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?
 A: $$ \int_0^1 e^{-t} = [-e^{-t}]_0^1 = -\frac{1}{e} + 1$$
A: I assume you are integrating over the $t$ variable. $1/e^t$ is a continuous function, and you are integrating over a bounded interval, so the integral is well defined. An antiderivative of $1/e^t=e^{-t}$ is equal to $-e^{-t}$. So the integral equals $1-1/e$
A: 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 $1-\frac{1}{e}$.
A: The integration you have shown is not correct. Exponential functions are quite different to power functions and have very different antiderivatives. The correct integration is as follows:
$$\int_0^1 e^{-t} \, dt = -e^{-t}|_0^1 = 1-\frac{1}{e}.$$
A: Others have given a correct way to evaluate the integral. But even without evaluating the integral, you can see that it converges. After all, how can an integral fail to converge? It can fail to converge if one or both of the limits of integration is infinite, e.g., $\int_1^{\infty}x^{-1}\,dx$ fails to converge. It can fail to converge if the integrand is unbounded, e.g., $\int_0^1x^{-1}\,dx$ fails to converge. But if the limits of integration are bounded, and the integrand is bounded, the integral has no way to diverge - it must converge. 
In your example, the limits of integration are bounded, and the integrand satisfies $1/e\le e^{-t}\le1$ for $t$ in the interval of integration ($0\le t\le1$), so convergence is immediate. 
A: If you mean: How do you know the integral exists and has a finite value, the answer is simply that it's the integral of a continuous function over a bounded interval.
