Key Concepts and Formulas

• King Property of Definite Integrals: For a continuous function $F(x)$ on $[0, a]$, $\int\limits_0^a F(x) dx = \int\limits_0^a F(a-x) dx$.
• Linearity of Definite Integrals: For continuous functions $h_1(x)$ and $h_2(x)$ and constants $A$ and $B$, $\int\limits_0^a (A \cdot h_1(x) \pm B \cdot h_2(x)) dx = A \int\limits_0^a h_1(x) dx \pm B \int\limits_0^a h_2(x) dx$.

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Step-by-Step Solution

Step 1: Define the Integral and Apply the King Property Let the given integral be $I$. $I = \int\limits_0^a f(x) g(x) dx \quad \text{(Equation 1)}$ We apply the King Property of definite integrals, $\int_0^a F(x)dx = \int_0^a F(a-x)dx$, with $F(x) = f(x)g(x)$. This substitution is crucial because the given properties of $f(x)$ and $g(x)$ are in terms of $(a-x)$. $I = \int\limits_0^a f(a-x) g(a-x) dx \quad \text{(Equation 2)}$

Step 2: Utilize the Given Function Properties We are given $f(x) = f(a-x)$ and $g(x) + g(a-x) = 4$. From the second property, we can express $g(a-x)$ as $g(a-x) = 4 - g(x)$. Now, substitute $f(a-x) = f(x)$ and $g(a-x) = 4 - g(x)$ into Equation 2. $I = \int\limits_0^a f(x) (4 - g(x)) dx$ This step incorporates the specific characteristics of the functions into the integral, transforming it into a form that can be simplified.

Step 3: Simplify the Integrand and Separate the Integral Expand the integrand: $I = \int\limits_0^a (4f(x) - f(x)g(x)) dx$ Using the linearity property of definite integrals, we split the integral into two parts: $I = \int\limits_0^a 4f(x) dx - \int\limits_0^a f(x)g(x) dx$ The first integral can be simplified by taking the constant 4 outside: $I = 4 \int\limits_0^a f(x) dx - \int\limits_0^a f(x)g(x) dx$ Notice that the second integral, $\int\limits_0^a f(x)g(x) dx$, is precisely our original integral $I$ (from Equation 1). This is a common outcome when applying the King Property and allows us to form an equation for $I$.

Step 4: Solve for $I$ Substitute $I$ back into the equation from Step 3: $I = 4 \int\limits_0^a f(x) dx - I$ Now, we solve this algebraic equation for $I$. Add $I$ to both sides: $I + I = 4 \int\limits_0^a f(x) dx$ $2I = 4 \int\limits_0^a f(x) dx$ Divide by 2 to find the value of $I$: $I = \frac{4}{2} \int\limits_0^a f(x) dx$ $I = 2 \int\limits_0^a f(x) dx$ This final step isolates $I$ and expresses it in terms of $\int_0^a f(x)dx$, matching one of the given options.

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Common Mistakes & Tips

• Incorrect application of the King Property: Ensure you substitute $(a-x)$ for every $x$ in the integrand when applying $\int_0^a F(x)dx = \int_0^a F(a-x)dx$.
• Missing one of the function properties: Both $f(x) = f(a-x)$ and $g(x) + g(a-x) = 4$ are essential. A common error is to only use one of them.
• Algebraic errors: Be careful when rearranging the equation to solve for $I$. Ensure you correctly add or subtract terms from both sides.

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Summary

This problem is solved by strategically applying the King Property of definite integrals. By transforming the integrand using $\int_0^a F(x)dx = \int_0^a F(a-x)dx$, we create an opportunity to use the given properties of $f(x)$ and $g(x)$. Substituting $f(a-x)=f(x)$ and $g(a-x)=4-g(x)$ into the transformed integral leads to an equation where the original integral $I$ appears on both sides. Solving this equation algebraically yields the final result.

The final answer is \boxed{\text{2\int\limits_0^a \, f(x)dx}}.