Question:

Step 1: Differentiate the given integral equation. The given equation is $$ \int_0^{f(x)}t^2\,dt=\int_0^x t^2 f(t)\,dt $$ First, evaluate the left-hand side integral: $$ \int_0^{f(x)}t^2\,dt = \left[\frac{t^3}{3}\right]_0^{f(x)} = \frac{(f(x))^3}{3} $$ So the equation becomes: $$ \frac{(f(x))^3}{3} = \int_0^x t^2 f(t)\,dt $$ Now, differentiate both sides with respect to $x$. Using the chain rule for the left-hand side and the Fundamental Theorem of Calculus for the right-hand side: $$ \frac{d}{dx}\left(\frac{(f(x))^3}{3}\right) = \frac{d}{dx}\left(\int_0^x t^2 f(t)\,dt\right) $$ $$ \frac{1}{3} \cdot 3(f(x))^2 f'(x) = x^2 f(x) $$ $$ (f(x))^2 f'(x) = x^2 f(x) $$ Step 2: Solve the resulting differential equation. Since $f: (0,\infty)\to(0,\infty)$, we know that $f(x) > 0$. Therefore, we can divide both sides by $f(x)$: $$ f(x) f'(x) = x^2 $$ This is a separable differential equation. We can write $f'(x)$ as $\frac{df}{dx}$: $$ f \frac{df}{dx} = x^2 $$ $$ f\,df = x^2\,dx $$ Integrate both sides: $$ \int f\,df = \int x^2\,dx $$ $$ \frac{f(x)^2}{2} = \frac{x^3}{3} + C $$ Step 3: Determine the constant of integration. We are given the initial condition $f(1)=3$. Substitute $x=1$ and $f(1)=3$ into the equation: $$ \frac{3^2}{2} = \frac{1^3}{3} + C $$ $$ \frac{9}{2} = \frac{1}{3} + C $$ Solve for $C$: $$ C = \frac{9}{2} - \frac{1}{3} = \frac{27 - 2}{6} = \frac{25}{6} $$ Substitute $C$ back into the equation for $f(x)^2$: $$ \frac{f(x)^2}{2} = \frac{x^3}{3} + \frac{25}{6} $$ $$ f(x)^2 = \frac{2x^3}{3} + \frac{25}{3} $$ Step 4: Calculate $8f(2)$. First, find the value of $f(2)$ by substituting $x=2$ into the equation for $f(x)^2$: $$ f(2)^2 = \frac{2(2^3)}{3} + \frac{25}{3} = \frac{2 \cdot 8}{3} + \frac{25}{3} = \frac{16}{3} + \frac{25}{3} = \frac{41}{3} $$ Since $f(x) > 0$, we take the positive square root: $$ f(2) = \sqrt{\frac{41}{3}} $$ Finally, calculate $8f(2)$: $$ 8f(2) = 8\sqrt{\frac{41}{3}} $$