Question

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:

• A. 3
• B. 1
• C. 6
• D. 2

Hint:

Recognize that \( \frac{d}{dx} \int_0^x f(t) \, dt = f(x) \).

Answer 1

The Correct Option is B

To determine \( f(6) \), we first analyze the integral condition: \(\int_0^x t f(t) \, dt = x^2 f(x)\) for \(x > 0\). Differentiate both sides with respect to \(x\). Applying the Leibniz rule to the left side yields \(\frac{d}{dx}\left(\int_0^x tf(t) \, dt\right) = x f(x) + \int_0^x t \frac{d}{dx}f(t) \, dt\). Differentiating the right side gives \(\frac{d}{dx}(x^2 f(x)) = 2x f(x) + x^2 f'(x)\). Equating these derivatives results in \(x f(x) + \int_0^x t f'(t) \, dt = 2x f(x) + x^2 f'(x)\). By comparing, we see that \(\int_0^x t f'(t) \, dt\) must satisfy \(x f(x) = 2x f(x) + x^2 f'(x)\). This simplifies to \(0 = x f(x) + x^2 f'(x)\). Rearranging gives \(x f(x) + x^2 f'(x) = 0\). For \(x > 0\), we can divide by \(x\) to obtain \(f(x) + x f'(x) = 0\), which is the differential equation \(f'(x)/f(x) = -1/x\).

This differential equation is solved by separating variables: \(\frac{d}{dx}\ln|f(x)| = -\frac{1}{x}\). Integrating both sides yields \(\ln|f(x)| = -\ln|x| + C\), where \(C\) is the constant of integration. This simplifies to \(|f(x)| = \frac{e^C}{x}\). Thus, the function is of the form \(f(x) = \frac{k}{x}\), where \(k = e^C\) is a constant.

Given \( f(2) = 3 \), we substitute this into the function: \(f(2) = \frac{k}{2} = 3\), implying \(k = 6\). Therefore, the function is \(f(x) = \frac{6}{x}\).

Finally, we compute \( f(6) \): \(f(6) = \frac{6}{6} = 1\).