$v(x)$ given $a(x)$, where $v_0 = 0$ and $x_0 = 0$
I'm clueless. This is what I thought:
$dv = v, dx = x$
$a(x) = \frac{dv}{dt} \frac{dx}{dx} = \frac{dx}{dt} \frac{dv}{dx} = v\frac{dv}{dx} = \frac{v^2}{x}$
$v = \sqrt{a(x)x}$
But that makes no sense. I don't know why. It doesn't make any sense that I wouldn't have to integrate anything. But I don't get what's wrong with that algebra.
$$a(x)=\dfrac{dv}{dt}=\dfrac{dv}{dx}\dfrac{dx}{dt}=v \dfrac{dv}{dx}$$
$$a(x)dx=vdv$$
$$\int_{x_0=0}^{x} a(\bar{x}) d\bar{x}=\frac{1}{2}(v(x)^2-v_0^2)$$
$$\int_{0}^{x} a(\bar{x}) d\bar{x}=\frac{1}{2}v(x)^2$$
$$v(x)=\pm\sqrt{2\int_{0}^{x} a(\bar{x}) d\bar{x}}$$