In Velocity and Position from Acceleration the acceleration was a function of time, so integrating over $t$ worked. But forces often depend on where an object is: a spring pulls harder the more it stretches, and gravity weakens with distance. Then $a$ is a function of $x$, and you can't integrate it over $t$ without first knowing $x(t)$, which is what you're trying to find.
Why it works
The chain rule rescues us. Velocity is $v = dx/dt$, so $$a = \dfrac{dv}{dt} = \dfrac{dv}{dx}\cdot\dfrac{dx}{dt} = v\,\dfrac{dv}{dx}.$$ Now both sides can be written with $x$ alone: $v\,dv = a(x)\,dx$. Integrate from the starting point to the end point: $$\tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 = \int_{x_0}^{x} a(x')\,dx'.$$ Time has disappeared. (Multiply by the mass and this becomes the work-energy theorem, which you'll derive in The Work-Energy Theorem from Newton's Second Law.)
Key idea
When the acceleration depends on position, write $a = v\,\dfrac{dv}{dx}$ and integrate over $x$: the area under the $a$–$x$ graph equals the change in $\tfrac{1}{2}v^2$. The same trick works when $a$ depends on $v$: then $\dfrac{dv}{dx} = \dfrac{a(v)}{v}$.
Worked example: A spring stops a glider
A glider has $a = -4x$ (m/s², m) and passes $x = 0$ at $6\ \text{m/s}$. How far does it go before stopping?
- $v\,dv = -4x\,dx$.
- Integrate from $(x, v) = (0, 6)$ to $(x_{\max}, 0)$: $\tfrac{1}{2}(0 - 36) = -2x_{\max}^2$.
- $x_{\max}^2 = 9$, so $x_{\max} = 3\ \text{m}$.
Answer: $3\ \text{m}$.
Worked example: Drag that depends on speed
A boat coasting at $10\ \text{m/s}$ has $a = -0.5v$ (m/s², m/s). How far does it go while slowing to $4\ \text{m/s}$?
- $v\,\dfrac{dv}{dx} = -0.5v$, so $\dfrac{dv}{dx} = -0.5$: the speed drops 0.5 m/s per meter traveled.
- $\Delta x = \dfrac{10 - 4}{0.5} = 12\ \text{m}$.
- Notice how much simpler this is than solving for $v(t) = 10e^{-0.5t}$ first.
Answer: $12\ \text{m}$.
The same method handles gravity far from Earth, where $a = -GM/r^2$ depends on the distance $r$ from the planet's center. Integrating $v\,dv = -\dfrac{GM}{r^2}\,dr$ gives the speed at any height, and in Gravitational Potential Energy and Escape Speed it gives the launch speed that never comes back.
Common mistake
Integrating $a(x)$ with respect to $t$, as in $v = \int (-4x)\,dt = -4xt$. That treats $x$ as a constant while it is changing; the result has no meaning. Check the variables: the function you integrate and the differential must match. A function of $x$ goes with $dx$, and the chain rule tells you it pairs with $v\,dv$.