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

$$6 \int_{1}^{x} f(t) \, dt = 3xf(x) + x^3 - 4$$

Differentiating with respect to $$x$$:

$$6f(x) = 3f(x) + 3xf'(x) + 3x^2$$

$$3f(x) - 3xf'(x) = 3x^2 \implies f(x) - xf'(x) = x^2$$

Rearranging into a recognizable quotient rule form:

$$\frac{xf'(x) - f(x)}{x^2} = -1 \implies \frac{d}{dx}\left(\frac{f(x)}{x}\right) = -1$$

Integrating both sides:

$$\frac{f(x)}{x} = -x + C \implies f(x) = -x^2 + Cx$$

Substitute $$x = 1$$ into the original integral equation (since $$\int_1^1 = 0$$):

$$0 = 3f(1) + 1 - 4 \implies f(1) = 1$$

Using $$f(1) = 1$$ in our function formula:

$$1 = -1 + C \implies C = 2 \implies f(x) = -x^2 + 2x$$

• $$f(2) = -(2)^2 + 2(2) = 0$$

• $$f(3) = -(3)^2 + 2(3) = -3$$

$$f(2) - f(3) = 0 - (-3) = 3$$