Integration Mastery Reading Time: 15 min JEE Essential

The King's Property Demystified: ∫_a^b f(x) dx = ∫_a^b f(a+b-x) dx

Master this powerful definite integrals property that solves complex JEE problems in seconds.

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Why This Property is Called "King's Property"

The King's Property is one of the most elegant and powerful tools in definite integration. It's called the "King" because it rules over complex integrals and can solve problems that appear impossible at first glance.

The Royal Property

$$ \int_a^b f(x) dx = \int_a^b f(a+b-x) dx $$

This single property can reduce minutes of computation to seconds of insight.

👑 Navigate the Kingdom

Mathematical Proof Key Applications Solved Examples JEE Practice

The Royal Proof: Why This Property Works

Step-by-Step Proof

Step 1: Start with the right-hand side

$$ I = \int_a^b f(a+b-x) dx $$

Step 2: Use substitution

Let $t = a + b - x$

Then $dt = -dx$ or $dx = -dt$

Step 3: Change limits

When $x = a$, $t = a + b - a = b$

When $x = b$, $t = a + b - b = a$

Step 4: Substitute in the integral

$$ I = \int_b^a f(t) (-dt) = \int_a^b f(t) dt $$

Step 5: Rename variable

$$ I = \int_a^b f(x) dx $$

✅ Proof Complete!

We have shown: $$\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$$

Royal Applications: When to Use the King's Property

Application 1: Trigonometric Simplification

The King's Property works wonders with trigonometric functions, especially when you see patterns like:

$$ \int_0^{\pi/2} f(\sin x) dx = \int_0^{\pi/2} f(\cos x) dx $$

This is a special case where $a=0$, $b=\pi/2$, so $a+b-x = \pi/2 - x$

Application 2: Creating Symmetry

When an integral looks complicated but has symmetric limits or the function has special properties:

$$ \int_a^b \frac{f(x)}{f(x) + f(a+b-x)} dx = \frac{b-a}{2} $$

This magical result comes directly from the King's Property!

Application 3: Solving Definite Integrals with Parameters

When integrals contain both $f(x)$ and $f(a+b-x)$ terms:

$$ I = \int_a^b \frac{f(x)}{f(x) + f(a+b-x)} dx $$

Adding this to its "twin" obtained by King's Property often gives the solution instantly.

Royal Examples: The King in Action

Example 1: Classic Trigonometric Application

Evaluate $I = \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx$

Step 1: Apply King's Property

Here $a=0$, $b=\pi/2$, so $a+b-x = \pi/2 - x$

$$ I = \int_0^{\pi/2} \frac{\sin(\pi/2 - x)}{\sin(\pi/2 - x) + \cos(\pi/2 - x)} dx $$

Step 2: Simplify using trigonometric identities

$$ I = \int_0^{\pi/2} \frac{\cos x}{\cos x + \sin x} dx $$

Step 3: Add the original and transformed integrals

$$ 2I = \int_0^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x} dx = \int_0^{\pi/2} 1 dx $$

Step 4: Solve

$$ 2I = \frac{\pi}{2} \Rightarrow I = \frac{\pi}{4} $$

Example 2: JEE Advanced Problem

Evaluate $I = \int_0^1 \frac{\ln(1+x)}{1+x^2} dx$

Step 1: Apply King's Property

Here $a=0$, $b=1$, so $a+b-x = 1 - x$

$$ I = \int_0^1 \frac{\ln(1+(1-x))}{1+(1-x)^2} dx = \int_0^1 \frac{\ln(2-x)}{1+(1-x)^2} dx $$

Step 2: Add original and transformed integrals

$$ 2I = \int_0^1 \frac{\ln(1+x) + \ln(2-x)}{1+x^2} dx $$

Step 3: Simplify the numerator

$$ \ln(1+x) + \ln(2-x) = \ln[(1+x)(2-x)] = \ln(2 + x - x^2) $$

Further manipulation leads to the solution $I = \frac{\pi}{8} \ln 2$

Royal Insights: Special Cases and Pro Tips

👑 Special Case 1: Symmetric Limits

When $a = -b$, then $a + b - x = -x$

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

This is particularly useful for even and odd functions!

👑 Special Case 2: Standard Trigonometric Limits

For $a=0$, $b=\pi/2$:

$$ \int_0^{\pi/2} f(\sin x) dx = \int_0^{\pi/2} f(\cos x) dx $$

This appears frequently in JEE problems!

💡 Pro Tip: When to Suspect King's Property

Integrals with $f(x)$ and $f(a+b-x)$ type terms
Trigonometric integrals with symmetric limits
Integrals where direct integration seems messy
Problems asking to prove integral equalities

Royal Practice: JEE Level Problems

Test Your Royal Skills

1. Evaluate $\int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$

Hint: Use King's Property and add the integrals

2. Prove that $\int_0^1 \frac{\ln(1+x)}{1+x^2} dx = \frac{\pi}{8} \ln 2$

Hint: Apply King's Property with a=0, b=1

3. Evaluate $\int_0^{\pi} \frac{x \sin x}{1 + \cos^2 x} dx$

Hint: Use King's Property with a=0, b=π

Strategy: For each problem, write down what $a+b-x$ becomes, apply King's Property, and look for patterns when you add the original and transformed integrals.

Avoid These Royal Blunders

❌ Common Mistake 1: Wrong Limits

Forgetting to change limits properly when using substitution in the proof.

Fix: Always write limits explicitly when $x=a$ and $x=b$.

❌ Common Mistake 2: Misapplying the Property

Using $f(a-b+x)$ instead of $f(a+b-x)$.

Fix: Remember the pattern: $a+b-x$ (addition in the middle).

✅ Success Formula

Step 1: Identify $a$ and $b$ in $\int_a^b f(x) dx$

Step 2: Write $f(a+b-x)$ explicitly

Step 3: Add original and transformed integrals if needed

Step 4: Look for simplification opportunities

Royal Summary: Key Takeaways

🎯 What to Remember

$\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$ always holds
Particularly powerful for trigonometric integrals
Often used by adding original and transformed integrals
Special cases with symmetric limits are very common

🚀 JEE Strategy

Recognize patterns that suggest King's Property
Practice the standard trigonometric applications
Use it to simplify before attempting integration
Combine with other integration techniques

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