Answer
\[
f(x)=-1+2 x, \quad a=1
\]
Work Step by Step
\[
\mathrm{A}(\mathrm{x})=-x+x^{2}
\]
$\mathrm{A}(\mathrm{x})$ is the area under the graph of $\mathrm{f}$ and over the interval $[a, x]$
$ \quad f(x)=A^{\prime}(x)=-1+2 x$
$A(a)=-a+a^{2}=0 \Rightarrow a=1$ or $a=0$
$\because f(1)>0$
$\therefore \quad a=1$
Note, we assume that $a\ne 0$, since we assume that the graph remains above the x-axis.
