I have a little discussion with my friends about my "resolution" and calculation of $$\int_{-\infty}^1 e^{4x} \, dx.$$ I did $$\int_{-\infty}^1 e^{4x} \, dx =\int_{-1}^{\infty} e^{-4x} \, dx = \lim_{b\rightarrow \infty}\frac{-e^{-4b}}{4} + \frac{e^4}{4} = \frac{e^4}{4}.$$ What's my error on this calculation? Or i'm doing right? Thanks for your points of views friends!
P.D.: My friends tell me that I'm using in a wrong way Barrow's rule.
The substitution "rule" may not have been proved for improper integrals. But you applied it correctly, doing two steps in one. The minus sign was omitted, but the limits were interchanged to compensate. The calculation is correct. It would have been preferable to use a new dummy variable symbol, letting $x=-t$.
More directly, we can calculate $$\lim_{b\to\infty} \int_{-b}^1 e^{4x}\,dx.$$ An antiderivative is $\frac{1}{4}e^{4x}$, so we want $$\lim_{b\to\infty} \left(\frac{1}{4}e^{4}-\frac{1}{4}e^{-4b}\right).$$