A sledge is pushed in a straight line. Assume surface is smooth. When the sledge is x distance away from the start the magnitude of its acceleration is given by $0.08e^{-4x}$ and is going in the same direction of the sledge's motion.
I need to show that $v = 0.2 \sqrt{1-e^{-4x}}$
Now I 'know' that to get $v$ from a you need to integrate a. So I did that and I got: $v=-0.02e^{-4x}$.
Are these two things one and the same? If not, can someone explain why?
How did you integrate acceleration to get velocity? Note that
$\Delta v = \int_{t_i}^{t_f} a(t) dt$
But you have an acceleration that is a function of position, not time. So you can't naively integrate this and get velocity. There is a trick. Notice that you can rewrite acceleration as
$a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}$
You can use this to set up your integral so that you are able to integrate a(x) w.r.t. x. But notice the "v" hanging around. You won't just get v on the other side of the equation.
Next, be careful of initial conditions. You haven't told us what the initial conditions are. But I infer that the sledge must start from rest because it starts at $x=0$ and if you plug $x=0$ into the expression you are supposed to prove you get $v(0) = 0$. So you need to pay attention to the fact that you are doing a definite integral, not simply an antiderivative.
To get velocity from acceleration, you need to integrate with respect to time. But your expression of acceleration is given with respect to position. Thus, your current calculation is not correct. You need to figure out how to convert the position-dependent information to time-dependent information.
Since they give you the solution and you just have to prove it is correct, you should start from the velocity information and see if you can get from there to acceleration - noting that
$$a=\frac{dv}{dt} = \frac{dv}{dx}\cdot\frac{dx}{dt}=\frac{dv}{dx}\cdot v$$
Now we know $v(x)$ so we can differentiate that with respect to $x$; then we multiply by $v$ and should get the expression for $a$ that was given.
$$\begin{align}\\ v &= 0.2\sqrt{1-e^{-4x}}\\ a &= 0.2\cdot \frac12 \cdot\left(1-e^{-4x}\right)^{-\frac12}\cdot 4e^{-4x}\cdot 0.2\sqrt{1-e^{-4x}}\\ &= 0.08 e^{-4x}\end{align}$$
Q.E.D. - no integration required.
