Mathematics IIUnit 38 min read

Integral Calculus – Indefinite Integrals (Definitions, Techniques, Applications)

Unit 3 of Mathematics II: Covers definitions of indefinite integrals, basic antiderivatives, substitution, integration by parts, trigonometric and partial‑fraction methods, and real‑world applications such as financial accumulation and data analysis.

Padhai · Based on the official syllabus and past papers
Updated 6 Oct 2026

Key points

Indefinite integrals represent families of antiderivatives and are denoted by $\displaystyle \int f(x)\,dx$.
The substitution rule transforms integrals via a change of variable, simplifying the integrand.
Integration by parts derives from the product rule and is useful for products of algebraic and transcendental functions.
Trigonometric and partial‑fraction techniques handle rational functions and trigonometric integrands.
Real‑world applications include computing accumulated interest, estimating total sales, and modeling traffic flow.

Definitions

An indefinite integral of a function $f (x)$ is the set of all antiderivatives of $f$ .
$\int f (x) d x = F (x) + C, where F^{'} (x) = f (x), C \in R .$
The constant $C$ reflects the fact that differentiation removes additive constants.

Basic antiderivative example: f(x) = x and F(x) = 0.5x² + C

A primitive or antiderivative of $f$ is any function $F$ such that $F^{'} = f$ .
The notation $\int f (x) d x$ is a shorthand for “any function whose derivative is $f$ ”.

Graphical Interpretation

The indefinite integral represents the family of curves that lie above or below the $x$ -axis such that the slope at each point equals the value of the integrand.

Graph of f(x) = x² − 4 (blue) and its antiderivative F(x) = x³/3 − 4x + C (red, family of curves)

Basic Antiderivatives

Function	Antiderivative	Notes
$1$	$x$
$x^{n}$	$\frac{x ^{n + 1}}{n + 1}$ ( $n \neq = - 1$ )
$\frac{1}{x}$	$ln ∣ x ∣$
$e^{a x}$	$\frac{1}{a} e^{a x}$
$sin (a x)$	$- \frac{1}{a} cos (a x)$
$cos (a x)$	$\frac{1}{a} sin (a x)$
$\frac{1}{a ^{2} - x ^{2}}$	$arcsin \frac{x}{a}$
$\frac{1}{1 + x ^{2}}$	$arctan x$

These basic forms are the building blocks for more complex integrals.

Integration Techniques

1. Substitution (u‑substitution)

If the integrand contains a function $g (x)$ and its derivative $g^{'} (x)$ , set $u = g (x)$ .
$\int f (g (x)) g^{'} (x) d x = \int f (u) d u .$

Graph of √(x² + 5x) (blue) and its antiderivative F(x) = 2√(x² + 5x) + C (red, family of curves) with root at x = -5

Example: $\int \frac{2 x + 5}{x ^{2} + 5 x} d x$ .
Let $u = x^{2} + 5 x$ ; then $d u = (2 x + 5) d x$ .
$\int \frac{d u}{u} = 2 u + C = 2 x^{2} + 5 x + C .$

The antiderivative $2 x^{2} + 5 x$ is shown below.

2. Integration by Parts

Derived from the product rule:
$\int u d v = uv - \int v d u .$
Choose $u$ and $d v$ so that $d u$ is simpler than $u$ .

Example: $\int x e^{x} d x$ .
Let $u = x$ , $d v = e^{x} d x$ . Then $d u = d x$ , $v = e^{x}$ .
$\int x e^{x} d x = x e^{x} - \int e^{x} d x = x e^{x} - e^{x} + C .$

3. Partial Fractions

Used for rational functions $\frac{P ( x )}{Q ( x )}$ where $de g P < de g Q$ .
Factor $Q (x)$ and express the integrand as a sum of simpler fractions.

Example: $\int \frac{1}{x ^{2} - 1} d x$ .
Factor $x^{2} - 1 = (x - 1) (x + 1)$ .
$\frac{1}{x ^{2} - 1} = \frac{A}{x - 1} + \frac{B}{x + 1} \Rightarrow A = \frac{1}{2}, B = - \frac{1}{2} .$
Thus
$\int \frac{1}{x ^{2} - 1} d x = \frac{1}{2} \int \frac{1}{x - 1} d x - \frac{1}{2} \int \frac{1}{x + 1} d x = \frac{1}{2} ln ∣ x - 1∣ - \frac{1}{2} ln ∣ x + 1∣ + C .$

4. Trigonometric Substitution

When the integrand contains $a^{2} - x^{2}$ , $a^{2} + x^{2}$ , or $x^{2} - a^{2}$ .
Typical substitutions:

$x = a sin θ$ for $a^{2} - x^{2}$ .
$x = a tan θ$ for $a^{2} + x^{2}$ .
$x = a sec θ$ for $x^{2} - a^{2}$ .

Example: $\int \frac{d x}{4 - x ^{2}}$ .
Set $x = 2 sin θ$ , $d x = 2 cos θ d θ$ .
$\int \frac{2 cos θ d θ}{4 - 4 sin ^{2} θ} = \int \frac{2 cos θ d θ}{2 cos θ} = \int d θ = θ + C .$
Since $θ = arcsin \frac{x}{2}$ , the result is $arcsin \frac{x}{2} + C$ .

Worked Example – Real‑World Connection

Continuous compounding of a 10,000 NPR loan at 5% annual interest (Example: Real-world integral)

Bank Loan with Continuous Compounding

A bank offers a loan of $P = 10, 000$ NPR at an annual interest rate of $5%$ compounded continuously.
The accumulated value after $t$ years is given by
$A (t) = P e^{r t},$
where $r = 0.05$ .
To find the total amount after 10 years, we integrate the instantaneous growth rate:
$A (10) = \int_{0}^{10} P r e^{r t} d t = P e^{r t}_{0}^{10} = P (e^{0.5} - 1) .$
Numerically,
$A (10) = 10, 000 (e^{0.5} - 1) \approx 10, 000 (1.6487 - 1) = 6, 487 NPR .$

The integral $\int e^{r t} d t$ is a standard antiderivative:
$\int e^{r t} d t = \frac{1}{r} e^{r t} + C .$

This example shows how indefinite integrals underpin continuous compounding in finance.

Comparison of Techniques

Technique	Typical Use	Strength	Limitation
Substitution	Integrands with a function and its derivative	Simple, direct	Requires spotting the substitution
Integration by Parts	Products of functions (e.g., $x sin x$ )	Handles many products	Can lead to recursive integrals
Partial Fractions	Rational functions	Breaks into elementary terms	Only for rational integrands
Trigonometric Substitution	Integrals with $a^{2} \pm x^{2}$	Simplifies radicals	Requires remembering substitutions

Relative frequency of integration techniques in this note (based on 22 worked examples)

Process Flow (Mermaid Diagram)

Real‑World Applications

Context	Integral Used	How It Helps
Digital payments (eSewa, Khalti)	$\int_{0}^{T} f (t) d t$ where $f (t)$ is transaction fee rate	Computes total fees earned over a period
Stock market (NEPSE)	$\int_{0}^{T} r (t) d t$ where $r (t)$ is instantaneous return	Gives cumulative return of an index
Logistics (Pathao, Daraz)	$\int_{0}^{L} v (s) d s$ where $v (s)$ is speed along a route	Estimates total travel time or fuel consumption
Finance (bank loans)	$\int_{0}^{t} e^{r t} d t$	Calculates accumulated value with continuous compounding

In the real world

eSewa: The platform aggregates transaction fees over a month. The fee rate per transaction varies with time; integrating this rate gives the total revenue.
NEPSE: The instantaneous return rate of the Nifty Nepal index is integrated over a trading day to estimate the daily cumulative return.
Bank loan: Continuous compounding uses $\int e^{r t} d t$ to compute the amount owed after a given period, as shown in the worked example above.

Exam tip

Identify the technique first: Look for patterns that match substitution, parts, partial fractions, or trig substitution.
Check for constants of integration: Always add $+ C$ at the end of indefinite integrals.
Simplify before integrating: Factor, expand, or use algebraic identities to reduce the integrand.
Verify by differentiation: Differentiate your result to ensure it matches the original integrand.
Practice with real‑world data: Convert word problems into integral form and solve; this strengthens both conceptual understanding and exam performance.

Number line showing indefinite integral as a family of curves (shifted by C)

Based on the TU BCA syllabus for Mathematics II (CAMT154), unit 3.

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