Let `f:[0,oo)->R` be a continuous strictly increasing function, such that `f^3(x)=int_0^x t*f^2(t)dt` for every `xgeq0.` Then value of `f(6)` is_______

Explanation

To solve the problem, we need to find the function \( f(x) \) given the equation:

\[ f^3(x) = \int_0^x t \cdot f^2(t) \, dt \]

for every \( x \geq 0 \), and then evaluate \( f(6) \).

### Step 1: Differentiate both sides with respect to \( x \)

Using the Fundamental Theorem of Calculus and the chain rule, we differentiate the left-hand side:

\[ \frac{d}{dx}[f^3(x)] = 3f^2(x)f'(x) \]

Now, we differentiate the right-hand side using Leibniz's rule:

\[ \frac{d}{dx}\left[\int_0^x t \cdot f^2(t) \, dt\right] = x \cdot f^2(x) \]

Equating both derivatives gives us:

\[ 3f^2(x)f'(x) = x \cdot f^2(x) \]

### Step 2: Simplify the equation

Assuming \( f^2(x) \neq 0 \) (which is valid since \( f(x) \) is strictly increasing and continuous), we can divide both sides by \( f^2(x) \):

\[ 3f'(x) = x \]

### Step 3: Solve for \( f'(x) \)

Rearranging gives:

\[ f'(x) = \frac{x}{3} \]

### Step 4: Integrate to find \( f(x) \)

Now we integrate \( f'(x) \):

\[ f(x) = \int \frac{x}{3} \, dx = \frac{x^2}{6} + C \]

where \( C \) is the constant of integration.

### Step 5: Determine the constant \( C \)

To find \( C \), we can use the condition at \( x = 0 \):

Substituting \( x = 0 \) into the original equation:

\[ f^3(0) = \int_0^0 t \cdot f^2(t) \, dt = 0 \]

This implies:

\[ f(0) = 0 \implies \frac{0^2}{6} + C = 0 \implies C = 0 \]

Thus, we have:

\[ f(x) = \frac{x^2}{6} \]

### Step 6: Evaluate \( f(6) \)

Now we substitute \( x = 6 \):

\[ f(6) = \frac{6^2}{6} = \frac{36}{6} = 6 \]

### Final Answer

The value of \( f(6) \) is \( 6 \).

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