If f(a+b-x) = f(x) then ∫ₐᵇ xf(x) dx is equal to
(a) (a+b)/2 ∫ₐᵇ f(a+b+x) dx
(b) (a+b)/2 ∫ₐᵇ f(b-x)

Q: Wait, why do we use the King's Property here? Can't we just integrate by parts?

Integration by parts is a valid technique, but it requires knowing the form of $f(x)$. Here, $f(x)$ is an arbitrary function satisfying a symmetry condition. The King's Property is specifically designed to exploit this symmetry, making it the most elegant and efficient path to the solution.

Q: Why does $f(a+b-x)$ become $f(x)$? Is that always true?

It is not always true! It is true *only* because the problem statement explicitly gives us the condition $f(a+b-x) = f(x)$. Always check the problem statement for such constraints; they are the keys to unlocking the problem.

Q: Why did we add $I$ to itself at the end?

This is the 'aha!' moment of the problem. By applying the King's Property, we transformed the integral into a new form that, when expanded, contained the original integral $I$ as a term. Adding $I$ to both sides allows us to isolate it algebraically, effectively solving the integral without ever needing to know the explicit form of $f(x)$.

Question

If $f(a+b-x) = f(x)$ then $\int_{a}^{b} xf(x) dx$ is equal to

Options:

(A) $\frac{a+b}{2} \int_{a}^{b} f(a+b+x) dx$

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

(C) $\frac{a+b}{2} \int_{a}^{b} f(x) dx$

(D) $\frac{b-a}{2} \int_{a}^{b} f(x) dx$

Sigma Percentile · Accepted Answer

### The Symphony of Symmetry: Unlocking the Definite Integral

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving an integral; we are uncovering a hidden harmony.

When you encounter a problem like $\int_{a}^{b} xf(x) dx$ with the condition $f(a+b-x) = f(x)$, it is easy to feel intimidated. You might reach for integration by parts, or perhaps you will try to guess the form of $f(x)$.

But stop. Take a breath. The beauty of this problem lies not in brute force, but in the elegant application of symmetry.

### Phase 1: The Setup

First, let us give our integral a name. We define $I = \int_{a}^{b} xf(x) dx$.

By naming it, we transform it from a daunting expression into a manageable entity. We are essentially looking for the 'weighted average' of the function $f(x)$ over the interval $[a, b]$.

### Phase 2: The King's Property

Whenever you see a definite integral with limits $a$ and $b$, your mind should immediately jump to the 'King's Property'. This is one of the most powerful tools in your JEE toolkit.

It states that for any integrable function $g(x)$, the following holds true:

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

Why does this work? Imagine the graph of $g(x)$ from $a$ to $b$. Replacing $x$ with $a+b-x$ is equivalent to reflecting the interval $[a, b]$ about its midpoint, $\frac{a+b}{2}$.

The area under the curve remains unchanged because we are simply traversing the same interval in reverse. It is a geometric truth that simplifies algebraic nightmares.

### Phase 3: The Transformation

Let us apply this property to our integral $I$. We replace every instance of $x$ with $(a+b-x)$:

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

At first glance, this looks more complicated. But look closely at the condition provided in the problem: $f(a+b-x) = f(x)$.

This is the key that unlocks the door. The function is symmetric about the midpoint of the interval. Substituting this condition back into our integral, we get:

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

### Phase 4: The Algebraic Dance

Now, let us expand the integrand. We distribute $f(x)$ across the terms $(a+b)$ and $-x$:

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

Because integration is a linear operator, we can split this into two separate integrals. And here is the magic: the second integral, $\int_{a}^{b} xf(x) dx$, is exactly the original integral $I$ we started with!

So, our equation becomes:

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

### Phase 5: The Final Result

We have arrived at a simple algebraic equation. By adding $I$ to both sides, we get:

$$2I = (a+b) \int_{a}^{b} f(x) dx$$

Dividing by 2, we reach our destination:

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

This result is profound. It tells us that the integral of $x f(x)$ is simply the average of the limits, $\frac{a+b}{2}$, multiplied by the integral of $f(x)$.

The symmetry of the function effectively 'centers' the $x$ term at the midpoint of the interval. You have just solved a complex integral without ever needing to know the explicit form of $f(x)$.

That is the power of symmetry. Keep this tool sharp, and you will conquer any integral the JEE throws your way.

Animated Solution for Mathematics - Definite Integration: If $f (a + b - x) = f (x)$ then $\int_{a}^{b} x f (x) d x$ is equal to

Select Answer:

Visualized Solution

Define the Integral $I$

The King's Property

Applying the Property

Utilizing Symmetry

Splitting the Integral

Identifying the Original Integral

Solving for $I$

Final Result

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Symphony of Symmetry

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(a+b−x)=f(x) then ∫ab​xf(x)dx is equal to