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.

Note that $$(f(0))^{101}=1+\int_{0}^{0}f(t)\mathrm{d}t  \implies (f(0))^{100}=1$$ 
Differentiating each side of  $$ (f(x))^{101}  =  1 + \int_{0}^x f(t)\mathrm{d}t $$ with respect to $x$ gives us that $$101(f(x))^{100}f'(x)=f(x) \iff (f(x))^{99}f'(x)=\frac{1}{101} \tag{1}$$
Note that $$g(x)=(f(x))^{100}\Rightarrow g'(x)=\frac{100}{101}  \Rightarrow g(x)=\frac{100}{101}x+C$$ 
From $(1)$. Thus, we have that $(f(x))^{100}=\frac{100}{101}x+1$ from the fact that $(f(0))^{100}=1$. 
So, since the value is $101$, the answer is $2$.  
