MHB Evaluating the Improper Integral (IV)

[shamieh]

Hey, its me again, just needing someone to verify my findings. Thanks in advance.

$$
\int ^{\infty}_0 \frac{1}{e^{3x}} \, dx = \lim_{a\to\infty} \frac{1}{3} \int ^{3a}_0 e^{-u} \, dx$$

u = 3x ,,,,,, du/3 = dx

skipping a few steps...

$$\lim_{a\to\infty} -\frac{1}{3}e^{-u} |^{3a}_0 = 0 + \frac{1}{3} $$ thus: t < 1 so: as $$\lim_{a\to\infty}$$ converges

 

[Prove It]

Hey, its me again, just needing someone to verify my findings. Thanks in advance.

$$
\int ^{\infty}_0 \frac{1}{e^{3x}} \, dx = \lim_{a\to\infty} \frac{1}{3} \int ^{3a}_0 e^{-u} \, dx$$

u = 3x ,,,,,, du/3 = dx

skipping a few steps...

$$\lim_{a\to\infty} -\frac{1}{3}e^{-u} |^{3a}_0 = 0 + \frac{1}{3} $$ thus: t < 1 so: as $$\lim_{a\to\infty}$$ converges

It does converge, but not because t < 1. It converges because the integral is a finite number.

 

[shamieh]

Awesome, thank you for clarifying , that's what I've been a little confused about. :D