Q.If 𝑓(𝑎 + 𝑏 − 𝑥) = 𝑓(𝑥), then ∫ 𝑥 𝑓(𝑥)𝑑𝑥 𝑏 𝑎 is equal to
(A) 𝑎+𝑏 2 ∫ 𝑓(𝑏 − 𝑥)𝑑𝑥 𝑏 𝑎
(B) 𝑎+𝑏 2 ∫ 𝑓(𝑎 − 𝑥)𝑑𝑥 𝑏 𝑎
(C) 𝑏−𝑎 2 ∫ 𝑓(𝑥) 𝑑𝑥 𝑏 𝑎
(D) 𝑎+𝑏 2 ∫ 𝑓(𝑥)𝑑𝑥 𝑏 𝑎

Concept: King Property of Definite Integrals — when $f(a+b-x)=f(x)$, the integral remains unchanged after replacing $x$ with $a+b-x$.

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

Step 1: Apply the substitution $x \to a+b-x$.

Since $f(a+b-x)=f(x)$, we get:

Step 2: Add the two expressions for $I$: …

The King Property is one of those beautiful symmetries in definite integrals. It says:

Why? Because the substitution $x \to a+b-x$ simply reverses the interval — the limits swap, but the minus sign from $dx$ flips them back. It’s a pure renaming of the variable of integration.

Now, the problem gives us an extra condition: $f(a+b-x) = f(x)$. That means the function is symmetric about the midpoint of $[a,b]$. When that happens, the King Property becomes even more powerful — it lets us relate integrals of $x f(x)$ to integrals of $f(x)$ itself.

Let’s work through it.

1. Start with the integral we want:

2. Apply the King substitution: let $t = a+b-x$. Then $x = a+b-t$, and $dx = -dt$. When $x=a$, $t=b$; when $x=b$, $t=a$. So:

The two minus signs cancel (one from $dt$, one from swapping limits), giving:

3. Use the given condition: $f(a+b-t) = f(t)$. So:

4. Now split the integral:

But the second term is exactly $I$ again (just with the dummy variable $t$ instead of $x$). So:

$I = (a+b)\int_a^b f(t)\,dt - I$ …

Method: King Property applied to $\int_a^b x\,f(x)\,dx$

Use this whenever a definite integral carries an extra factor of $x$ (or $a+b-x$) multiplying a function that behaves nicely under the reflection $x\mapsto a+b-x$.

Steps

Step 1: Write the integral and reflect it.

For $I=\displaystyle\int_a^b g(x)\,dx$, the King Property gives a second, equal expression by replacing every $x$ with $a+b-x$ (the limits stay $a$ and $b$):

Step 2: Simplify the reflected integrand using the given symmetry.

Here $g(x)=x\,f(x)$, so $g(a+b-x)=(a+b-x)\,f(a+b-x)$. Any condition on $f$ (such as $f(a+b-x)=f(x)$) is what collapses the reflected form — always substitute it in at this step.

Step 3: Add the two forms of $I$. …

Common Mistakes

Mistake 1: Changing the limits when applying the King Property.

Why it's wrong: the substitution $x\to a+b-x$ leaves the limits as $a$ and $b$ — only the argument of the function changes. Students who rewrite them as $\int_{a+b-a}^{a+b-b}$ end up with swapped or altered bounds. Correct approach: keep $\int_a^b$; replace only $x$ inside the integrand.

Mistake 2: Forgetting to reflect the extra $x$ factor.

Why it's wrong: the reflection must be applied to the whole integrand $x f(x)$, giving $(a+b-x)f(a+b-x)$, not just to $f$. Dropping the $(a+b-x)$ is what makes the $2I$ trick fail. Correct approach: reflect every occurrence of $x$, then add. …