Your integrand, $|x(1-x)|$, is positive everywhere except at the two points $x=0$, $x=1$. That means an integral over any interval of positive length will be positive and over any interval of negative length will be negative.
So, the interval of integration must be of zero length, meaning $c=0$.
ADDED:
Your formula for the integral from $1$ to $c$ is wrong: the correct formula depends on whether $c<0$, $0≤c≤1$, or $c>1$, due to the absolute value in the integrand. Therefore your overall formula is also wrong, in general. (Your formula is correct for $c>1$, however.) That absolute value is what makes things tricky, and any correct answer must recognize that absolute value.
Here is more detail on my answer: let $a<b$ and let $f(x)$ be continuous on $[a,b]$ and positive everywhere in that interval except for finitely many values of $x$. (Continuity says $f(x)=0$ at those finitely many values of $x$.) We can then find $p$ and $q$ such that $a<p<q<b$ and none of those exceptional values of $x$ lie in $[p,q]$. We then get
$$\int_a^b f(x)\;dx=\int_a^p f(x)\;dx+\int_p^q f(x)\;dx+\int_q^b f(x)\;dx$$
$$\ge 0+\int_p^q f(x)\;dx+0$$
$$>0 + 0 + 0=0$$
The first inequality is due to $f(x)\ge 0$ in $[a,p]$ and in $[q,b]$, and the second is due to $f(x)>0$ in $[p,q]$. Therefore, the integral is positive.
If $a>b$ similar steps will show us that the integral is negative. So the only way the integral can be zero is for $a=b$.
If you don't like all that theory, you could come up with a correct formula for the integral--the formula would have three cases, for $c<0$, $0≤c≤1$, and $c>1$. The correct formula will show that only $c=0$ allows the integral to be zero. The multiple-case formula is not hard when you set the integrand to $-x(1-x)$ for $c<0$ and for $c>1$ and to $x(1-x)$ for $0≤c≤1$. I'll leave the formula as an exercise for the reader.
