(x_0) = F'_-(x_0) = f(x_0)$, so $F'(x_0) = f(x_0)$. Since $x_0 \in [a,b]$ was arbitrary, $F'(x) = f(x)$ on $[a,b]$. For the endpoints $x_0 = a$ or $x_0 = b$, the one-sided limit is the relevant definition of the derivative. Therefore $F$ is a primitive of $f$. $\blacksquare$

Corollary. If $f$ is continuous on $[a,b]$, then $f$ has primitives on $[a,b]$ — Theorem 1899 produces one explicitly.

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3.9 Theorem 1903 — Linearity of primitives / indefinite integrals

Statement. Let $I \subseteq \mathbb{R}$ be an interval and $f, g : I \to \mathbb{R}$ be functions that have primitives on $I$. Then for any $\alpha, \beta \in \mathbb{R}$, the linear combination $\alpha f + \beta g$ also has primitives on $I$, and $\int (\alpha f + \beta g)(x)\,dx \;=\; \alpha \int f(x)\,dx \;+\; \beta \int g(x)\,dx \;+\; K \qquad \text{for some } K \in \mathbb{R}.$

Source: Proofs - Integral Calculus.pdf p.15 (handwritten theorem 1903).

Proof.

Let $P_f, P_g$ be primitives of $f, g$ respectively on $I$, so $P_f' = f$ and $P_g' = g$. Compute the derivative of the linear combination $\alpha P_f + \beta P_g$: $\bigl[\alpha P_f(x) + \beta P_g(x)\bigr]' \;=\; \alpha\bigl[P_f(x)\bigr]' \;+\; \beta\bigl[P_g(x)\bigr]' \;=\; \alpha f(x) + \beta g(x) \;=\; (\alpha f + \beta g)(x).$ The first equality is the linearity of the derivative.

Therefore $\alpha P_f + \beta P_g$ is a primitive of $\alpha f + \beta g$. Since any two primitives of the same function differ by a constant, we get $\int (\alpha f + \beta g)(x)\,dx \;=\; \alpha P_f(x) + \beta P_g(x) + K \;=\; \alpha \int f(x)\,dx + \beta \int g(x)\,dx + K. \qquad \blacksquare$

Corollary (linearity of definite integrals). Combined with Theorem 1894, we obtain $\int_a^b [\alpha f(x) + \beta g(x)]\,dx \;=\; \alpha \int_a^b f(x)\,dx + \beta \int_a^b g(x)\,dx.$

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3.10 Proposition 1905 — Integration by parts

Statement. Let $f, g : I \to \mathbb{R}$ be differentiable functions on an interval $I$. Then $\int f'(x) g(x)\,dx + \int f(x) g'(x)\,dx \;=\; f(x) g(x) + K \qquad \text{for some } K \in \mathbb{R},$ equivalently, $\boxed{\int f'(x) g(x)\,dx \;=\; f(x) g(x) - \int f(x) g'(x)\,dx.}$

Source: Proofs - Integral Calculus.pdf p.16 (handwritten).

Proof.

Start from the product rule (Leibniz) for the derivative: $\bigl[f(x) \cdot g(x)\bigr]' \;=\; f'(x) g(x) + f(x) g'(x).$

Integrate both sides on the interval $I$. The left-hand side integrates to a primitive of itself, so $f(x) \cdot g(x) \;=\; \int \bigl[f'(x) g(x) + f(x) g'(x)\bigr]\,dx.$

By linearity (Theorem 1903): $f(x) \cdot g(x) \;=\; \int f'(x) g(x)\,dx + \int f(x) g'(x)\,dx + h \qquad \text{for some } h \in \mathbb{R}.$

Setting $K := -h$ and rearranging: $f(x) \cdot g(x) + K \;=\; \int f'(x) g(x)\,dx + \int f(x) g'(x)\,dx. \qquad \blacksquare$

Practical formula. The version used in practice is $\int f'g\,dx = fg - \int fg'\,dx$. The strategy is to choose $g$ (the factor to be differentiated, so $g \to g'$) to be the one that becomes simpler after differentiation, and $f'$ (the factor to be integrated, so $f' \to f$) to be the one with a known antiderivative. Use the LIATE heuristic (see M-IC-1 in §5).

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3.11 Proposition 1908 — Integration by substitution

Statement. Let $I, J \subseteq \mathbb{R}$ be intervals. Assume (1) $f : I \to \mathbb{R}$ is continuous; (2) $\ell : [c, d] \subseteq J \to \mathbb{R}$ is $C^{(1)}$, strictly increasing, with $\ell(c) = a$ and $\ell(d) = b$ (so $\ell([c,d]) = [a,b] \subseteq I$).

Then: (i) $\ell$ is invertible on $[c, d]$ with inverse $\ell^{-1}$, and $c = \ell^{-1}(a)$, $d = \ell^{-1}(b)$; (ii) $(f \circ \ell) \cdot \ell' : [c, d] \to \mathbb{R}$ is integrable; (iii) the substitution formula holds: $\int_a^b f(x)\,dx \;=\; \int_c^d f(\ell(t)) \cdot \ell'(t)\,dt \;=\; \int_{\ell^{-1}(a)}^{\ell^{-1}(b)} f(\ell(t)) \cdot \ell'(t)\,dt.$

Source: Proofs - Integral Calculus.pdf p.17 (handwritten).

Proof.

(i) $\ell$ is strictly increasing, hence injective; on the domain $[c, d]$ it is bijective onto its image $[a, b]$, and so admits an inverse $\ell^{-1} : [a, b] \to [c, d]$. From $\ell(c) = a$ and $\ell(d) = b$ we read off $c = \ell^{-1}(a)$ and $d = \ell^{-1}(b)$.

(ii) Consider the integral function $F(x) \;=\; \int_a^x f(t)\,dt, \qquad x \in [a, b].$ By the First FTC (Theorem 1894), $\int_a^b f(x)\,dx = F(b) - F(a) = F(b)$ (since $F(a) = 0$).

Applying the chain rule to $F \circ \ell$: $\bigl[F \circ \ell\bigr]'(t) \;=\; \bigl[F' \circ \ell\bigr](t) \cdot \ell'(t) \;=\; F'(\ell(t)) \cdot \ell'(t).$ By the Second FTC (Theorem 1899) — which applies because $f$ is continuous — $F'(x) = f(x)$ on $[a, b]$. Substituting: $\bigl[F \circ \ell\bigr]'(t) \;=\; f(\ell(t)) \cdot \ell'(t) \;=\; (f \circ \ell)(t) \cdot \ell'(t).$

This expression is integrable on $[c, d]$: $f \circ \ell$ is continuous (composition of continuous functions), and $\ell' \in C^{([c,d])}$, so their product is continuous, hence integrable by Theorem 1875.

(iii) Apply the First FTC to the continuous integrand $(f \circ \ell) \cdot \ell'$ on $[c, d]$, with primitive $F \circ \ell$: $\int_c^d (f \circ \ell)(t) \cdot \ell'(t)\,dt \;=\; \bigl[F \circ \ell\bigr]_c^d \;=\; F(\ell(d)) - F(\ell(c)) \;=\; F(b) - F(a) \;=\; F(b) \;=\; \int_a^b f(x)\,dx. \qquad \blacksquare$

Mnemonic. The substitution $x = \ell(t)$ with $dx = \ell'(t)\,dt$ transforms the integral: "replace $x$ by $\ell(t)$, and the measure $dx$ by $\ell'(t)\,dt$; change the bounds via $\ell^{-1}$." The strict monotonicity hypothesis guarantees the bounds are transported correctly (no folding). For a general substitution (not necessarily increasing), one has the same formula but with absolute value bars on $\ell'$ if the domains overlap.

Indefinite version. $\int f'(x) g[f(x)]\,dx = G[f(x)] + C$, where $G$ is an antiderivative of $g$ — this is the most commonly used form on the exam, as a pattern-matching shortcut.

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3.12 Theorem 1928 — Monotonicity of improper integrals

Statement. Let $f, g : [a, +\infty) \to \mathbb{R}$ be integrable on every $[a, t] \subseteq [a, +\infty)$, and assume $f(x) \le g(x)$ on $[a, +\infty)$. Then $\int_a^{+\infty} f(x)\,dx \;\le\; \int_a^{+\infty} g(x)\,dx,$ whenever both sides exist (possibly $\pm\infty$). In particular: (1) if $\int_a^{+\infty} g$ converges to a finite $L$, then $\int_a^{+\infty} f \le L$; (2) if $\int_a^{+\infty} f = +\infty$, then $\int_a^{+\infty} g = +\infty$.

Source: Integral Calculus.pdf p.46 (Sicconi "Comparison Test") and Proofs - Integral Calculus.pdf p.19 (comparison criterion, Theorem 1932). Cited on May 2024 MCQ3 solution as "theorem 1928".

Proof.

For every $t \ge a$, the (ordinary) definite-integral monotonicity (Prop. 1882) applied on $[a,t]$ gives $\int_a^t f(x)\,dx \;\le\; \int_a^t g(x)\,dx.$ Take $t \to +\infty$: if both limits exist (the hypothesis "integrable on every $[a,t]$" plus eventual positivity ensures monotone convergence gives a limit in $[0, +\infty]$), then passing to the limit in the above inequality preserves $\le$, yielding $\int_a^{+\infty} f(x)\,dx \;\le\; \int_a^{+\infty} g(x)\,dx. \qquad \blacksquare$

Application to May 2024 MCQ3 (Mode A). The integrand $\frac{x + \ln x}{x^3}$ is positive on $[1, +\infty)$. Asymptotically $\frac{x + \ln x}{x^3} \sim \frac{x}{x^3} = \frac{1}{x^2}$ at $+\infty$, so by asymptotic comparison the integral converges (since $\int_1^{+\infty} \frac{1}{x^2}\,dx = 1$ is finite). Moreover, on $[1, +\infty)$ we have $\ln x \ge 0$, so $\frac{x + \ln x}{x^3} \ge \frac{x}{x^3} = \frac{1}{x^2}$. By Theorem 1928, $\int_1^{+\infty} \frac{x + \ln x}{x^3}\,dx \;\ge\; \int_1^{+\infty} \frac{1}{x^2}\,dx.$ Answer: D — "finite, but greater or equal than $\int_1^{+\infty} \frac{1}{x^2}\,dx$".

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3.13 Theorem 1932 — Comparison criterion for improper integrals

Statement. Let $f, g : [a, +\infty) \to \mathbb{R}$ be positive and integrable on every $[a, b] \subseteq [a, +\infty)$. Assume $f(x) \le g(x)$ on $[a, +\infty)$. Then: (1) if $\int_a^{+\infty} g(x)\,dx = L \ge 0$ is finite, then $\int_a^{+\infty} f(x)\,dx = K \le L$ is also finite; (2) if $\int_a^{+\infty} f(x)\,dx = +\infty$, then $\int_a^{+\infty} g(x)\,dx = +\infty$.

Source: Proofs - Integral Calculus.pdf p.19 (handwritten theorem).

Proof.

By the "existence of the improper integral" theorem for positive functions (integrability on every $[a, t]$ plus positivity gives a monotone-increasing integral function that has a limit in $[0, +\infty]$), both improper integrals $\int_a^{+\infty} f$ and $\int_a^{+\infty} g$ exist and are $\ge 0$.

By the monotonicity theorem (Theorem 1928), since $f \le g$, $\int_a^{+\infty} f(x)\,dx \;\le\; \int_a^{+\infty} g(x)\,dx.$

Therefore: (1) If $\int_a^{+\infty} g$ converges (to a finite $L$), then $\int_a^{+\infty} f \le L < +\infty$ too, hence it converges. (2) If $\int_a^{+\infty} f = +\infty$, then $\int_a^{+\infty} g \ge +\infty = +\infty$. $\blacksquare$

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3.14 Asymptotic comparison criterion (Theorem 1933)

Statement. Let $f, g : [a, +\infty) \to \mathbb{R}$ be positive and integrable on every $[a, b] \subseteq [a, +\infty)$. Then: (1) if $f \sim g$ as $x \to +\infty$ (i.e., $\lim_{x \to +\infty} f(x) / g(x) = 1$), then $\int_a^{+\infty} f$ and $\int_a^{+\infty} g$ have the same convergence behaviour: they both converge to positive numbers, or both diverge to $+\infty$; (2) if $f = o(g)$ as $x \to +\infty$ (i.e., $\lim_{x \to +\infty} f/g = 0$), then: (a) if $\int_a^{+\infty} g = L \ge 0$ finite, then $\int_a^{+\infty} f = K \le L$ finite; (b) if $\int_a^{+\infty} f = +\infty$, then $\int_a^{+\infty} g = +\infty$.

Source: Proofs - Integral Calculus.pdf p.20 (handwritten theorem 1933).

Proof.

(1) If $f \sim g$, then $\lim_{x \to +\infty} f(x)/g(x) = 1$. By the definition of limit, for $\varepsilon = 1/2$ there exists $B > 0$ such that for every $x \ge B$: $\frac{1}{2} \;\le\; \frac{f(x)}{g(x)} \;\le\; \frac{3}{2},$ hence (since $g > 0$): $\frac{1}{2} g(x) \;\le\; f(x) \;\le\; \frac{3}{2} g(x) \qquad \forall\, x \ge B.$

By the comparison criterion (Theorem 1932) applied on $[B, +\infty)$ (and noting that the behaviour on the bounded $[a, B]$ does not affect convergence at infinity):

• If $\int_B^{+\infty} \tfrac{3}{2} g = \tfrac{3}{2} \int_B^{+\infty} g$ converges (so $\int_B^{+\infty} g$ converges), then $\int_B^{+\infty} f$ converges.
• If $\int_B^{+\infty} \tfrac{1}{2} g = \tfrac{1}{2} \int_B^{+\infty} g$ diverges to $+\infty$ (so $\int_B^{+\infty} g$ diverges), then $\int_B^{+\infty} f$ also diverges.

So $\int f$ and $\int g$ converge or diverge together.

(2) If $f = o(g)$, then $\lim f/g = 0$. For $\varepsilon = 1$ there exists $B > 0$ such that $0 \le f(x)/g(x) \le 1$ for $x \ge B$ (using that $f, g$ are positive, so the ratio is $\ge 0$). Hence $0 \;\le\; f(x) \;\le\; g(x) \qquad \forall\, x \ge B.$ The conclusions (a), (b) follow directly from the comparison criterion. $\blacksquare$

Practical recipe. To determine if $\int_a^{+\infty} f(x)\,dx$ converges for a positive $f$:

1. Find the leading asymptotic behaviour of $f$ at $+\infty$: usually $f(x) \sim C / x^p$ for some $p$.
2. Apply the harmonic rule: $\int_1^{+\infty} 1/x^p\,dx$ converges iff $p > 1$ (diverges iff $p \le 1$).
3. Conclude via asymptotic comparison.

For a singularity at a finite endpoint (second kind), the harmonic rule flips: $\int_0^1 1/x^p\,dx$ converges iff $p < 1$.

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3.15 Theorem 1937 — Absolute convergence for improper integrals

Statement. Let $f : [a, +\infty) \to \mathbb{R}$ be integrable on every $[a, b] \subseteq [a, +\infty)$. If $\int_a^{+\infty} |f(x)|\,dx$ converges (we say $f$ is absolutely integrable), then $\int_a^{+\infty} f(x)\,dx$ also converges, and $\Bigl|\int_a^{+\infty} f(x)\,dx\Bigr| \;\le\; \int_a^{+\infty} |f(x)|\,dx.$

Source: Proofs - Integral Calculus.pdf p.22 (handwritten theorem 1937).

Proof.

Starting from the pointwise bound $-|f(x)| \le f(x) \le |f(x)|$, we have $0 \;\le\; f(x) + |f(x)| \;\le\; 2|f(x)| \qquad \forall\, x.$ Since $\int_a^{+\infty} |f|$ converges, so does $\int_a^{+\infty} 2|f| = 2 \int_a^{+\infty} |f|$. By the comparison criterion (applied to the non-negative function $f + |f|$ bounded above by $2|f|$), $\int_a^{+\infty} [f(x) + |f(x)|]\,dx$ also converges.

Writing $f(x) = [f(x) + |f(x)|] - |f(x)|$ (the $|f|$'s cancel): $\int_a^{+\infty} f(x)\,dx \;=\; \int_a^{+\infty} [f(x) + |f(x)|]\,dx \;-\; \int_a^{+\infty} |f(x)|\,dx,$ which is a difference of two convergent integrals, hence converges.

For the inequality: $-|f(x)| \le f(x) \le |f(x)|$ combined with monotonicity gives $-\int_a^{+\infty} |f| \le \int_a^{+\infty} f \le \int_a^{+\infty} |f|$, i.e., $\Bigl|\int_a^{+\infty} f(x)\,dx\Bigr| \;\le\; \int_a^{+\infty} |f(x)|\,dx. \qquad \blacksquare$

Use case. Absolute convergence is the standard tool for oscillating integrands like $\int_1^{+\infty} \sin x / x^2\,dx$: since $|\sin x / x^2| \le 1/x^2$, absolute convergence holds by comparison, hence the integral converges. Note the theorem is one-way: there is no comparable "absolute divergence" criterion.

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3.16 Theorem 2030 — Barrow–Torricelli theorem

Statement. Let $g : [a, b] \to \mathbb{R}$ with $g(a) = 0$. The following are equivalent: (1) $g$ is continuously differentiable on $[a, b]$ (i.e., $g \in C^{(1)}_0([a,b])$, where $C^{(1)}_0$ means $C^{(1)}$ with value $0$ at $a$); (2) there exists a unique function $\gamma : [a, b] \to \mathbb{R}$, continuous on $[a,b]$, such that $g(x) \;=\; \int_a^x \gamma(t)\,dt \qquad \forall\, x \in [a, b].$

Consequently the map $T : C^{(1)}_0([a, b]) \to C([a, b])$ defined by $T(g) = g'$ is a bijection, with inverse $T^{-1}(\gamma)(x) = \int_a^x \gamma(t)\,dt$.

Source: Proofs - Integral Calculus.pdf p.14 (handwritten theorem 2030).

Proof.

($\Leftarrow$) Assume there exists $\gamma \in C([a, b])$ with $g(x) = \int_a^x \gamma(t)\,dt$ for every $x \in [a, b]$.

• By the Second FTC (Theorem 1899, applied to the continuous $\gamma$), the integral function $G(x) := \int_a^x \gamma(t)\,dt$ is differentiable with $G'(x) = \gamma(x)$. Since $g = G$, we have $g' = \gamma$.
• $g' = \gamma$ is continuous by hypothesis, so $g \in C^{(1)}([a,b])$.
• $g(a) = \int_a^a \gamma = 0$ $\checkmark$.

Hence $g \in C^{(1)}_0([a, b])$.

($\Rightarrow$) Assume $g \in C^{(1)}_0([a, b])$, so $g$ is $C^{(1)}$ and $g(a) = 0$. By the First FTC (Theorem 1894) applied to $g'$ on $[a, x]$: $g(x) - g(a) \;=\; \int_a^x g'(t)\,dt \qquad \Longrightarrow \qquad g(x) \;=\; \int_a^x g'(t)\,dt,$ since $g(a) = 0$. Set $\gamma := g' \in C([a,b])$. Then $g(x) = \int_a^x \gamma(t)\,dt$. $\checkmark$

Uniqueness of $\gamma$. Suppose $\gamma_1, \gamma_2 \in C([a,b])$ both satisfy $g(x) = \int_a^x \gamma_i(t)\,dt$. By the Second FTC, $g' = \gamma_1$ and $g' = \gamma_2$, so $\gamma_1 = \gamma_2$.

Bijectivity of $T$.

• Injectivity. If $T(g_1) = T(g_2) = \gamma$, then both $g_1(x) = \int_a^x \gamma$ and $g_2(x) = \int_a^x \gamma$ (using $g_i \in C^{(1)}_0$ and the $\Rightarrow$ direction), so $g_1 = g_2$.
• Surjectivity. For any $\gamma \in C([a,b])$, define $g(x) = \int_a^x \gamma(t)\,dt$. Then $g \in C^{(1)}_0$ by the $\Leftarrow$ direction, and $T(g) = g' = \gamma$.

So $T$ is a bijection with inverse $T^{-1}(\gamma) = \int_a^{\cdot} \gamma(t)\,dt$. $\blacksquare$

Interpretation. Differentiation and integration (with a fixed base point) are inverse operations on the appropriate function spaces. This is the conceptual core of the FTC and of the Newton–Leibniz identity.

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§4. Worked Examples

Example 4.1 — Standard antiderivatives table

The table below summarises the essential antiderivatives that every student should know by heart. Each entry can be verified by differentiating the right-hand side (by the uniqueness theorem, if $P'(x) = f(x)$ on an interval, then $\int f\,dx = P(x) + C$). Source: Integral Calculus.pdf p.15.

| $f(x)$ | $\int f(x)\,dx$ | Notes | |---|---|---| | $k$ (constant) | $kx + C$ | e.g., $\int 5\,dx = 5x + C$ | | $x^a$ $(a \neq -1)$ | $\dfrac{x^{a+1}}{a+1} + C$ | e.g., $\int x^2\,dx = x^3/3 + C$ | | $\dfrac{1}{x}$ | $\ln|x| + C$ | on $(0, +\infty)$: $\ln x + C$; on $(-\infty, 0)$: $\ln(-x) + C$ | | $e^x$ | $e^x + C$ | | | $a^x$ | $\dfrac{a^x}{\ln a} + C$ | e.g., $\int 2^x\,dx = 2^x / \ln 2 + C$ | | $\sin x$ | $-\cos x + C$ | | | $\cos x$ | $\sin x + C$ | | | $\dfrac{1}{\cos^2 x}$ | $\tan x + C$ | | | $\dfrac{1}{\sqrt{1 - x^2}}$ | $\arcsin x + C$ | | | $\dfrac{1}{1 + x^2}$ | $\arctan x + C$ | |

Composition shortcuts (by the chain rule):

• $\int f'(x) \cdot [f(x)]^a\,dx = \dfrac{[f(x)]^{a+1}}{a+1} + C$, $a \neq -1$.
• $\int \dfrac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C$.
• $\int f'(x) e^{f(x)}\,dx = e^{f(x)} + C$.
• $\int f'(x) \sin[f(x)]\,dx = -\cos[f(x)] + C$.
• $\int f'(x) \cos[f(x)]\,dx = \sin[f(x)] + C$.
• $\int \dfrac{f'(x)}{1 + [f(x)]^2}\,dx = \arctan[f(x)] + C$.

Linear exponent shortcut. $\int e^{ax + b}\,dx = \dfrac{1}{a} e^{ax+b} + C$.

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Example 4.2 — Integration by parts on $\int x^2 \ln x\,dx$ (May 2024 Q9)

Source: General_24524_ENG_SOL.pdf Q9. This is the flagship computation of integral calculus — the exam explicitly requires it.

Goal. Compute $\int x^2 \ln x\,dx$ and then the definite integral $\int_1^e x^2 \ln x\,dx$.

Step 1 — Apply integration by parts. By the LIATE priority (see M-IC-1 in §5), we pick the logarithm as the factor to differentiate (highest priority) and the algebraic factor $x^2$ as the factor to integrate:

• Let $g(x) = \ln x$ (differentiate): $g'(x) = 1/x$.
• Let $f'(x) = x^2$ (integrate): $f(x) = x^3 / 3$.

Step 2 — Apply the formula $\int f' g = f g - \int f g'$. $\int x^2 \ln x\,dx \;=\; \underbrace{\frac{x^3}{3} \ln x}_{fg} \;-\; \int \underbrace{\frac{x^3}{3}}_{f} \cdot \underbrace{\frac{1}{x}}_{g'}\,dx \;=\; \frac{x^3}{3} \ln x \;-\; \int \frac{x^2}{3}\,dx.$

Step 3 — Compute the remaining integral. $\int \frac{x^2}{3}\,dx \;=\; \frac{1}{3} \cdot \frac{x^3}{3} + C \;=\; \frac{x^3}{9} + C.$

Step 4 — Assemble. $\boxed{\int x^2 \ln x\,dx \;=\; \frac{x^3}{3} \ln x \;-\; \frac{x^3}{9} \;+\; C.}$

Step 5 — Definite integral from 1 to $e$ (apply Theorem 1894, FTC I). $\int_1^e x^2 \ln x\,dx \;=\; \Bigl[\frac{x^3}{3} \ln x - \frac{x^3}{9}\Bigr]_1^e.$ Evaluate at $x = e$: $\ln e = 1$, so $\frac{e^3}{3} \cdot 1 - \frac{e^3}{9} \;=\; \frac{3 e^3}{9} - \frac{e^3}{9} \;=\; \frac{2 e^3}{9}.$ Evaluate at $x = 1$: $\ln 1 = 0$, so $\frac{1}{3} \cdot 0 - \frac{1}{9} \;=\; -\frac{1}{9}.$

Subtract (upper minus lower): $\int_1^e x^2 \ln x\,dx \;=\; \frac{2 e^3}{9} - \Bigl(-\frac{1}{9}\Bigr) \;=\; \boxed{\frac{2}{9} e^3 \;+\; \frac{1}{9}.}$

Sanity check. $\tfrac{2}{9} e^3 \approx \tfrac{2}{9} \cdot 20.09 \approx 4.46$ plus $\tfrac{1}{9} \approx 0.11$ gives $\approx 4.58$. The integrand $x^2 \ln x$ is positive on $(1, e]$ and zero at $x = 1$, with maximum value $e^2 \cdot 1 \approx 7.39$ at $x = e$. On the interval of length $e - 1 \approx 1.72$, an average value around $\approx 2.67$ gives area $\approx 4.58$. $\checkmark$

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Example 4.3 — Integration by substitution: $\int x e^{x^2}\,dx$

Source: Integral Calculus.pdf p.20 (shortcut version) and p.24 (substitution version).

Using the shortcut. Since $(x^2)' = 2x$, we have $x = \frac{1}{2} \cdot 2x$, so $x e^{x^2} = \frac{1}{2}(2x)e^{x^2} = \frac{1}{2}(x^2)' \cdot e^{x^2}$. By the composition shortcut $\int f'(x) e^{f(x)}\,dx = e^{f(x)} + C$: $\int x e^{x^2}\,dx \;=\; \frac{1}{2} \int 2x \cdot e^{x^2}\,dx \;=\; \frac{1}{2} e^{x^2} + C.$

Using substitution (mechanical). Let $u = x^2$, so $du = 2x\,dx$, i.e., $x\,dx = \frac{1}{2}\,du$. Substituting: $\int x e^{x^2}\,dx \;=\; \int e^u \cdot \frac{1}{2}\,du \;=\; \frac{1}{2} e^u + C \;=\; \frac{1}{2} e^{x^2} + C. \qquad \checkmark$

Lesson. Always try the composition shortcut first; substitution is the safety net when the shortcut is not obvious.

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Example 4.4 — Partial fractions: $\int \dfrac{1}{(x - 1)(x + 2)}\,dx$

Adapted from Integral Calculus.pdf pp.21–22 (Sicconi's "Algebraic fractions").

Step 1 — Decompose. The denominator has two distinct linear factors. Write $\frac{1}{(x - 1)(x + 2)} \;=\; \frac{A}{x - 1} \;+\; \frac{B}{x + 2}.$

Step 2 — Find $A$ and $B$. Multiply by $(x - 1)(x + 2)$: $1 \;=\; A(x + 2) + B(x - 1).$

• Plug $x = 1$: $1 = A \cdot 3 + B \cdot 0 \Rightarrow A = 1/3$.
• Plug $x = -2$: $1 = A \cdot 0 + B \cdot (-3) \Rightarrow B = -1/3$.

Step 3 — Integrate. $\int \frac{1}{(x-1)(x+2)}\,dx \;=\; \int \Bigl[\frac{1/3}{x - 1} + \frac{-1/3}{x + 2}\Bigr]\,dx \;=\; \frac{1}{3}\ln|x - 1| - \frac{1}{3}\ln|x + 2| + C \;=\; \frac{1}{3}\ln\Bigl|\frac{x - 1}{x + 2}\Bigr| + C.$

Domain note. On an interval not containing $x = 1$ or $x = -2$, the argument of the log has constant sign and the absolute values can be dropped (with the sign chosen to make it positive).

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Example 4.5 — Improper integral comparison: $\int_1^{+\infty} \dfrac{x + \ln x}{x^3}\,dx$ (May 2024 MCQ3)

Source: General_24524_ENG_SOL.pdf MCQ3.

Step 1 — Asymptotic behaviour at $+\infty$. Since $\ln x = o(x)$, we have $x + \ln x \sim x$, hence $\frac{x + \ln x}{x^3} \;\sim\; \frac{x}{x^3} \;=\; \frac{1}{x^2} \qquad \text{as } x \to +\infty.$

Step 2 — Harmonic rule. $\int_1^{+\infty} 1/x^2\,dx$ converges (exponent $p = 2 > 1$), in fact $= 1$.

Step 3 — Asymptotic comparison (Theorem 1933). The integrand is positive on $[1, +\infty)$ and $\sim 1/x^2$. So $\int_1^{+\infty} \frac{x + \ln x}{x^3}\,dx$ converges (finite).

Step 4 — Direct comparison for the sharper bound. On $[1, +\infty)$ we have $\ln x \ge 0$, so $\frac{x + \ln x}{x^3} \;\ge\; \frac{x}{x^3} \;=\; \frac{1}{x^2} \qquad \forall\, x \ge 1.$ By Theorem 1928 (monotonicity of improper integrals): $\int_1^{+\infty} \frac{x + \ln x}{x^3}\,dx \;\ge\; \int_1^{+\infty} \frac{1}{x^2}\,dx.$

Conclusion. The integral is finite AND greater or equal to $\int_1^{+\infty} 1/x^2\,dx$. Answer D (Mode A) / C (Mode B). $\checkmark$

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Example 4.6 — Principal value: $\operatorname{PV}\int_{-\infty}^{+\infty}(2x - 3)\,dx$ (Final exercises Q4)

Source: Final exercises solution.pdf Q4.

Claim (a). $\int_{-\infty}^{+\infty}(2x - 3)\,dx = -\infty$? FALSE.

By definition, split at $0$: $\int_{-\infty}^{+\infty}(2x - 3)\,dx \;=\; \lim_{s \to -\infty} \int_s^0 (2x - 3)\,dx \;+\; \lim_{t \to +\infty} \int_0^t (2x - 3)\,dx.$ A primitive of $2x - 3$ is $x^2 - 3x$. Compute each piece: $\lim_{s \to -\infty} \bigl[x^2 - 3x\bigr]_s^0 \;=\; \lim_{s \to -\infty} \bigl[0 - (s^2 - 3s)\bigr] \;=\; \lim_{s \to -\infty} \bigl(-s^2 + 3s\bigr) \;=\; -\infty - \infty \;=\; -\infty,$ $\lim_{t \to +\infty} \bigl[x^2 - 3x\bigr]_0^t \;=\; \lim_{t \to +\infty} \bigl(t^2 - 3t\bigr) \;=\; +\infty.$ Sum: $-\infty + (+\infty)$, an indeterminate form. The classical improper integral does not exist — not $-\infty$, not $+\infty$, and not any finite number. Hence (a) is FALSE.

Claim (b). $\operatorname{PV}\int_{-\infty}^{+\infty}(2x - 3)\,dx = 0$? FALSE.

By definition, the PV uses a single symmetric limit: $\operatorname{PV}\int_{-\infty}^{+\infty}(2x - 3)\,dx \;=\; \lim_{K \to +\infty} \int_{-K}^{K} (2x - 3)\,dx \;=\; \lim_{K \to +\infty} \bigl[x^2 - 3x\bigr]_{-K}^K.$ Compute: $\bigl[x^2 - 3x\bigr]_{-K}^K \;=\; (K^2 - 3K) - (K^2 + 3K) \;=\; -6K.$ Take $K \to +\infty$: $-6K \to -\infty$. Hence $\operatorname{PV}\int_{-\infty}^{+\infty}(2x - 3)\,dx \;=\; -\infty.$ So (b) "$= 0$" is FALSE.

Claim (c). $\operatorname{PV}\int_{-\infty}^{+\infty}(2x - 3)\,dx = -\infty$? TRUE.

As computed in (b), the PV equals $-\infty$. $\checkmark$

Instructive remark. Note the contrast:

• For a pure odd function like $f(x) = 2x$, the PV is $0$ (the symmetric cancellation), while the classical integral $\int_{-\infty}^{+\infty} 2x\,dx$ is indeterminate.
• Here $f(x) = 2x - 3$ has an even part $-3$ that contributes $\int_{-K}^K (-3)\,dx = -6K \to -\infty$. The odd part $2x$ cancels but the even part blows up, hence the PV diverges to $-\infty$.
• If a function is odd (so $f(-x) = -f(x)$), its PV is always $0$ by symmetry. If it is even and integrable, its PV equals twice $\int_0^{+\infty} f$.

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Example 4.7 — Continuous compounding via integral: $M(t) = 1 + \int_0^t f'(s)\,ds$ (Final exercises Q5)

Source: Final exercises solution.pdf Q5.

Setup. A capital $C = 1$ is available today ($t = 0$). Let $f(t)$ denote the future value at time $t \ge 0$, with $f(0) = 1$. The instantaneous rate of change of capital (interest) is $f'(t)$. The total interest accumulated between $0$ and $t$ is the area under $f'(s)$: $I(t) \;=\; \int_0^t f'(s)\,ds.$ By the First FTC (Theorem 1894), $I(t) = f(t) - f(0) = f(t) - 1$. Hence the available amount of money at time $t$ is $M(t) \;=\; C + I(t) \;=\; 1 + \int_0^t f'(s)\,ds.$

Compute with $f'(t) = (1.02)^t \ln(1.02)$, $t = 2$ years, $C = 1$.

Notice that $f'(t) = (1.02)^t \ln(1.02)$ is exactly the derivative of $(1.02)^t$ (since $\frac{d}{dt}(1.02)^t = (1.02)^t \ln(1.02)$). Hence a primitive is $P(t) = (1.02)^t$: $M(2) \;=\; 1 + \int_0^2 (1.02)^s \ln(1.02)\,ds \;=\; 1 + \bigl[(1.02)^s\bigr]_0^2 \;=\; 1 + \bigl[(1.02)^2 - 1\bigr] \;=\; (1.02)^2 \;=\; 1.0404.$

For $C = 100$: scaling by $100$, $M(2) = 100 \cdot (1.02)^2 = 104.04$.

Remark. This is precisely the discrete compounding formula $M = C(1 + r)^t$ with $r = 0.02$, $t = 2$ — which should be! We have built it from the differential law $f'(t) = f(t) \ln(1 + r)$ via the FTC.

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Example 4.8 — $\int_0^2 2t^3 e^{t^2}\,dt$ with substitution + by parts (TA41)

Source: TA41_clac_int1.pdf Q3.

Use the substitution $\varphi(t) = t^2$, so $d\varphi = 2t\,dt$ and the bounds $t = 0, 2$ become $\varphi = 0, 4$. Write $2 t^3 = 2t \cdot t^2 = (2t)\cdot \varphi$: $\int_0^2 2 t^3 e^{t^2}\,dt \;=\; \int_0^2 \bigl[2 t \cdot t^2\bigr] e^{t^2}\,dt \;=\; \int_0^4 \varphi \cdot e^{\varphi}\,d\varphi.$

Now integrate by parts on $\int \varphi e^{\varphi}\,d\varphi$ with $g(\varphi) = \varphi$ (differentiate), $f'(\varphi) = e^{\varphi}$ (integrate): $\int \varphi e^{\varphi}\,d\varphi \;=\; \varphi e^{\varphi} - \int e^{\varphi}\,d\varphi \;=\; \varphi e^{\varphi} - e^{\varphi} + C \;=\; e^{\varphi}(\varphi - 1) + C.$

Evaluate bounds: $\int_0^4 \varphi e^{\varphi}\,d\varphi \;=\; \bigl[e^{\varphi}(\varphi - 1)\bigr]_0^4 \;=\; e^4(4 - 1) - e^0(0 - 1) \;=\; 3 e^4 - (-1) \;=\; 3 e^4 + 1.$

Answer: $3 e^4 + 1$ (MCQ option A in TA41 Q3).

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§5. Solution Methods

Each method is a named algorithm with Input → Steps → Output → Pitfalls, with cross-references to May 2024 / Final-exercises problems.

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M-IC-1 — Integration by parts (LIATE priority)

Used on: May 2024 Q9 (antiderivative of $x^2 \ln x$), TA41 Q3, any product of "incompatible" factors.

Input. An integral $\int u(x) \cdot v(x)\,dx$ where $u, v$ are differentiable, and no shortcut applies.

Steps.

1. LIATE heuristic — choose $g =$ the factor whose type comes earliest in this priority list (so that $g \to g'$ simplifies it), and $f' =$ the remaining factor (which must have a known antiderivative):
   • Logarithms: $\ln x$, $\log_a x$
   • Inverse trig: $\arctan x, \arcsin x$
   • Algebraic: polynomials $x^n$
   • Trigonometric: $\sin x, \cos x$
   • Exponentials: $e^x, a^x$
2. Compute $g'(x)$ (differentiate) and $f(x) = \int f'(x)\,dx$ (integrate, no $+C$ needed here).
3. Apply the formula: $\int f'(x) g(x)\,dx \;=\; f(x) g(x) \;-\; \int f(x) g'(x)\,dx.$
4. Compute the remaining integral $\int f g'\,dx$. It should be simpler than the original.
5. Add $+ C$ if indefinite, or evaluate bounds if definite (via FTC I).

Output. A closed-form antiderivative, or a definite-integral value.

Pitfalls.

• Wrong sign in $fg - \int fg'\,dx$. Memorise the formula carefully: the minus is on $\int fg'$, not on $fg$.
• Infinite loop: if after applying IBP the new integral equals the original one (up to sign), add them both to one side and solve. E.g., $\int e^x \sin x\,dx$ cycles back after two IBPs.
• Special case $\int \ln x\,dx$: write $\ln x = 1 \cdot \ln x$, take $f' = 1$, $g = \ln x$: $\int \ln x\,dx = x \ln x - \int x \cdot (1/x)\,dx = x \ln x - x + C = x(\ln x - 1) + C$.
• Forgetting $+C$ on indefinite — this is a grader pet peeve.

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M-IC-2 — Integration by substitution

Used on: Many TA problems, compounding-style integrals, and whenever the integrand contains a composite function with a visible derivative factor.

Input. An integral $\int f(x)\,dx$ where $f$ contains a composite pattern $g[h(x)] \cdot h'(x)$ or close variant.

Steps.

1. Try the shortcut first (see the composition-shortcut table in §4, Example 4.1). If one of them matches, apply it directly:
   • $\int f' \cdot (f)^a\,dx = \frac{f^{a+1}}{a+1} + C$ ($a \neq -1$), $\int f'/f\,dx = \ln|f| + C$, $\int f' e^f\,dx = e^f + C$, etc.
   • Multiplication trick (Integral Calculus.pdf p.19): if the derivative factor is off by a constant, factor it out: $\int x e^{x^2}\,dx = \frac{1}{2}\int 2x e^{x^2}\,dx = \frac{1}{2} e^{x^2} + C$.
2. If shortcuts fail, substitute. Pick $u = g(x)$ (usually the "inside" of a composition). Compute $du = g'(x)\,dx$, i.e., $dx = du / g'(x)$.
3. Rewrite the integrand entirely in terms of $u$. If some $x$'s remain, express them via $x = g^{-1}(u)$ and substitute.
4. For definite integrals, change the bounds: new lower bound = $g(\text{old lower})$, new upper bound = $g(\text{old upper})$. Important: this only works if $g$ is monotonic on the interval (otherwise split the domain or use $|g'|$).
5. Integrate in the $u$ variable.
6. Either back-substitute $u = g(x)$ (indefinite case), or evaluate the new bounds directly (definite case — no back-substitution needed).

Output. Closed-form antiderivative in $x$ (indefinite) or numerical value (definite).

Pitfalls.

• Forgetting to change the bounds when evaluating a definite integral. If you keep the old bounds, you must first back-substitute to $x$.
• Non-monotone substitution (e.g., $u = x^2$ on an interval containing $0$). Split into monotonic pieces or introduce $|du|$.
• Leaving residual $x$'s in the integrand. Every $x$ must be expressed via $u$ before you integrate.

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M-IC-3 — Integration by partial fractions for $P(x)/Q(x)$

Used on: Rational-function integrands whose denominator is quadratic (factorisable).

Input. An integral $\int P(x)/Q(x)\,dx$ where $P, Q$ are polynomials, $\deg P < \deg Q$, and $Q$ factors over $\mathbb{R}$.

Steps.

1. Reduce first if $\deg P \ge \deg Q$: polynomial-divide $P$ by $Q$ to get $P = \text{quotient} \cdot Q + \text{remainder}$, so $P/Q = \text{quotient} + \text{remainder}/Q$ with $\deg(\text{remainder}) < \deg Q$. Integrate the quotient separately (easy polynomial).
2. Factor $Q$. For a quadratic $Q(x) = (x - r_1)(x - r_2)$ with distinct real roots, or $Q(x) = (x - r)^2$ for a double root.
3. Decompose:
   • Distinct linear factors. $\dfrac{P(x)}{(x - r_1)(x - r_2)} = \dfrac{A}{x - r_1} + \dfrac{B}{x - r_2}$.
   • Repeated linear factor. $\dfrac{P(x)}{(x - r)^2} = \dfrac{A}{(x - r)^2} + \dfrac{B}{x - r}$.
4. Solve for $A, B$ by multiplying through by $Q$ and either plugging in the root values or matching coefficients.
5. Integrate term-by-term: $\int \frac{A}{x - r}\,dx = A \ln|x - r| + C$; $\int \frac{A}{(x - r)^2}\,dx = -\frac{A}{x - r} + C$.

Output. A sum of logarithms and/or simple rational terms, plus $C$.

Pitfalls.

• Wrong decomposition when a factor is repeated: must include both $\frac{A}{(x-r)^2}$ and $\frac{B}{x-r}$.
• Absolute value in the log: $\ln|x - r|$, not $\ln(x - r)$ — crucial when $r$ is inside the integration interval of a definite integral (though in that case the integrand is unbounded and you are in the improper regime).
• Forgetting the addition/algebraic trick for $\int \frac{2x - 3}{x + 2}\,dx$-style problems: add and subtract the constant in the denominator at the numerator to simplify (see Integral Calculus.pdf pp.20–21).

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M-IC-4 — Convergence of an improper integral by asymptotic comparison

Used on: May 2024 MCQ3, TA41 TA2 Qs 17–19, every "does this integral exist?" question.

Input. An improper integral $\int_a^{+\infty} f(x)\,dx$ (first kind) or $\int_a^b f(x)\,dx$ with a singularity at $b^-$ (second kind), $f$ positive and eventually monotonic.

Steps.

1. Identify where the problem is. For first kind, study $x \to +\infty$. For second kind, study $x \to b^-$ (or wherever $f$ blows up).
2. Find the asymptotic order. Compute $\lim_{x \to \text{problem}} f(x) \cdot x^p$ (first kind) or $\lim_{x \to b^-} f(x) \cdot (b - x)^p$ (second kind); find the critical $p$ such that $f(x) \sim C / x^p$ (resp. $\sim C/(b-x)^p$).
3. Apply the harmonic rule.
   • First kind, singularity at $+\infty$: $\int_1^{+\infty} 1/x^p\,dx$ converges iff $p > 1$.
   • Second kind, singularity at finite point: $\int_0^1 1/x^p\,dx$ converges iff $p < 1$.
4. Conclude via asymptotic comparison (Theorem 1933): if $f \sim C/x^p$ at $+\infty$ and $C > 0$, then $\int_a^{+\infty} f$ converges iff $\int 1/x^p$ converges iff $p > 1$.
5. If $f$ is not positive (e.g., oscillating like $\sin x / x^2$): apply the absolute convergence test (Theorem 1937) — show $\int |f|$ converges, hence so does $\int f$.

Output. "Converges" or "diverges".

Pitfalls.

• Not checking positivity before applying comparison/asymptotic tests. Those theorems require $f, g$ positive (or eventually positive).
• Forgetting the harmonic flip between first and second kind: at $+\infty$, convergence wants $p > 1$; at a finite singularity, $p < 1$.
• Mistaking $\ln x$ for a polynomial: $\ln x = o(x^\varepsilon)$ for any $\varepsilon > 0$, i.e., a log factor never changes the asymptotic order.
• Using the absolute convergence test in the wrong direction: $\int f$ converging does not imply $\int |f|$ converging (counterexample $\int \sin x / x\,dx$ converges but $\int |\sin x / x|$ diverges). The theorem is one-way.

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M-IC-5 — Cauchy principal value (PV) when $\int$ diverges

Used on: Final exercises Q4, oscillating or odd-plus-even integrands on $(-\infty, +\infty)$.

Input. An integrand $f : \mathbb{R} \to \mathbb{R}$ where $\int_{-\infty}^{+\infty} f\,dx$ does not converge (but possibly has a finite PV).

Steps.

1. Check if $f$ is odd. If $f(-x) = -f(x)$, then for every $K > 0$, $\int_{-K}^K f(x)\,dx = 0$. Hence $\operatorname{PV}\int_{-\infty}^{+\infty} f = 0$.
2. If $f$ is even and integrable on $[0, \infty)$, $\operatorname{PV}\int_{-\infty}^{+\infty} f = 2 \int_0^{+\infty} f$ (if convergent).
3. General case: find a primitive $P$ of $f$, then $\operatorname{PV}\int_{-\infty}^{+\infty} f(x)\,dx \;=\; \lim_{K \to +\infty} \bigl[P(K) - P(-K)\bigr].$
4. For a PV at a finite singularity $c \in (a, b)$: $\operatorname{PV}\int_a^b f = \lim_{\varepsilon \to 0^+} \bigl[\int_a^{c - \varepsilon} f + \int_{c + \varepsilon}^b f\bigr]$ using the same $\varepsilon$ on both sides.

Output. A finite number, or $\pm\infty$, or "does not exist" (if the symmetric limit itself oscillates).

Pitfalls.

• Confusing PV with the classical improper integral. They agree when the latter converges; the PV may be finite even when the classical is indeterminate (odd-function example), but the classical divergence can "win" (Example 4.6: PV is $-\infty$, not $0$).
• Using two independent limits instead of one symmetric $K$. The whole point of PV is the single-parameter limit.

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M-IC-6 — Applying the FTC ("primitive first, then evaluate bounds")

Used on: May 2024 Q9 (b2), every definite-integral computation.

Input. A definite integral $\int_a^b f(x)\,dx$ with $f$ integrable and having a primitive on $[a, b]$.

Steps.

1. Find a primitive $P$ of $f$ using any of M-IC-1, M-IC-2, M-IC-3, or the antiderivatives table (Example 4.1). Drop the $+C$ — any one primitive suffices.
2. Verify that the hypotheses of Theorem 1894 hold: $f$ integrable on $[a, b]$ (e.g., continuous), and $P$ a genuine primitive (check $P'(x) = f(x)$).
3. Evaluate $\bigl[P(x)\bigr]_a^b = P(b) - P(a)$ (Newton–Leibniz).

Output. The numerical value of the definite integral.

Pitfalls.

• Evaluating before dropping $+C$. The $+C$ cancels in $P(b) - P(a)$, so you do not need it, but do not write it and then add it to the answer.
• Wrong sign: always upper minus lower, $P(b) - P(a)$.
• Swapped bounds: if you end up with $a > b$, flip and add a minus sign: $\int_a^b = -\int_b^a$.

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M-IC-7 — Verify $F$ is a primitive of $f$ (differentiate F, check = f)

Used on: Multiple-choice "which of these is an antiderivative of $f$?" questions (TA41 Q7).

Input. A function $f(x)$ and a candidate $F(x)$.

Steps.

1. Differentiate $F(x)$, applying product/chain/quotient rules as needed.
2. Simplify $F'(x)$.
3. Check whether $F'(x) = f(x)$ pointwise on the interval.
4. If yes, $F$ is a primitive. If not, rule out.

Output. "Yes, $F$ is a primitive" or "no".

Pitfalls.

• On MCQs, always try this method first before attempting the much harder forward computation. Differentiation is mechanical; integration can be hard.
• Multiple correct answers. By Theorem 1903, any two primitives differ by a constant. On an MCQ, two candidate primitives that differ by a constant are both correct if the problem does not pin down the constant. In that case look for additional info (e.g., value at a specific point).

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§6. Practice Problems with Solutions

Problem 6.1 — TA41 #13: $\int_1^{+\infty} \dfrac{1}{(3x + 2)^2}\,dx$

Source: TA41_calc_int2.pdf Q13.

Behaviour. The integrand is positive and $\sim \frac{1}{9 x^2}$ at $+\infty$; by asymptotic comparison with $\int 1/x^2$ (convergent), the integral converges.

Value. Rewrite using the shortcut $\int f'(x) \cdot f(x)^{-2}\,dx = -1/f(x) + C$ with $f(x) = 3x + 2$, $f'(x) = 3$: $\int \frac{1}{(3x + 2)^2}\,dx \;=\; \frac{1}{3} \int \frac{3}{(3x + 2)^2}\,dx \;=\; \frac{1}{3} \cdot \frac{-1}{3x + 2} + C \;=\; -\frac{1}{3(3x + 2)} + C.$

Compute the improper integral: $\int_1^{+\infty} \frac{1}{(3x+2)^2}\,dx \;=\; \lim_{K \to +\infty}\Bigl[-\frac{1}{3(3x+2)}\Bigr]_1^K \;=\; \lim_{K \to +\infty}\Bigl[-\frac{1}{3(3K+2)} + \frac{1}{3 \cdot 5}\Bigr] \;=\; 0 + \frac{1}{15} \;=\; \boxed{\frac{1}{15}}.$

Answer: A ($1/15$).

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Problem 6.2 — TA41 #17: $\int_1^{+\infty} \dfrac{x^5 \sqrt{x} + 2}{x^7 \sqrt[3]{x} + 7}\,dx$

Source: TA41_calc_int2.pdf Q17.

Asymptotically at $+\infty$: $\frac{x^5 \sqrt{x} + 2}{x^7 \sqrt[3]{x} + 7} \;\sim\; \frac{x^{5 + 1/2}}{x^{7 + 1/3}} \;=\; \frac{x^{11/2}}{x^{22/3}} \;=\; \frac{1}{x^{22/3 - 11/2}} \;=\; \frac{1}{x^{44/6 - 33/6}} \;=\; \frac{1}{x^{11/6}}.$

Since $11/6 > 1$, the integral $\int 1/x^{11/6}\,dx$ converges. By asymptotic comparison, the original integral converges (exists and is finite). Answer: B.

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Problem 6.3 — TA41 #19: For which $\alpha > 0$ does $\int_2^{+\infty} \dfrac{(x^2 - 2x)^3}{(x^2 - 3)^2 \sqrt{x^\alpha - 1}}\,dx$ converge?

Source: TA41_calc_int2.pdf Q19.

Asymptotically at $+\infty$: $\frac{(x^2 - 2x)^3}{(x^2 - 3)^2 \sqrt{x^\alpha - 1}} \;\sim\; \frac{(x^2)^3}{(x^2)^2 \cdot \sqrt{x^\alpha}} \;=\; \frac{x^6}{x^4 \cdot x^{\alpha / 2}} \;=\; \frac{1}{x^{\alpha/2 - 2}}.$

Convergence requires $\alpha/2 - 2 > 1$, i.e., $\alpha > 6$ (strict inequality!). Answer: D — "none of the preceding" (since "b $\alpha \ge 6$" allows $\alpha = 6$ which gives $p = 1$, the boundary case which diverges).

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Problem 6.4 — May 2024 Q9 (full)

Source: General_24524_ENG_SOL.pdf Q9.

(a) State and prove the First Fundamental Theorem of Calculus. See §3.7 above — reproduce the statement and the 5-step proof verbatim.

(b1) Antiderivative of $f(x) = x^2 \ln x$ on $(0, +\infty)$.

Integrate by parts with $g = \ln x$ (LIATE priority — log), $f'_1 = x^2$: $\int x^2 \ln x\,dx \;=\; \frac{x^3}{3}\ln x - \int \frac{x^3}{3} \cdot \frac{1}{x}\,dx \;=\; \frac{x^3}{3}\ln x - \frac{x^3}{9} + C.$

(b2) Definite integral $\int_1^e x^2 \ln x\,dx$.

Apply FTC I (Theorem 1894): $\int_1^e x^2 \ln x\,dx \;=\; \Bigl[\frac{x^3}{3}\ln x - \frac{x^3}{9}\Bigr]_1^e \;=\; \Bigl(\frac{e^3}{3} - \frac{e^3}{9}\Bigr) - \Bigl(0 - \frac{1}{9}\Bigr) \;=\; \frac{2 e^3}{9} + \frac{1}{9}.$

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Problem 6.5 — Final exercises Q4 (full true/false)

Source: Final exercises solution.pdf Q4. See Example 4.6 for the full solution. Answers: (a) False, (b) False, (c) True.

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Problem 6.6 — Final exercises Q5 (continuous compounding)

Source: Final exercises solution.pdf Q5. See Example 4.7. Answer: $M(2) = (1.02)^2 = 1.0404$ for $C = 1$; $M(2) = 100 \cdot (1.02)^2 = 104.04$ for $C = 100$.

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Problem 6.7 — TA41 Q2: $\int_0^1 7 x \ln(x^2 + 1)\,dx$ via substitution + by-parts

Source: TA41_clac_int1.pdf Q2. Use the substitution $t = x^2 + 1$, $dt = 2x\,dx$, so $x\,dx = \frac{1}{2}\,dt$. Bounds: $x = 0 \Rightarrow t = 1$; $x = 1 \Rightarrow t = 2$. $\int_0^1 7 x \ln(x^2 + 1)\,dx \;=\; \int_1^2 7 \cdot \frac{1}{2} \ln t\,dt \;=\; \frac{7}{2} \int_1^2 \ln t\,dt.$

Using $\int \ln t\,dt = t(\ln t - 1) + C$ (from Example 4.2's special case of by-parts): $\frac{7}{2} \bigl[t(\ln t - 1)\bigr]_1^2 \;=\; \frac{7}{2}\bigl[2(\ln 2 - 1) - 1 \cdot (0 - 1)\bigr] \;=\; \frac{7}{2}\bigl[2 \ln 2 - 2 + 1\bigr] \;=\; \frac{7}{2}(2 \ln 2 - 1) \;=\; 7 \ln 2 - \frac{7}{2}.$

Answer: A ($7 \ln 2 - 7/2$).

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Problem 6.8 — TA41 Q7: Antiderivative of $f(x) = \dfrac{5 \ln x}{x(\ln x + 2)}$

Source: TA41_calc_int2.pdf Q7. Use the substitution $t = \ln x + 2$, $dt = dx/x$, and $\ln x = t - 2$: $\int \frac{5 \ln x}{x(\ln x + 2)}\,dx \;=\; \int \frac{5(t - 2)}{t}\,dt \;=\; 5 \int \Bigl(1 - \frac{2}{t}\Bigr)\,dt \;=\; 5 t - 10 \ln|t| + C \;=\; 5(\ln x + 2) - 10 \ln|\ln x + 2| + C.$

Absorb the constant $5 \cdot 2 = 10$ into $C$ (the constant is arbitrary): $= 5 \ln x + 10 - 10 \ln|\ln x + 2| + C \;=\; 5 \ln x - 10 \ln(\ln x + 2) + C' \qquad \text{on } (1, +\infty),$ where we dropped absolute values since on $(1, +\infty)$, $\ln x + 2 > 2 > 0$. Answer: B ($5 \ln x - 10 \ln(\ln x + 2)$).

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§7. Common Pitfalls

1. Forgetting the $+C$ on indefinite integrals. An indefinite integral is a family of primitives; without $+C$ it is incomplete. Points are routinely deducted for this on partial-exam questions. (Definite integrals, however, are numbers — no $+C$.)

2. Sign error in integration by parts. The formula is $\int f'g\,dx = fg - \int fg'\,dx$, minus on $\int fg'$. A very common slip is writing $fg + \int fg'$. Double-check by differentiating your answer: $(fg)' = f'g + fg'$ should recover the integrand $f'g$.

3. Wrong bounds after substitution. If you substitute $u = g(x)$ in a definite integral $\int_a^b$, you must change the bounds to $g(a), g(b)$ — or, alternatively, back-substitute to $x$ before evaluating. Leaving the original $x$-bounds with the $u$-antiderivative gives wrong answers.

4. Divergent ≠ PV divergent. A classically divergent integral can have a finite PV (Cauchy odd-cancellation); a classically convergent integral always equals its PV. Never assume PV = 0 unless you verify the symmetric limit explicitly. (See Example 4.6: $\operatorname{PV}\int (2x - 3)\,dx = -\infty$, not $0$.)

5. Confusing definite integral (a number) vs. indefinite (a function family).

   • $\int f(x)\,dx$: a family of primitives parametrised by $C \in \mathbb{R}$ — a set of functions of $x$.
   • $\int_a^b f(x)\,dx$: a single real number that depends on $f, a, b$ but not on any "$x$". So the expression "$x = 2$ in $\int_1^3 x\,dx$" makes no sense — the integration variable is a dummy.

6. Improper at both ends requires TWO independent limits, not one. For $\int_{-\infty}^{+\infty} f(x)\,dx$, the classical improper integral is $\lim_{s \to -\infty} \int_s^c f + \lim_{t \to +\infty} \int_c^t f$ where $s, t$ are independent. Collapsing into a single symmetric $K$-limit gives the PV, which is different and weaker.

7. Using Theorem 1894 without verifying hypotheses. FTC I requires (i) $f$ integrable on $[a,b]$ and (ii) $f$ admits a primitive $P$. If $f$ has a jump discontinuity, $f$ does not have a primitive on any interval crossing the jump, so the FTC fails. E.g., $\operatorname{sgn}(x)$ is integrable on $[-1, 1]$ but has no primitive through $x = 0$ — no function $P$ has $P'(0^-) = -1$, $P'(0^+) = +1$ and $P$ differentiable at $0$.

8. Asymptotic comparison without positivity. Theorems 1932, 1933 require $f, g > 0$ (or at least eventually positive). For oscillating integrands like $\sin x / x$, first apply the absolute convergence test (Theorem 1937): if $\int |f|$ converges, so does $\int f$.

9. Harmonic rule flipped. At $+\infty$, $\int 1/x^p\,dx$ converges iff $p > 1$ (large $p$ makes $f$ decay faster). At a finite singularity like $x \to 0^+$, $\int_0^1 1/x^p\,dx$ converges iff $p < 1$ (small $p$ makes $f$ blow up more slowly). Memorise both: the rule inverts.

10. Assuming $F(x) = \int_a^x f(t)\,dt$ is always differentiable. Theorem 1899 requires $f$ continuous. If $f$ has a jump, $F$ is still Lipschitz/continuous, but $F'$ may not exist at the jump — so $F$ is not a primitive of $f$ there. Example: $f$ piecewise constant with a jump at $0$ has $F$ with a corner at $0$ — $F'_-(0) \neq F'_+(0)$.

11. Losing the modulus $|f|$ in $\int f'/f\,dx = \ln|f|$. The log shortcut requires the absolute value when $f$ can be negative. On an interval where $f$ has constant sign, the absolute value can be dropped with the appropriate sign, but writing $\ln(\text{negative})$ is undefined.

12. Not recognising a shortcut-applicable pattern. Before reaching for by-parts, always scan for:

    • $f' \cdot f^a$ pattern ($\rightarrow f^{a+1}/(a+1)$),
    • $f'/f$ pattern ($\rightarrow \ln|f|$),
    • $f' \cdot e^f$ ($\rightarrow e^f$),
    • $e^{ax+b}$ (linear exponent, $\rightarrow \frac{1}{a}e^{ax+b}$).

    These avoid unnecessary work. Example: $\int x e^{x^2}\,dx$ is a 3-second shortcut, not a 30-second by-parts problem.

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End of 03 — Integral Calculus. Cross-links: §01 Linear Algebra (eigenvalue/Sylvester signing reused for Hessians); §02 Differential Calculus (Lagrange MVT used inside the proof of FTC I, §3.7); §04 Probability (continuous RV expectation $E[X] = \int x f_X(x)\,dx$); §05 Mathematical Finance (continuous compounding, duration as integral).