The equality you write in your post is most certainly not true. Arguably the only pair for which equality holds is:
$$\dot{a} = \frac{\mathrm{d}a}{\mathrm{d}t}$$
Which is true by definition of the "dot" notation.
The next closest is $a' = \dot{a}$ but in most cases there is a different connotation to these, at least in physics. In physics, the "prime" notation almost always denotes differentiation with respect to a position variable, i.e.
$$a'\equiv{a}'(x)\equiv\frac{\mathrm{d}a}{\mathrm{d}x}$$
The $\Delta$ notation is only tangentially related to differentiation. In particular, you can say that:
$$\frac{\mathrm{d}a}{\mathrm{d}t}=\lim_{\Delta{t}\to{0}}~\frac{\Delta{a}}{\Delta{t}}$$
You might recognize that this is basically just the limit definition of a derivative. A small change in $a$ would be denoted:
$$\Delta{a}=a_{\text{final}}-a_{\text{initial}}=a(t_0+\Delta{t})-a(t_0)$$
Then we can see that the definition of the derivative is:
$$\frac{\mathrm{d}a}{\mathrm{d}t}\equiv\lim_{\Delta{t}\to{0}}~\frac{a(t+\Delta{t})-a(t)}{\Delta{t}}$$
The integrals are where this goes really wrong. In particular $\int_i^fa$ doesn't have obvious meaning or utility since it doesn't have a differential like $\mathrm{d}t$. The final integral:
$$\int_i^f~a~\mathrm{d}a$$
Is, by the fundamental theorem of calculus, equal to the anti-derivative of $a$ with respect to $a$ (i.e., this is asking "what function $A$ is such that if I differentiate it with respect to $a$ as $A'(a)$ returns the function $a$?") evaluated from $a = i$ to $a =f$. So:
$$\int_i^fa~\mathrm{d}a=A(f)-A(i)$$
The anti-derivative function $A$ is itself easy to calculate, it can be checked to be $a^2/2$, because:
$$\frac{d}{da}~(\frac{a^2}{2}) = \frac{2a^{2-1}}{2} = a$$
So finally:
$$\int_i^fa~\mathrm{d}a=\frac{f^2}{2}-\frac{i^2}{2}$$
Hope this answer helps. As far as there being a table or something similar, you could probably find one online. However, as you go forward in physics and math, this sort of thing will become second nature to you. Just stick with it!