Why does $\int \limits_{0}^{1} \frac{1}{e^t}dt $ converge? I'd like your help with see why does $$\int^1_0 \frac{1}{e^t} \; dt $$ converge?
As I can see this it is suppose to be:
$$\int^1_0 \frac{1}{e^t}\;dt=|_{0}^{1}\frac{e^{-t+1}}{-t+1}=-\frac{e^0}{0}+\frac{e}{1}=-\infty+e=-\infty$$
Thanks a lot?

Your calculation is incorrect. Instead, you should have
$$\int_0^1\frac{1}{e^t}dt=\int_0^1e^{-t}dt=-e^{-t}\big|_0^1=(-e^{-1}-(-e^{0}))=1-\frac{1}{e}$$
You got mixed up with the rule for powers, 
$$\int x^n\,dx=\frac{x^{n+1}}{n+1}+C$$
but for exponentials we have
$$\int e^x\,dx=e^x+C$$
You can also see that $e^{-t}$ is bounded above by the constant function $1$ on the interval $\[0,1\]$:
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ 
so that the area underneath the curve $e^{-t}$ from $0$ to $1$ has to be less than the area underneath the curve $1$ from $0$ to $1$ (which is $1$). So this picture tells you that the value of $\int_0^1\frac{1}{e^t}dt$ is less than $1$, which we've confirmed by showing that is in fact equal to $1-\frac{1}{e}$.
