Key Concepts and Formulas

• Leibniz Integral Rule: If $F(x) = \int_{a(x)}^{b(x)} g(t) dt$, then $F'(x) = g(b(x)) \cdot b'(x) - g(a(x)) \cdot a'(x)$. When $a(x) = 0$ and $b(x) = x$, this simplifies to $\frac{d}{dx} \int_0^x g(t) dt = g(x)$.
• Product Rule: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$.
• Separable Differential Equations: If $\frac{dy}{dx} = G(x)H(y)$, then $\int \frac{dy}{H(y)} = \int G(x) dx$.

Step-by-Step Solution

1. Differentiate the Given Integral Equation

We are given $\int_0^x t f(t) dt = x^2 f(x)$ for $x>0$. We want to differentiate both sides with respect to $x$ to eliminate the integral.

• Differentiate the Left Hand Side (LHS): The LHS is $\int_0^x t f(t) dt$. Why we do this: Applying Leibniz's rule will allow us to eliminate the integral. Using the Leibniz Integral Rule, $\frac{d}{dx} \int_0^x t f(t) dt = x f(x)$.

• Differentiate the Right Hand Side (RHS): The RHS is $x^2 f(x)$. Why we do this: We need to use the product rule since both $x^2$ and $f(x)$ are functions of $x$. Using the Product Rule, $\frac{d}{dx} (x^2 f(x)) = 2x f(x) + x^2 f'(x)$.

• Equate the Differentiated Sides: We have $x f(x) = 2x f(x) + x^2 f'(x)$.

2. Form a Separable Differential Equation

We have the equation $x f(x) = 2x f(x) + x^2 f'(x)$. We need to rearrange this into a separable differential equation.

• Rearrange the equation: Why we do this: We want to isolate terms with $f(x)$ on one side and terms with $f'(x)$ on the other. Subtracting $2x f(x)$ from both sides, we get $-x f(x) = x^2 f'(x)$.

• Separate variables: Why we do this: To integrate, we need to isolate $f(x)$ and $f'(x)$ on one side and $x$ terms on the other. Recall that $f'(x) = \frac{df}{dx}$. $-x f(x) = x^2 \frac{df}{dx}$. Dividing both sides by $x^2 f(x)$ and multiplying by $dx$, we get $\frac{df}{f(x)} = -\frac{x}{x^2} dx = -\frac{1}{x} dx$.

3. Integrate to Find the General Solution for $f(x)$

Now that the variables are separated, we integrate both sides of the equation.

• Integrate both sides: $\int \frac{1}{f(x)} df = \int -\frac{1}{x} dx$

• Integrate the LHS: $\int \frac{1}{f(x)} df = \ln|f(x)|$

• Integrate the RHS: $\int -\frac{1}{x} dx = -\ln|x| + C_1$ Since $x > 0$, we have $-\ln x + C_1$.

• Combine and Solve for $f(x)$: $\ln|f(x)| = -\ln x + C_1$ Why we use $\ln|C|$ for the constant: It simplifies the algebra of logarithmic terms. Let $C_1 = \ln|C|$. Then $\ln|f(x)| = -\ln x + \ln|C| = \ln\left|\frac{C}{x}\right|$. Exponentiating both sides, we get $|f(x)| = \left|\frac{C}{x}\right|$, so $f(x) = \frac{C}{x}$.

4. Use the Initial Condition to Determine the Constant

We are given $f(2) = 3$. We use this to find the value of $C$.

• Substitute the initial condition: Why we do this: This gives us a specific point that the solution must pass through. Substituting $x=2$ and $f(2)=3$ into $f(x) = \frac{C}{x}$, we get $3 = \frac{C}{2}$.

• Solve for $C$: $C = 3 \times 2 = 6$.

• Write the particular solution: $f(x) = \frac{6}{x}$

5. Calculate $f(6)$

We want to find $f(6)$.

• Substitute $x=6$ into the particular solution: Why we do this: To find the value of the function at $x=6$. $f(6) = \frac{6}{6} = 1$.

Common Mistakes & Tips

• Leibniz Rule: Ensure you correctly apply Leibniz's rule, especially when the limits of integration are functions of $x$.
• Product Rule: Do not forget the product rule when differentiating products of functions.
• Constant of Integration: Always add the constant of integration after performing indefinite integration.

Summary

We started with an integral equation and transformed it into a differential equation using Leibniz's rule. After separating variables and integrating, we obtained a general solution for $f(x)$. We then used the initial condition $f(2) = 3$ to find the specific solution $f(x) = \frac{6}{x}$. Finally, we calculated $f(6)$ to be 1.

The final answer is \boxed{1}, which corresponds to option (A).