Key Concepts and Formulas

1. King's Property of Definite Integrals: For a continuous function $g(x)$ and constants $a, b$, $\int_a^b g(x) dx = \int_a^b g(a+b-x) dx$. This property is invaluable for integrals where the integrand exhibits symmetry or can be simplified by replacing $x$ with $a+b-x$.

2. Functional Equations and Substitution: Understanding how to manipulate given functional equations by substituting variables is crucial for simplifying expressions involving the function.

3. Change of Variables in Definite Integrals: When performing a substitution $x = \phi(t)$, it is essential to change the differential $dx = \phi'(t) dt$ and update the limits of integration according to the new variable $t$.

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Step-by-Step Solution

Step 1: Define the Integral and Apply King's Property

Let the given integral be $I$. $I = \frac{1}{a+b}\int_a^b x \left( f(x) + f(x+1) \right) dx \quad \ldots(1)$ We apply the King's Property to the integral part of $I$. Let $g(x) = x(f(x) + f(x+1))$. According to King's Property, $\int_a^b g(x) dx = \int_a^b g(a+b-x) dx$. Replacing $x$ with $(a+b-x)$ in the integral: $I = \frac{1}{a+b}\int_a^b (a+b-x) \left( f(a+b-x) + f(a+b-x+1) \right) dx \quad \ldots(2)$

Step 2: Utilize the Given Functional Equation

We are given the functional equation $f(a+b+1-x) = f(x)$ for all $x$. This equation implies a symmetry. We use this to simplify the terms $f(a+b-x)$ and $f(a+b-x+1)$ in equation (2).

• To simplify $f(a+b-x)$: From the given equation $f(a+b+1-x) = f(x)$, let's replace $x$ with $(x+1)$. $f(a+b+1 - (x+1)) = f(x+1)$ $f(a+b-x) = f(x+1) \quad \ldots(3)$

• To simplify $f(a+b-x+1)$: From the given equation $f(a+b+1-x) = f(x)$, let's replace $x$ with $(x-1)$. $f(a+b+1 - (x-1)) = f(x-1)$ $f(a+b+2-x) = f(x-1)$ This is not directly $f(a+b-x+1)$. Let's consider the relation from (3): $f(a+b-x) = f(x+1)$. If we replace $x$ with $(x-1)$ in equation (3), we get: $f(a+b-(x-1)) = f((x-1)+1)$ $f(a+b-x+1) = f(x) \quad \ldots(4)$

Substitute the simplified terms from (3) and (4) back into equation (2): $I = \frac{1}{a+b}\int_a^b (a+b-x) \left( f(x+1) + f(x) \right) dx \quad \ldots(5)$

Step 3: Combine the Integrals and Simplify

We now have two expressions for $I$: Equation (1): $I = \frac{1}{a+b}\int_a^b x \left( f(x) + f(x+1) \right) dx$ Equation (5): $I = \frac{1}{a+b}\int_a^b (a+b-x) \left( f(x) + f(x+1) \right) dx$

Add equation (1) and equation (5): $2I = \frac{1}{a+b}\int_a^b \left[ x \left( f(x) + f(x+1) \right) + (a+b-x) \left( f(x) + f(x+1) \right) \right] dx$ Factor out the common term $(f(x) + f(x+1))$: $2I = \frac{1}{a+b}\int_a^b \left[ x + (a+b-x) \right] \left( f(x) + f(x+1) \right) dx$ Simplify the term in the square brackets: $2I = \frac{1}{a+b}\int_a^b (a+b) \left( f(x) + f(x+1) \right) dx$ Since $(a+b)$ is a constant, it cancels out: $2I = \int_a^b \left( f(x) + f(x+1) \right) dx$ Divide by 2 to get an expression for $I$: $I = \frac{1}{2}\int_a^b f(x) dx + \frac{1}{2}\int_a^b f(x+1) dx \quad \ldots(6)$

Step 4: Manipulate the Second Integral Term

Let's focus on the second integral in equation (6): $\frac{1}{2}\int_a^b f(x+1) dx$. We want to transform this integral to match one of the options. Option (A) has the form $\int_{a-1}^{b-1} f(x+1)dx$. This suggests a substitution that shifts the limits by $-1$ and results in $f(x+1)$.

Let's consider the integral $\int_a^b f(x+1) dx$. Perform a change of variable: Let $u = x+1$. Then $x = u-1$, and $dx = du$. When $x = a$, $u = a+1$. When $x = b$, $u = b+1$. So, $\int_a^b f(x+1) dx = \int_{a+1}^{b+1} f(u) du = \int_{a+1}^{b+1} f(x) dx$.

Substituting this back into equation (6): $I = \frac{1}{2}\int_a^b f(x) dx + \frac{1}{2}\int_{a+1}^{b+1} f(x) dx \quad \ldots(7)$ This expression for $I$ does not directly match any of the options. Let's re-examine the problem and the options.

The question asks for the value of the expression, and the options are in the form of definite integrals. We need to show that $I$ is equal to one of the given options.

Let's go back to the expression $I = \frac{1}{2}\int_a^b f(x) dx + \frac{1}{2}\int_a^b f(x+1) dx$. Consider the second term: $\frac{1}{2}\int_a^b f(x+1) dx$. Let's apply a substitution to change the limits. Let $y = x+1$. Then $x = y-1$ and $dx=dy$. When $x=a$, $y=a+1$. When $x=b$, $y=b+1$. So, $\int_a^b f(x+1) dx = \int_{a+1}^{b+1} f(y) dy$. Thus, $I = \frac{1}{2}\int_a^b f(x) dx + \frac{1}{2}\int_{a+1}^{b+1} f(x) dx$.

Let's consider option (A): $\int_{a - 1}^{b - 1} {f(x+1)dx}$. Let's evaluate this option using a substitution. Let $t = x+1$. Then $x = t-1$ and $dx = dt$. When $x = a-1$, $t = (a-1)+1 = a$. When $x = b-1$, $t = (b-1)+1 = b$. So, $\int_{a - 1}^{b - 1} {f(x+1)dx} = \int_a^b f(t) dt = \int_a^b f(x) dx$. This means option (A) simplifies to $\int_a^b f(x) dx$.

Now, let's see if our $I$ can be simplified to $\int_a^b f(x) dx$. From Step 3, we had $2I = \int_a^b (f(x) + f(x+1)) dx$. We also know from Step 2 that $f(x+1) = f(a+b-x)$. So, $2I = \int_a^b (f(x) + f(a+b-x)) dx$. Using King's Property on $\int_a^b f(a+b-x) dx$: $\int_a^b f(a+b-x) dx = \int_a^b f(x) dx$. Therefore, $2I = \int_a^b f(x) dx + \int_a^b f(x) dx = 2\int_a^b f(x) dx$. This implies $I = \int_a^b f(x) dx$.

So, $I = \int_a^b f(x) dx$. We showed in the evaluation of option (A) that $\int_{a - 1}^{b - 1} {f(x+1)dx} = \int_a^b f(x) dx$. Therefore, $I = \int_{a - 1}^{b - 1} {f(x+1)dx}$.

Step 5: Verification of the Correct Option

We have found that $I = \int_a^b f(x) dx$. Let's re-evaluate the options based on this finding.

Option (A): $\int_{a - 1}^{b - 1} {f(x+1)dx}$ Let $t = x+1$. Then $x = t-1$ and $dx = dt$. Limits: $x=a-1 \implies t=a$, $x=b-1 \implies t=b$. Integral becomes $\int_a^b f(t) dt = \int_a^b f(x) dx$. This matches our result for $I$.

Option (B): $\int_{a + 1}^{b + 1} {f(x + 1)dx}$ Let $t = x+1$. Then $x = t-1$ and $dx = dt$. Limits: $x=a+1 \implies t=a+2$, $x=b+1 \implies t=b+2$. Integral becomes $\int_{a+2}^{b+2} f(t) dt$. This does not match.

Option (C): $\int_{a - 1}^{b - 1} {f(x)dx}$ This does not match our result for $I$.

Option (D): $\int_{a + 1}^{b + 1} {f(x)dx}$ This does not match our result for $I$.

Thus, the correct option is (A).

Common Mistakes & Tips

• Incorrect Substitution in Functional Equation: Carefully check the variable substitutions when using the functional equation to ensure the correct form of $f$ is obtained.
• Forgetting to Change Limits: When performing a change of variables in a definite integral, always remember to update the limits of integration according to the new variable.
• Algebraic Errors: Double-check all algebraic manipulations, especially when combining integrals or simplifying expressions.

Summary

The problem requires the application of King's Property and careful manipulation of the given functional equation. By applying King's Property to the original integral and using the functional equation to simplify the integrand, we obtain a new expression for the integral. Adding the original and the transformed integral leads to a significant simplification, revealing that the integral is equivalent to $\int_a^b f(x) dx$. Finally, by performing a change of variables on the options, we identify that option (A) is the only one that simplifies to $\int_a^b f(x) dx$.

The final answer is \boxed{\int_{a - 1}^{b - 1} {f(x+1)dx}}. This corresponds to option (A).