Let f:[1,) R be a differentiable function.

Question

Let $f:[1,\infty)	o R$ 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 :

Accepted Answer

$\begin{gathered} 6{\int }_{x}^{1}f(t)dt=3xf(x)+{x}^{2}-4 \ \ 6f(x)=3f(x)+3x{f}^{'}(x)+3{x}^{2} \ \frac{f(x)-x{f}^{'}(x)}{{x}^{2}}=1 \ \frac{d}{dx}(\frac{f(x)}{x})=-1f(x)=-{x}^{2}+kx \ 1=-1+k(f(1)=1) \ f(x)=-{x}^{2}+2x \ f(2)-f(3)=3 \end{gathered}$

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