derivative function of area function $1-(1-x)^s$

Question

I'm looking for a function $f(z)$ with $\int \limits^x_0 f(z)dz=1-(1-x)^s$ where $0<s=\log_t2<1$ ($t>2$). The derivative of the integral leads to $f(x)=s(1-x)^{s-1}$, but if one integrates this function, the original area function couldn't be calculated. What's wrong here with my considerations?

Answer 1

Integrate within limits $0$ to $x$;

$ \int_0^x f(z)dz=\int_0^x s(1-t)^{s-1}dt$

$=\left(-(1-t)^s\right )_0^x$

$=-(1-x)^s+1$