Improper Integrals: Converge or Diverge: int[1/rad(x-1)]dx

[skyblue]

Can someone please help me with these? I need some guidance. I would really appreciate it.

1. int {1/[radical(x-1)]} from 2 to infinity. 2 is the lower limit, infinity is the upper limit. i put it converges because i looked at the graph and i think it has a limit?

2. int {e^2x} dx from -infinity to 0

i put it diverges because the limit does not exist. i got [(e^2x)/2] 0 to N. then i put (e^0 / 2) - (e^2N / 2) and then i looked at the graph of the integral.

 

[morson]

There are ways of proving convergence other than just "looking at the graph"! That probably isn't enough if that is a homework question.

 

[galactus]

#2:

Remember, \(\displaystyle e^{0}=1\), not 0.

\(\displaystyle \L\\\int{e^{2x}}dx=\frac{e^{2x}}{2}\)

\(\displaystyle \L\\\lim_{l\rightarrow{-}\infty}\frac{e^{2x}}{2}\left|_{l}^{0}=\lim_{l\rightarrow{-}\infty}\left[\frac{1}{2}-\frac{1}{2}e^{2l}\right]=\frac{1}{2}\)

It converges

 

[Opalg]

#1:

\(\displaystyle \L \int_2^{\;\;\;\;\;X} \frac1{\sqrt{x-1}}dx = {\huge[}2\sqrt{x-1}{\huge]}_2^X = 2\sqrt{X-1} - 2 \to\infty\) as \(\displaystyle X\to\infty\). So the integral diverges.

It's true that \(\displaystyle 1/\sqrt{x-1}\) converges to zero as x goes to infinity. But it converges "slowly" (in some intuitive sense) so that the area under the curve becomes infinite.