How to solve $a$ in $\int_a^xf\left(t\right)dt=x^2-2x-3$ How do you solve for $a$ in $\int_a^xf\left(t\right)dt=x^2-2x-3$?
I have differentiated both sides with respect to $x$, but do not know how to continue after this.
 A: Method 1 (your approach)
Differentiating both sides with respect to $x$, $f(x) = 2x-2$.
And thus, $\int_a^xf\left(t\right)dt=\int_a^x\left(2t-2\right)dt=x^2-2x-(a^2-2a)$
Comparing this to the original equation, $x^2-2x-3=x^2-2x-(a^2-2a)$
You can solve it from here.
A: Well after differentiating you know:
$$f(x) = 2x-2$$ and therefore $\int_a^x 2t-2 dt = t^2-2t|_a^x = x^2-2x-a^2+2a$.
From this immediately follows: $a^2-2a = 3$ with the solutions $a=-1$ and $a=3$
A: Method 2(MUST SEE!)
Substitute $x=a$ onto both sides of your original equation.
$\int_a^xf\left(t\right)dt=x^2-2x-3$ becomes $\int_a^af\left(t\right)dt=a^2-2a-3$.
From this, you can IMMEDIATELY notice that $0=a^2-2a-3$, providing you the solutions.
A: Put $F(x) = \displaystyle \int_{a}^x f(t)dt=x^2-2x-3$. Since $F(a) = 0$, $a^2-2a-3 = 0 \implies (a-1)^2 = 4\implies a - 1 = \pm 2 \implies a = -1, 3$. Thus there are $2$ answers: $a = -1,3$.
A: $$F(x)-F(a)=x^2-2x-3\implies f(x)=2x-2$$
$$\int_a^xf\left(t\right)dt=[t^2-2t]_a^x=x^2-2x-a^2+2a$$
$$2a-a^2=-3\implies a^2-2a-3=0\implies a=1,-3$$
A: Note that your constant of $-3$ in $$\int_a^xf\left(t\right)dt=x^2-2x-3$$ is indeed             $-(a^2-2a).$ 
Thus $$ a^2-2a =3$$ which implies $a=3$ or $a=-1$
