Animated Solution for Mathematics - Definite Integration: If $f(a+b+1-x)=f(x)$, for all $x$, where $a$ and $b$ are fixed positive real numbers, then $\frac{1}{(a+b)}{\int }_a^{b}x(f(x)+f(x+1))dx$ is equal to

Imagine you are standing before a complex integral, one that seems designed to intimidate. You see an $x$ sitting outside a sum of functions, and your first instinct might be to panic.

But in the world of JEE Advanced, intimidation is often just a mask for a hidden, beautiful symmetry. Let us embark on a journey to solve the integral:

Our Secret Weapon

When you see a definite integral with limits $a$ and $b$, and an $x$ that refuses to be integrated easily, your mind should immediately jump to the King's Property:

This property is not just a formula; it is a geometric transformation. It tells us that the area under the curve remains invariant if we flip the variable $x$ relative to the midpoint of the interval $[a,b]$.

Let us apply this to our integral $I$. By replacing $x$ with $(a+b-x)$, we transform our integral into:

Now, we look at the functional equation provided: $f(a+b+1-x)=f(x)$. This is our key.

The second term in our new integral, $f(a+b+1-x)$, is exactly what the problem gives us; it is simply $f(x)$. But what about $f(a+b-x)$?

If we take the given condition $f(a+b+1-x)=f(x)$ and replace $x$ with $(x+1)$, we get:

This simplifies to $f(a+b-x)=f(x+1)$. The puzzle pieces are falling into place.

Now, we have two versions of $I$. Let us call the original one Equation (1) and the transformed one Equation (2). When we add them together, something magical happens:

Look closely at the term in the square brackets: $x+a+b-x$. The $x$ and $-x$ cancel out, leaving only the constant $(a+b)$.

This constant $(a+b)$ in the numerator perfectly cancels the $(a+b)$ in the denominator. We are left with:

We can split this into two integrals: ${\int }_a^{b}f(x)dx+{\int }_a^{b}f(x+1)dx$. Using the King's property in reverse on the second integral, we find that ${\int }_a^{b}f(x+1)dx$ is actually equal to ${\int }_a^{b}f(x)dx$.

Thus, $2I=2{\int }_a^{b}f(x)dx$, which means:

By testing the options, specifically Option (4), we perform a simple substitution $t=x+1$ and find that it matches our result perfectly. We have navigated the complexity and found the simple truth hidden within.