I want to ask a question about the Steady State Approximation integral.
I understand the steady state integral works in Leyman's terms as the intermediate is formed as fast as it is consumed so the intermediate's concentration is really low and nearly always the same.
If I had a reaction:
$$\ce{A->I->P}$$
Then essentially,
$$\frac{d[I]}{d t} \approx 0$$
and therefore using this information and previous knowledge of rate
$$\begin{array}{l}{\frac{d[I]}{d t}=k_{a}[A]-k_{b}[I]} \\ {\frac{d[I]}{d t}=k_{a}[A]-k_{b}[I] \approx 0}\end{array}$$
which results after rearrangement:
$$[I] \approx \frac{k_{a}[A]}{k_{b}}$$
and hence $$\frac{d[P]}{d t}=k_{b}[I] \approx k_{b} \frac{k_{a}[A]}{k_{b}} \approx k_{a}[A]$$
The trouble is, I couldn't understand how:
$$\frac{d[P]}{d t}=k_{a}[A]$$
can be solved to give
$$[P]=\left(1-e^{-k_{a} t}\right)[A]_{0}$$
I recognise from previous work that:
$$\frac{d[A]}{d t}=-k_{a}[A]$$ gives
$$A =[A]_{0}e^{-k_{a} t}$$
from my final year at school, but I couldn't seem to apply this understanding to the final step of the integral above.
How can this be shown?