Identity and relationship between integrals and antiderivatives If I have that 
$$G(ab)=\int_1^{ab}\frac{dt}{t}=\int_1^a \frac{dt}{t}+\int_a^{ab}\frac{dt}{t}.$$
$$G(ab)=G(a)+\int_a^{ab}\frac{dt}{t}.$$
How is that true? Specifically this line is reasoning (link to discussion):
$$G(\textrm{banana})=\int_1^{\textrm{banana}} \frac{dt}{t}$$
As clever as banana man is he forgot to explain why that is true and I can't find it in any of my books or wikipedia entries. What does it mean?
Does this mean that the area under the graph of 2 from 1 to 2 is given as 
$$\int_1^2 1dx$$ ?
 A: It seems that you are confused as to how they are defining $G(x)$.  That is, $G(x)$ is defined as follows:
$$G(x) = \int_1^x\frac{1}{t}\,dt$$
Thus, by definition of $G$, you have that:
$$G(\text{banana}) = \int_1^\text{banana}\frac{1}{t}\,dt$$
Now, for the if/then statement, we make use of the theorem that: $$\int_a^b f(x)\,dx + \int_b^c f(x)\,dx = \int_a^c f(x)\,dx$$
(If it helps, $G(t)$ is actually just the natural log, $\ln t$.)
EDIT: I'm going to try to further explain this, as it appears that your question wasn't really about the if/then statement, but rather about a sort of function called an "integral function."
In an integral function, the variable is not in the integrand, but rather in one of the limits.  That is, we can define an integral function $G(x)$ as:
$$G(x) = \int_a^xf(t)\,dt$$
Note that the variable of the function ($x$) is different than the variable of integration ($t$).
Thus, $G(1) = \int_a^1 f(t)\, dt$.
It is very important to note that this does not imply that $f(1) = \int_a^1 f(t)\, dt$. 
