I have the following piece wise function. Suppose that $f : \!R \rightarrow \!R$ has derivatives of all orders.
Also, $f(0)=0$ $$ g(t)= \begin{cases} f(t)/t \quad t\neq 0 \\ f'(0)\quad t=0\\ \end{cases} $$
Prove that $g(t) = \frac{f(t)}{t} = \int_{0}^{1} f'(tx)dx$
Below is what I have tried:
Using FTC, $\int_{0}^{1} f'(tx)dx = f(t \cdot 1) - f(t \cdot 0) = f(t) - f(0)$
Since it is provided that $f(0) = 0$, now I have just $f(t)$
And the proof becomes $g(t) = \frac{f(t)}{t} = f(t)$, for this to be true, $t$ need to be $1$. But I cannot find a reason to justify that.
Am I on the right track? Or did I mess up from the beginning?
Ok I figured it out. I need to substitute $u=tx$ as a function of $x$
Now I have $\frac{du}{t}=dx$, put it back into the integral $\int_{0}^{1}f'(tx)dx$. This gives $$\int_{0}^{1} \frac{f'(u)}{t}du$$
Evaluate this integral gives $\frac{f(t)}{t} - \frac{f(0)}{t}$
And since $f(0) = 0$, then $g(t) = \frac{f(t)}{t}$