If $\displaystyle 2 \int_{2}^xf(t)\,dt = xf(x) + x^3$ $\forall x \ge 1$ then find $f(2)$

Question

Let $f$ be a real valued function on $[1,\infty)$ such that $f(1) = 3.$

If $\displaystyle 2 \int_{2}^xf(t)\,dt = xf(x) + x^3$ $\forall x \ge 1$ then find $f(2).$

Here is my approach put $x =2$ to get

\begin{align*} 2f(2)+ 8 &= 0\\ f(2) &= -4. \end{align*}

But if I differentiate both sides I get

\begin{align*} 2f(x) &= f(x) + xf'(x) + 3x^2\\ xf'(x) - f(x) &= -3x^2. \end{align*}

Solving this gives $f(x) = -3x^2 + cx.$ Using the initial condition I get

$$f(x) = -3x^2 + 6x$$ or $$f(2) = 0.$$

By different methods I am getting different values of $f(2).$ Can anyone here please tell me which one is correct and why?

Thank you.

Answer 1

First assume that $f(x)$ is piecewise continuous, so the integral makes sense and one has by the equality of the integral that $$f(2)=-4.$$ Now assume that $f$ is continuous in order to apply FTC. Then following your work, one can solve and get that $$f(x)=-3x^2+4x.$$ This means that if $f$ is assumed to be continuous, then $f(1)=1$. As Greg Martin mentioned, the given conditions are inconsistent.