1. ## Convergent Improper integrals

So I set up everything to find the limit as b approaches infinite. After integration, I have;

$\displaystyle \frac{-e^{-4x}}{16}(4x+1)$

The problem I have is when applying the FTC with b. I know it ends up being zero, but I don't understand how. There are 2 x values there to replace with infinity, how can I evaluate the outcome?

Also, I think I'm doing something wrong with replacing the x values with 7 because the result is a complicated number; $\displaystyle -1.2532*10^{-12}$

2. Originally Posted by Archduke01

So I set up everything to find the limit as b approaches infinite. After integration, I have;

$\displaystyle \frac{-e^{-4x}}{16}(4x+1)$

Good. So now you must evaluate $\displaystyle \lim_{b\to\infty}\left[\frac{-e^{-4x}}{16}(4x+1)\right]^b_7=\lim_{b\to\infty}\frac{-e^{-4b}}{16}(4b+1)+\frac{-e^{-28}}{16}(29)=\frac{29}{16}\,e^{-28}$

Tonio

The problem I have is when applying the FTC with b. I know it ends up being zero, but I don't understand how. There are 2 x values there to replace with infinity, how can I evaluate the outcome?

Also, I think I'm doing something wrong with replacing the x values with 7 because the result is a complicated number; $\displaystyle -1.2532*10^{-12}$
.