Question about an integral.. Why is $a = $2? I found the item in monbukagakusho 2013 math B exam.

Consider the function
  $$F(x) = \int_a^x f(t)\ dt = x^3 - 2x^2 - x - a$$
  with $a \ne 0$. Find $a$.

I looked at the answer sheet and $a = 2$, but I ended up with weird equation (?)
$0 = a^3-2a^2-2a$
so what should I do to get $a = 2$?
Thank you 
 A: Differentiating both sides of the equation, with respect to $x$, you get 
$$f(x) = 3x^2 - 4x - 1$$
Now you want to integrate $f$ between $a$ and $x$; this yields
$$F(x) = \int_a^x 3t^2 - 4t - 1 \ dt = |t^3 - 2t^2 - t|_a^x = x^3 - 2x^2 - x - a^3 + 2a^2 + a$$
But you want $F(x) =  x^3 - 2x^2 - x  + a$; this implies $-a^3   + 2a^2 = 0$, and given that $a \neq 0$, the only solution is $a = 2$
A: If
$$
F(x) = \int_a^x f(t)\ dt = x^3 - 2x^2 - x - a
$$
then $F'(x) = f(x)$. So $f(x) = 3x^2 - 4x - 1$. Now then
$$
\int_a^x 3t^2 - 4t - 1\; dt = t^3 - 2t^2 - t\bigg]_a^x =  x^3 - 2x^2 - x - (a^3-2a^2 - a).
$$
So all you have to do is to solve
$$
a^3 - 2a^2 - a = a.
$$
That is
$$
a^3 - 2a^2 - 2a= 0
$$
As noted in above/and below comments, $a= 2$ is not a solution. But if you had $F(x) = x^3 -2x^2 -x \color{red}+ a$, then you would need to solve $a^3 -2a^2 = 0$ which indeed has solutions $0$ and $s$.
A: Since for all $x$ 
$$\int_a^x \mbox{something } dt = x^3- 2x^2 -x -a$$
hold, in particular this holds for $x=a$. So you get
$$0=\int_a^a \mbox{something } dt = a^3- 2a^2 -a -a$$
So your equation works, but the solution is not $a=2$.
But if you had 
$$F(x) = \int_a^x \mbox{something } dt = x^3- 2x^2 -x +a$$
then you get (substituting $x=a$) the equation
$$a^2(a-2)=0$$
so maybe there is some mistake on the exercise.
