TITLE: Is Integral is considered defined even when one method gives defined results and other undefined results
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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.