A rate law tells you the speed right now. Integrate it over a clock and you get the concentration ITSELF at any time — an integrated rate law. Three of them matter, and each is named by its order. In all three, $[\mathrm{A}]$ is the concentration left at time $t$ and $[\mathrm{A}]_0$ is the concentration at the start — the zero subscript always means “when the clock read zero”. $t$ is the elapsed time in seconds and $k$ is the rate constant.

Zero order: $[\mathrm{A}] = [\mathrm{A}]_0 - kt$. The rate never changes, so the concentration falls in a straight line and simply stops when the reactant runs out. Here $k$ wears $\mathrm{mol/(L \cdot s)}$ — it IS the rate. First order: $[\mathrm{A}] = [\mathrm{A}]_0\,e^{-kt}$, with $k$ in $\mathrm{s^{-1}}$. The rate is proportional to what is left, so the fall is exponential — forever approaching zero, never arriving. Second order: $\dfrac{1}{[\mathrm{A}]} = \dfrac{1}{[\mathrm{A}]_0} + kt$, with $k$ in $\mathrm{L/(mol \cdot s)}$, waiting in the lesson after next.

Here is the exam's favourite question, and the free gift inside it: each law is STRAIGHT when you plot the right thing. $[\mathrm{A}]$ against $t$ is straight for zero order; $\ln[\mathrm{A}]$ against $t$ for first order; $1/[\mathrm{A}]$ against $t$ for second. That is how an order is measured in the first place — try all three plots and see which one the data agrees to lie down on. And when you are handed $k$ instead of a plot, read its units: they confess the order before you compute a thing.