Let f:[1,)R be differentiable.

Question

Let $f:[1,\infty)	o\mathbb{R}$ be differentiable. If $6\displaystyle\int_1^x f(t)\,dt = 3xf(x) + x^2 - 4$ for all $x\ge1$, then the value of $f(2)-f(3)$ is:

Accepted Answer

Differentiating: $6f(x) = 3f(x)+3xf'(x)+2x \implies 3f(x)-2x = 3xf'(x)$, i.e.

$$\frac{d}{dx}\left(\frac{f(x)}{x}\right) = -1$$

(after simplification), so $\dfrac{f(x)}{x} = -x+k$. At $x=1$: from the original relation, $0 = 3f(1)+1-4 \implies f(1)=1$, giving $k=2$. So $f(x) = -x^2+2x$.

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

Solutions · 1