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(D) Given the equation: $6 \int_1^x f(t) dt = 3xf(x) + x^3 - 4$.Differentiating both sides with respect to $x$ using the Leibniz rule:$6f(x) = 3f(x) +

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$3$

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(D) Given the equation: $6 \int_1^x f(t) dt = 3xf(x) + x^3 - 4$.Differentiating both sides with respect to $x$ using the Leibniz rule:$6f(x) = 3f(x) + 3xf'(x) + 3x^2$.Simplifying the equation:$3f(x) = 3xf'(x) + 3x^2$.Dividing by $3$:$f(x) = xf'(x) + x^2$.Rearranging into a linear differential equation form $f'(x) - \frac{1}{x}f(x) = -x$:The integrating factor is $IF = e^{\int -\frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x}$.Multiplying by $IF$: $\frac{1}{x}f'(x) - \frac{1}{x^2}f(x) = -1$.Integrating both sides: $\frac{f(x)}{x} = -x + C$.So,$f(x) = -x^2 + Cx$.At $x=1$,the original equation gives $6 \int_1^1 f(t) dt = 3(1)f(1) + 1^3 - 4$,which implies $0 = 3f(1) - 3$,so $f(1) = 1$.Substituting $f(1)=1$ into $f(x) = -x^2 + Cx$: $1 = -1 + C$,so $C = 2$.Thus,$f(x) = -x^2 + 2x$.Now,$f(2) = -(2)^2 + 2(2) = 0$ and $f(3) = -(3)^2 + 2(3) = -9 + 6 = -3$.Therefore,$f(2) - f(3) = 0 - (-3) = 3$.

Let $f:[1, \infty) \rightarrow \mathbb{R}$ be a differentiable function. If $6 \int_{1}^{x} f(t) dt = 3xf(x) + x^{3} - 4$ for all $x \ge 1$,then the value of $f(2) - f(3)$ is

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