1. ## improper integrals

lim as 2 approaches infinity $\int_2^{inf} e^{-1.9x} dx$

the limit of integration infinity becomes b, and you have to apply the FTC to determine whether it's convergent/divergent etc.

My answer for the integration alone is $-1.9e^{-1.9x}$. When I replace the x by 2 and evaluate, I get -0.0425. However I don't know what to do when doing the same thing with B as my value.

This is what I'm left with; lim as 2 approaches infinity $-1.9e^{-1.9b} - 0.0425$

How can I follow this up?

2. Originally Posted by Archduke01
lim as 2 approaches infinity $\int_2^{inf} e^{-1.9x} dx$

the limit of integration infinity becomes b, and you have to apply the FTC to determine whether it's convergent/divergent etc.

My answer for the integration alone is $-1.9e^{-1.9x}$. When I replace the x by 2 and evaluate, I get -0.0425. However I don't know what to do when doing the same thing with B as my value.

This is what I'm left with; lim as 2 approaches infinity $-1.9e^{-1.9b} - 0.0425$

How can I follow this up?
how does "2" approach infinity?

learn the correct format for improper integrals ...

$\lim_{b \to \infty} \int_2^b e^{-1.9x} \, dx
$

$\lim_{b \to \infty} \left[-\frac{e^{-1.9x}}{1.9}\right]_2^b$

$\lim_{b \to \infty} -\frac{1}{1.9} \left[e^{-1.9b} - e^{-3.8}\right] = \frac{e^{-3.8}}{1.9} \approx 0.011774...$

3. Originally Posted by skeeter
how does "2" approach infinity?

learn the correct format for improper integrals ...

$\lim_{b \to \infty} \int_2^b e^{-1.9x} \, dx
$

$\lim_{b \to \infty} \left[-\frac{e^{-1.9x}}{1.9}\right]_2^b$

$\lim_{b \to \infty} -\frac{1}{1.9} \left[e^{-1.9b} - e^{-3.8}\right] = \frac{e^{-3.8}}{1.9} \approx 0.011774...$
Thank you!

What I don't understand is why $e^{-1.9b}$ is removed? It becomes infinity. What's the reasoning?

4. Originally Posted by Archduke01
Thank you!

What I don't understand is why $e^{-1.9b}$ is removed? It becomes infinity. What's the reasoning?
$e^{-1.9b} = \frac{1}{e^{1.9b}}$

care to re-evaluate what you just stated?