Let for some function $y = f(x)$, $\int_0^x t f(t) \, dt = x^2 f(x)$, $x > 0$ and $f(2) = 3$. Then $f(6)$ is equal to

Step 1: Differentiate both sides

We are given an integral equation. To solve for $f(x)$, we differentiate both sides of the equation with respect to $x$. On the left side, we use the Fundamental Theorem of Calculus. On the right side, we use the product rule for differentiation.

Step 2: Apply differentiation rules

Applying the Fundamental Theorem of Calculus to the left side, we get $x f(x)$. For the right side, using the product rule $\frac{d}{dx}(uv) = u'v + uv'$, where $u=x^2$ and $v=f(x)$, we get $2x f(x) + x^2 f'(x)$.

Step 3: Rearrange and separate variables

Now, we rearrange the equation to isolate $f'(x)$ and $f(x)$ terms. We move $2x f(x)$ to the left side, then divide by $x^2 f(x)$ to separate the variables, preparing for integration.

Step 4: Integrate both sides

We integrate both sides of the separated equation. The integral of $\frac{f'(x)}{f(x)}$ is $\ln|f(x)|$, and the integral of $-\frac{1}{x}$ is $-\ln|x|$. We add the constant of integration $C$. Since $x > 0$, we can drop the absolute value for $x$.

Step 5: Solve for $f(x)$

Using logarithm properties, $-\ln|x|$ can be written as $\ln|x^{-1}|$. Then, exponentiating both sides, we get $f(x) = e^{\ln|x^{-1}| + C} = e^{\ln|x^{-1}|} e^C$. Let $A = e^C$, which is a positive constant. So, $f(x) = \frac{A}{x}$.

Step 6: Find the constant A

We are given the condition $f(2) = 3$. We substitute $x=2$ into our derived function $f(x) = \frac{A}{x}$ and set it equal to 3 to find the value of the constant $A$.

Step 7: Calculate $f(6)$

Now that we have found $A=6$, our function is $f(x) = \frac{6}{x}$. We can substitute $x=6$ into this function to find the value of $f(6)$.

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