Q.Choose the correct answer: If $f(a+b-x)=f(x)$, then $\int_{a}^{b}x\,f(x)\,dx$ is equal to (A) $\frac{a+b}{2}\int_{a}^{b}f(b-x)\,dx$ (B) $\frac{a+b}{2}\int_{a}^{b}f(b+x)\,dx$ (C) $\frac{b-a}{2}\int_{a}^{b}f(x)\,dx$ (D) $\frac{a+b}{2}\int_{a}^{b}f(x)\,dx$

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

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

Apply the substitution $x = a+b-t$, so $dx = -dt$, and when $x=a$, $t=b$; when $x=b$, $t=a$. Then

since $f(a+b-t)=f(t)$.

Thus

$I = \int_a^b (a+b)\, f(t)\, dt - \int_a^b t\, f(t)\, dt = (a+b)\int_a^b f(x)\,dx - I.$ …

The King Property of definite integrals is one of those elegant symmetries that turns a seemingly messy problem into a clean result. It says:

Why? Because substituting $x = a+b-t$ reverses the limits but the variable name is just a dummy — the function's shape over the interval is mirrored. When the function itself satisfies $f(a+b-x) = f(x)$, this symmetry becomes even more powerful: the function is symmetric about the midpoint of $[a,b]$.

Here, we have $\int_a^b x\,f(x)\,dx$ with that condition. The trick is to apply the substitution $x \to a+b-x$ to the integral itself, then add the original and transformed versions. Because $f$ is symmetric, the $x$ in the integrand gets replaced by $a+b-x$, and averaging the two forms eliminates the $x$ in favour of the constant midpoint.

Let's work through it.

1. Apply the King substitution. Let $I = \int_a^b x\,f(x)\,dx$. Put $t = a+b-x$. Then $dx = -dt$, and when $x=a$, $t=b$; when $x=b$, $t=a$. So

2. Use the given symmetry. We know $f(a+b-t) = f(t)$. Therefore

3. Rename the dummy variable. Since $t$ is just a placeholder, write it as $x$ again:

4. Add the two expressions for $I$. We have:

Adding them:

5. Factor out the constant. …

Method: King property to remove a linear weight $x$

Use this for $\int_a^b x\,f(x)\,dx$ when $f$ is symmetric about the midpoint, i.e. $f(a+b-x)=f(x)$: the reflection $x\to a+b-x$ replaces the weight by the constant midpoint.

Steps

Step 1: Apply the King substitution.

Apply it to $g(x)=x f(x)$, turning $x$ into $a+b-x$ and $f(a+b-x)$ into $f(x)$ by the given symmetry.

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

Common Mistakes

Mistake 1: Confusing $f(b-x)$ with $f(a+b-x)$.

Why it's wrong: the King property uses the shift $a+b-x$, matching the given condition; options built on $f(b-x)$ or $f(b+x)$ are wrong unless $a=0$. Correct approach: apply $x\to a+b-x$.

Mistake 2: Forgetting to add the two forms of $I$.

Why it's wrong: the weight $x$ is removed only by averaging $I$ with its reflection, giving $2I=(a+b)\int_a^b f$. Correct approach: add the original and King-substituted integrals. …