Question

Let \( f : [0, \infty) \to [0, \infty) \) be continuous on \( [0, \infty) \) and differentiable on \( (0, \infty) \). If \[ f(x) = \int_0^x \sqrt{f(t)} \, dt, \text{ then } f(6) = \, \text{..........} \]

Hint:

When solving differential equations derived from integrals, first differentiate both sides, then solve the resulting equation.

Answer 1

Correct Answer: 9

Given the integral equation \( f(x) = \int_0^x \sqrt{f(t)} \, dt \), we aim to find \( f(6) \) and confirm it fits within the given range [9,9].

First, differentiate both sides with respect to \( x \) using the Fundamental Theorem of Calculus:

\( f'(x) = \sqrt{f(x)} \).

Introduce a substitution: let \( g(x) = \sqrt{f(x)} \). This implies \( f(x) = g(x)^2 \) and \( f'(x) = 2g(x)g'(x) \).

Given \( f'(x) = g(x) \), we equate:

\( 2g(x)g'(x) = g(x) \).

Assuming \( g(x) \neq 0 \), divide through by \( g(x) \) to obtain:

\( 2g'(x) = 1 \).

Integrating both sides with respect to \( x \), we have:

\( g(x) = \frac{x}{2} + C \).

Initially, \( f(0) = 0 = \int_0^0 \sqrt{f(t)} \, dt \), which implies \( g(0) = 0 \), giving \( 0 = 0 + C \), so \( C = 0 \).

Thus, \( g(x) = \dfrac{x}{2} \), and hence \( f(x) = g(x)^2 = \left(\dfrac{x}{2}\right)^2 = \dfrac{x^2}{4} \).

Now, calculate \( f(6) \):

\( f(6) = \dfrac{6^2}{4} = \dfrac{36}{4} = 9 \).