Let f:(0, +infty) to mathbb{R} and F(x) = int | vedclass

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(C) Given $F(x) = \int_0^{x^2} f(t) dt = x^2(1 + x)$.Applying the Leibniz rule for differentiation under the integral sign,we differentiate both sides

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$4$

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(C) Given $F(x) = \int_0^{x^2} f(t) dt = x^2(1 + x)$.Applying the Leibniz rule for differentiation under the integral sign,we differentiate both sides with respect to $x$:$\frac{d}{dx} \left( \int_0^{x^2} f(t) dt \right) = \frac{d}{dx} (x^2 + x^3)$.Using the chain rule,$f(x^2) \cdot \frac{d}{dx}(x^2) = 2x + 3x^2$.$f(x^2) \cdot 2x = 2x + 3x^2$.For $x > 0$,we can divide by $2x$:$f(x^2) = \frac{2x + 3x^2}{2x} = 1 + \frac{3}{2}x$.To find $f(4)$,we set $x^2 = 4$,which implies $x = 2$ (since $x > 0$):$f(4) = 1 + \frac{3}{2}(2) = 1 + 3 = 4$.

Let $f:(0, +\infty) \to \mathbb{R}$ and $F(x) = \int_0^{x^2} f(t) dt$. If $F(x) = x^2(1 + x)$,then $f(4)$ equals

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