TITLE: Find the integrable function $g:[0,b]\rightarrow \mathbb R$ in order to satisfy $\int_{0}^{x} t\cdot g(t) dt=x+x^2$
QUESTION [5 upvotes]: I have found $g(x)=\frac{1}{x}+2$ but it doesn't satisfy for $x=0$ , that is needed.
I also tried to define the function $g(x)=\frac{1}{x}+2$, with $ x>0$ and $ x\leq b$, $g(0)=0$.
Is that possible and if not how do i solve this problem?
Can we use Upper and Lower sums to solve it? (Riemann definition)

REPLY [3 votes]: If
$$
\int_0^xt\,g(t)\,\mathrm{d}t=x+x^2
$$
Then, for $x\ge0$,
$$
x\int_0^x|g(t)|\,\mathrm{d}t\ge x+x^2
$$
which implies
$$
\lim_{x\to0}\int_0^x|g(t)|\,\mathrm{d}t\ge\lim_{x\to0}\,(1+x)=1
$$
However, if $g(t)$ were integrable, 
$$
\lim_{x\to0}\int_0^x|g(t)|\,\mathrm{d}t=0
$$
Therefore, no integrable $g$ can satisfy the equation.