Is Integral is considered defined even when one method gives defined results and other undefined results something that I found confusing me.
Lets see for example the follow integral: 
$$\int_{A}^{B}e^{t\cdot x} \cdot dx=\left[\frac{e^{t\cdot x}}{t}\right]_{A}^B=\frac{e^{t\cdot B}-e^{t\cdot A}}{t}$$
From this results we may conclude that for $t=0$ , the integral is undefined:
$$\frac{e^{t\cdot B}-e^{t\cdot A}}{t}=\frac{1-1}{0}=\frac{0}{0}$$
but from the other hand, if we evaluate the integral again for this case of $t=0$, then we get a defined results:
$$\int_{A}^{B}e^{t\cdot x} \cdot dx=\int_{A}^{B}e^{0\cdot x} \cdot dx=\int_{A}^{B}1\cdot dx=B-A$$
So is this integral is defined for $t=0$ or not?
 A: In the $\frac{0}{0}$ case, we can apply L'Hôpital's rule to obtain
$$lim_{t\to 0}\frac{e^{tB}-e^{tA}}{t}=lim_{t\to 0}\frac{Be^{tB}-Ae^{tA}}{1}=B-A$$
so the two results agree. If we integral a continuous function of two variables with respect to one variable, the result should be  a continuous function of the other.
In general, if $F(t)=\int_a^b f(t,x)dx$, we would define, for fixed $t_0$, $F(t_0)=\int_a^b f(t_0,x)dx$ rather than evaluating for general $t$ and then setting $t=t_0$. But assuming $f$ is sufficiently nice, them the two will agree so long as we take limits
A: You are using the formula $\int e^{tx}\, dx = \frac{e^{tx}}{t} + C$, but this rule is not valid when $t=0$.  You should actually use
$$\int e^{tx}\, dx = \begin{cases}
 \frac{e^{tx}}{t} + C  &\text{ if } t \neq 0 \\ 
x + C &\text{ if } t = 0.\\ 
\end{cases}$$
Then the two results are the same.  In general, when you use a formula involving a division be aware that the formula will not be valid when the denominator is $0$.  This is often an unstated assumption, so you should check if both sides a given equation are undefined or not in that case.
