When should I use King's Property versus integrating directly using standard substitution?

Apply King's Property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$ whenever the denominator remains structurally invariant under $x \to a+b-x$ while the numerator changes to its co-function or complementary algebraic partner. If evaluating the antiderivative directly appears algebraically intractable or impossible, King's Property is almost universally the intended solution route.

How do I spot questions that demand the Newton-Leibniz Differentiation Rule?

Look for equations featuring variable upper or lower bounds of integration, such as $g(x) = \int_{0}^{x^2} f(t) \, dt$, or expressions asking for derivatives, monotonicity, or extremum values of an integral function $F(x)$. If the integration boundaries contain $x$ and the question involves $f'(x)$ or critical points, immediately apply Newton-Leibniz differentiation.

Definite Integration Notes, Formulas, Traps & PYQs | ExamiFly

Beta

Please wait while we prepare everything for you

Definite Integration Notes, Formulas, Traps & PYQs | ExamiFly | Examifly

Key Formulas & High-Yield Equations

King's Property (Reflection Property)

$\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$

Valid for any function

f(x)

continuous on

[a, b]

. Essential for eliminating non-symmetrical trigonometric or algebraic terms in the denominator without changing integration limits.

Queen's Property (Halving the Upper Limit)

$\int_{0}^{2a} f(x) \, dx = \int_{0}^{a} \left( f(x) + f(2a-x) \right) dx$

Requires

f(x)

to be continuous on

[0, 2a]

. Simplifies to

2\int_{0}^{a} f(x) \, dx

when

f(2a-x) = f(x)

, and vanishes to

0

when

f(2a-x) = -f(x)

Newton-Leibniz Differentiation Rule

$\frac{d}{dx}\int_{\phi(x)}^{\psi(x)} f(t) \, dt = f(\psi(x)) \cdot \psi'(x) - f(\phi(x)) \cdot \phi'(x)$

Strictly valid when

f(t)

is continuous on

[\phi(x), \psi(x)]

and limit functions

\phi(x)

\psi(x)

are differentiable. Integrand

f(t)

must be independent of

x

Limit of a Riemann Sum

$\lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{kn} f\left(\frac{r}{n}\right) = \int_{0}^{k} f(x) \, dx$

Requires mapping

\frac{r}{n} \to x

\frac{1}{n} \to dx

, with integration lower limit

\lim_{n \to \infty} \frac{1}{n} = 0

and upper limit

\lim_{n \to \infty} \frac{kn}{n} = k

Symmetric Interval Integration (Odd/Even)

$\int_{-a}^{a} f(x) \, dx = \begin{cases} 2\int_{0}^{a} f(x) \, dx & \text{if } f(-x) = f(x) \\ 0 & \text{if } f(-x) = -f(x) \end{cases}$

Strictly requires symmetric interval limits

[-a, a]

and defined

f(x)

across the complete interval domain.

Solved High-Yield Sample Questions

Practice Mode

Question #1JEE Main 2024

Evaluate the definite integral $I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^{2024} x}{\sin^{2024} x + \cos^{2024} x} \, dx$ .

Solution Locked

Step-by-Step Solution & Concept

Step 1

Apply King's Property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$ with $a = 0$ and $b = \frac{\pi}{2}$ . Substitute $x \to \frac{\pi}{2} - x$ to get $I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^{2024}\left(\frac{\pi}{2}-x\right)}{\sin^{2024}\left(\frac{\pi}{2}-x\right) + \cos^{2024}\left(\frac{\pi}{2}-x\right)} \, dx = \int_{0}^{\frac{\pi}{2}} \frac{\cos^{2024} x}{\cos^{2024} x + \sin^{2024} x} \, dx$ .

Step 2

Add the initial expression $I$ and transformed expression $I$ together: $2I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^{2024} x + \cos^{2024} x}{\sin^{2024} x + \cos^{2024} x} \, dx = \int_{0}^{\frac{\pi}{2}} 1 \, dx$ .

Step 3

Evaluate the simplified linear integral: $2I = [x]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} \implies I = \frac{\pi}{4}$ .

Select an option & click Check Answer to reveal solution

Question #2JEE Main 2023

Evaluate the limiting sum $L = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{r}{n^2 + r^2}$ .

Solution Locked

Step-by-Step Solution & Concept

Step 1

Express the series in canonical Riemann sum format by factoring $n^2$ from denominator: $L = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{\frac{r}{n}}{1 + \left(\frac{r}{n}\right)^2}$ .

Step 2

Convert the sum into a definite integral by substituting $\frac{r}{n} \to x$ , $\frac{1}{n} \to dx$ , lower boundary $\lim_{n \to \infty} \frac{1}{n} = 0$ , and upper boundary $\lim_{n \to \infty} \frac{n}{n} = 1$ : $L = \int_{0}^{1} \frac{x}{1 + x^2} \, dx$ .

Step 3

Integrate using substitution $u = 1 + x^2 \implies du = 2x \, dx$ : $L = \frac{1}{2} \left[ \ln(1 + x^2) \right]_{0}^{1} = \frac{1}{2}(\ln 2 - \ln 1) = \frac{1}{2}\ln 2$ .

Select an option & click Check Answer to reveal solution

If you want to explore more chapters of Mathematics then:

Click Here

Mastering Definite Integration: Formulas, Traps & Exam Strategy | ExamiFly

Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

King's Property (Reflection Property)

Queen's Property (Halving the Upper Limit)

Newton-Leibniz Differentiation Rule

Limit of a Riemann Sum

Symmetric Interval Integration (Odd/Even)

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

Master Definite Integration with 250+ Curated NEET & JEE Drills & AI Error-Correction