Key Concepts and Formulas

• The King Property of Definite Integrals: For a definite integral $\int_a^b f(x) \, dx$, the property states that $\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$. This is particularly useful when the integrand $f(x)$ or the sum $f(x) + f(a+b-x)$ simplifies.
• Greatest Integer Function: $[x]$ denotes the greatest integer less than or equal to $x$.
• Algebraic Simplification: Recognizing and simplifying perfect square trinomials.

Step-by-Step Solution

Let the given integral be $I$. $I = \int\limits_4^{10} {{{\left[ {{x^2}} \right]dx} \over {\left[ {{x^2} - 28x + 196} \right] + \left[ {{x^2}} \right]}}}$

Step 1: Simplify the quadratic expression in the denominator. The expression $x^2 - 28x + 196$ is a perfect square trinomial. $x^2 - 28x + 196 = x^2 - 2(x)(14) + 14^2 = (x-14)^2$ Reasoning: Simplifying this expression makes the integrand easier to work with and reveals the structure needed for applying integral properties. Substituting this back into the integral, we get: $I = \int\limits_4^{10} {{{\left[ {{x^2}} \right]dx} \over {\left[ {{{\left( {x - 14} \right)}^2}} \right] + \left[ {{x^2}} \right]}}} \quad \quad \ldots(1)$

Step 2: Apply the King Property. The limits of integration are $a=4$ and $b=10$. Thus, $a+b = 4+10 = 14$. We replace $x$ with $(a+b-x) = (14-x)$ in the integrand. Let $f(x) = \frac{\left[ {{x^2}} \right]}{\left[ {{{\left( {x - 14} \right)}^2}} \right] + \left[ {{x^2}} \right]}$. We need to find $f(14-x)$.

• The term $\left[ {{x^2}} \right]$ becomes $\left[ {{ (14-x)^2 }} \right]$. Since $(14-x)^2 = (x-14)^2$, this becomes $\left[ {{ (x-14)^2 }} \right]$.
• The term $\left[ {{ (x-14)^2 }} \right]$ becomes $\left[ {{ ((14-x)-14)^2 }} \right] = \left[ {{ (-x)^2 }} \right] = \left[ {{ x^2 }} \right]$.
• The term $\left[ {{ x^2 }} \right]$ becomes $\left[ {{ (14-x)^2 }} \right]$. Reasoning: The King Property allows us to transform the integral into an equivalent form by substituting $x$ with the sum of the limits minus $x$. This often leads to a simplification when combined with the original integral. Applying this transformation to the integral $I$: $I = \int\limits_4^{10} {{{\left[ {{ (x-14)^2 }} \right]dx} \over {\left[ {{ x^2 }} \right] + \left[ {{ (x-14)^2 }} \right]}}} \quad \quad \ldots(2)$

Step 3: Combine the original and transformed integrals. Add Equation (1) and Equation (2): $I + I = \int\limits_4^{10} {{{\left[ {{x^2}} \right]} \over {\left[ {{{\left( {x - 14} \right)}^2}} \right] + \left[ {{x^2}} \right]}}} \,dx + \int\limits_4^{10} {{{\left[ {{ (x-14)^2 }} \right]} \over {\left[ {{ x^2 }} \right] + \left[ {{ (x-14)^2 }} \right]}}} \,dx$ 2I = \int\limits_4^{10} \left( {{{\left[ {{x^2}} \right]} + \left[ {{ (x-14)^2 }} \right]} \over {\left[ {{ x^2 }} \right] + \left[ {{ (x-14)^2 }} \right]}}} \right) \,dx Reasoning: Adding the two forms of the integral is the standard method after applying the King Property. This step is designed to create a situation where the integrand simplifies significantly.

Step 4: Simplify the integrand and evaluate the integral. The numerator and denominator of the integrand are identical: $\left[ {{x^2}} \right] + \left[ {{ (x-14)^2 }} \right]$. Therefore, the integrand simplifies to 1. $2I = \int\limits_4^{10} 1 \, dx$ Now, we evaluate this simple definite integral: $2I = \left[ x \right]_4^{10}$ $2I = 10 - 4$ $2I = 6$ Reasoning: The simplification of the integrand to 1 makes the final integration trivial.

Step 5: Solve for $I$. Divide by 2 to find the value of the original integral: $I = \frac{6}{2}$ $I = 3$

Common Mistakes & Tips

• Incorrect application of the King Property: Ensure that $x$ is replaced by $(a+b-x)$ in every instance within the integrand.
• Algebraic errors: Mistakes in simplifying the quadratic expression or in the substitution of $(14-x)$ can lead to incorrect results.
• Handling the Greatest Integer Function: While the greatest integer function can complicate integrals, in this specific problem, the symmetry of the arguments $x^2$ and $(x-14)^2$ under the transformation $x \to 14-x$ made it straightforward. For other problems, one might need to consider intervals where the floor function's value changes.

Summary

The problem was solved by first simplifying the quadratic term in the denominator to $(x-14)^2$. Then, the King Property of definite integrals ($\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$) was applied with $a=4$ and $b=10$. This resulted in an integral that, when added to the original integral, yielded an integrand of 1. Evaluating $\int_4^{10} 1 \, dx$ gave $6$, and dividing by 2 provided the final answer of $3$.

The final answer is \boxed{3}, which corresponds to option (B).