Sum of digits of the $100$ th power of a continuous function in single variable 
Let $f(x)$  be a continuous function such that $f(x) > 0$ for all
  non-negative $x$
And,
$$ (f(x))^{101}  =  1 + \int_{0}^x f(t)dt $$
Then,
$$ (f(101))^{100} = 100A+10B+C $$
Where $A,B,C$  are integers from $[0,9]$  . So, $ A + B + C = ?$

I've tried using Newton-Leibneitz theorem on the initial statement but all I ended up at was that $f(101)$ is an integral multiple of $101$ .
I do not know how to proceed beyond that.

differentiating both sides you get,
$101 f(x)^{100} f'(x) = f(x)$
i.e $f(x)^{99} f'(x) = \frac{1}{101}$
so integrating both sides we get
$f(x)^{100} = \frac{100}{101}x + c$ where $c$ is some constant 
now $f(0) = 1$ from  the initial equation. hence $c = 1 $
Now put $x=101$ to get $f(101)^{100} = 101$
