If f(a + b + 1 - x) = f(x) ∀ x, where a and b are fixed positive real numbers, then 1/(a+b) ∫[a to b

Q: Wait, why do we replace $x$ with $x + 1$ in the first step? How did you know to do that?

That is a great observation! We look at the target integral, which contains both $f(x)$ and $f(x + 1)$. Since the given functional equation only gives us a symmetry for $f(x)$, we need a corresponding symmetry for $f(x + 1)$ to simplify the integrand. By shifting $x$ to $x + 1$, we bridge the gap between the two terms.

Q: Why does King's Property work here? I thought it was only for simple integrals.

King's Property, $\int_{a}^{b} g(x) dx = \int_{a}^{b} g(a + b - x) dx$, is a powerful tool precisely because it allows us to eliminate the variable $x$ when it is multiplied by a function. In this problem, the $x$ outside the bracket is the 'annoying' part. By applying the property, the $x$ becomes $(a + b - x)$, and when we add the original and transformed integrals, the $x$ and $-x$ cancel out, leaving only the constant $(a + b)$.

Q: Why is $\int_{a}^{b} f(x + 1) dx$ equal to $\int_{a}^{b} f(x) dx$?

Think of it as a horizontal shift. If you substitute $t = x + 1$, the limits change from $x = a$ to $t = a + 1$ and $x = b$ to $t = b + 1$. However, because of the symmetry $f(a + b + 1 - x) = f(x)$, the area under the curve $f(x + 1)$ between $a$ and $b$ is identical to the area under $f(x)$ between $a$ and $b$. It is essentially shifting the same area horizontally.

Question

If $f(a + b + 1 - x) = f(x) \forall x$, 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

Options:

(A) $\int_{a-1}^{b-1} f(x) dx$

(B) $\int_{a+1}^{b+1} f(x+1) dx$

(C) $\int_{a-1}^{b-1} f(x+1) dx$

(D) $\int_{a+1}^{b+1} f(x) dx$

Sigma Percentile · Accepted Answer

### The Symphony of Symmetry: Unlocking the Integral

Welcome, fellow traveler on the path of JEE Advanced mathematics. Today, we are going to dissect a problem that, at first glance, looks like a tangled mess of functional equations and integrals.

But I promise you, beneath the surface lies a beautiful, elegant symmetry waiting to be revealed. Let us embark on this journey together.

### Phase 1: The Master Key

We are given the functional equation $f(a + b + 1 - x) = f(x)$. This is our North Star.

In the world of competitive exams, whenever you see a functional equation involving $x$ and a sum of constants, your intuition should immediately pivot to symmetry. This equation tells us that the function $f$ is symmetric about the point $\frac{a + b + 1}{2}$.

This is our master key.

### Phase 2: The Hidden Relation

Our target integral is defined as:

$$I = \frac{1}{a + b} \int_{a}^{b} x(f(x) + f(x + 1)) dx$$

Notice the presence of $f(x + 1)$. We have a relation for $f(x)$, but what about $f(x + 1)$?

Let us perform a simple, yet powerful, substitution. Replace $x$ with $x + 1$ in our master equation: $f(a + b + 1 - (x + 1)) = f(x + 1)$.

Simplifying this, we get $f(a + b - x) = f(x + 1)$. This is a crucial piece of the puzzle. Keep it safe; we will need it soon.

### Phase 3: The King's Property

Now, look at the integral $I$. The $x$ multiplying the function is the obstacle.

Whenever you see an $x$ multiplied by a function inside a definite integral, your brain should scream: **King's Property!** Let us apply it: $\int_{a}^{b} g(x) dx = \int_{a}^{b} g(a + b - x) dx$.

Replacing $x$ with $(a + b - x)$ in our integral $I$, we get:

$$I = \frac{1}{a + b} \int_{a}^{b} (a + b - x) [f(a + b - x) + f(a + b - x + 1)] dx$$

Using the relations we derived earlier, $f(a + b - x) = f(x + 1)$ and $f(a + b + 1 - x) = f(x)$, the integrand transforms beautifully into $(a + b - x)(f(x + 1) + f(x))$.

### Phase 4: The Grand Cancellation

Now, let us add our original integral $I$ to this transformed integral $I$. On the left side, we have $2I$.

On the right side, we factor out the common term $(f(x) + f(x + 1))$:

$$2I = \frac{1}{a + b} \int_{a}^{b} [x + (a + b - x)] (f(x) + f(x + 1)) dx$$

Look at that! The $x$ and $-x$ cancel out perfectly, leaving us with $(a + b)$. The $(a + b)$ in the numerator and the $(a + b)$ in the denominator cancel out, leaving us with:

$$2I = \int_{a}^{b} (f(x) + f(x + 1)) dx$$

### Phase 5: The Final Reveal

We are almost there. We can split this into two integrals: $\int_{a}^{b} f(x) dx + \int_{a}^{b} f(x + 1) dx$.

By applying the symmetry property again, 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 simplifies to **$I = \int_{a}^{b} f(x) dx$**.

Looking at our options, we find that $\int_{a-1}^{b-1} f(x + 1) dx$ is the correct match, as it represents the same area shifted by one unit. We have conquered the monster!

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

Select Answer:

Visualized Solution

The Given Functional Equation

Shifting the Variable

The Target Integral

Applying King's Property

Transforming the Integrand

Using Symmetry Relations

The Transformed Integral

Summing the Integrals

Simplifying the Expression

Splitting the Integral

Proving Integral Equality

Final Result for $I$

Matching with Options

Conclusion

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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