$$\int_{-\infty}^1 e^x dx$$
=$$e^x\bigg|_{-\infty}^{1}$$
I know....

Question

$$\int_{-\infty}^1 e^x dx$$
=$$e^x\bigg|_{-\infty}^{1}$$
I know....

lgbasallote · Answer

improper integral huh

lgbasallote · Answer

hint $$e^{-\infty} = 0$$

lgbasallote · Answer

does that help?

anonymous · Answer

yep...but the set up is ok?

lgbasallote · Answer

yes @MathSofiya that's right

anonymous · Answer

I've been separating integrals for the past hour...so that was strange....lol

anonymous · Answer

I thought I had to do something like  $$\lim_{t \rightarrow \infty}$$

lgbasallote · Answer

well that's the proper way...

helder_edwin · Answer

actually
$$  \Large \int_{-\infty}^1e^x\,dx=\lim_{b\to-\infty}\int_b^1e^x\,dx=
     \lim_{b\to-\infty}(e^1-e^b) $$

anonymous · Answer

To be precise, one has to find
$$
\lim_{c\to -\infty} \int_c^1 e^x dx =\
\lim_{c\to -\infty} e^1 - e^c=e- 0=e
$$

lgbasallote · Answer

$$\int_{-\infty}^1 e^x dx$$
$$\lim_{t\rightarrow -\infty} \int_t^1 e^xdx$$
$$e^x|_t^1$$
$$e^1 - e^t$$
$$e - e^{-\infty}$$
$$e - 0$$
$$e$$

anonymous · Answer

thanks...that's more like it...the whole section was along the lines of $$\lim_{t \rightarrow \infty}$$ and this integral seemed too easy
Thanks again

lgbasallote · Answer

as long as you know $e^{-\infty} = 0$ it's easy

anonymous · Answer

yes sir!