Let f:[1, infty) rightarrow [2, infty) be a d | vedclass

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(A) Given the equation $6 \int_1^x f(t) dt = 3x f(x) - x^3$.Differentiating both sides with respect to $x$ using the Newton-Leibniz theorem:$6 f(x) =

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$6$

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(A) Given the equation $6 \int_1^x f(t) dt = 3x f(x) - x^3$.Differentiating both sides with respect to $x$ using the Newton-Leibniz theorem:$6 f(x) = 3 f(x) + 3x f'(x) - 3x^2$.Rearranging the terms:$3x f'(x) = 3 f(x) + 3x^2 \Rightarrow x f'(x) - f(x) = x^2$.Dividing by $x^2$ (assuming $x \geq 1$):$\frac{x f'(x) - f(x)}{x^2} = 1 \Rightarrow \frac{d}{dx} \left( \frac{f(x)}{x} \right) = 1$.Integrating both sides with respect to $x$:$\frac{f(x)}{x} = x + C$.Given $f(1) = 2$,we substitute $x=1$:$\frac{f(1)}{1} = 1 + C \Rightarrow 2 = 1 + C \Rightarrow C = 1$.Thus,$f(x) = x^2 + x$.Calculating $f(2)$:$f(2) = 2^2 + 2 = 4 + 2 = 6$.

Let $f:[1, \infty) \rightarrow [2, \infty)$ be a differentiable function such that $f(1)=2$. If $6 \int_1^x f(t) dt = 3x f(x) - x^3$ for all $x \geq 1$,then the value of $f(2)$ is

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