Question

using DCTT integral from 1 to inf of 1/(x+e^2x) is smaller than 1/(x+e^x)? or do i use 1/(e^2x) to compare?

Answer 1

LESS in denominator is always bigger..

Answer 2

you could also simply use 1/e^x to compare

Answer 3

is this true? \[\int\limits_{1}^{\infty} \frac{ dx }{ x+e^x } \le \frac{ 1 }{ e^x }\]

Answer 4

looks good except that integral symbols are missing on right side

Answer 5

lol yes your right xD

Answer 6

interesting... now if the limits of integration included negative numbers than i couldn't say the by removing x the denominator gets smaller

Answer 7

yes we're talking about a function that works only in the given interval,
if the interval changes, we need to look for a different function to compare

Answer 8

mmm k thank you, i was worried that i couldn't know for sure if the function gets smaller or bigger by removing a variable, only constants.

Answer 9

graphing both the functions gives you a better idea about the area under the curve..

Answer 10

ok thank you, (unfortunately i am bad at graphing these types of functions)

Answer 11

no, im saying to graph it using tool just to get some idea...

Answer 12

we already know that :
\[\large \dfrac{1}{x+e^x} \lt \dfrac{1}{e^x}\]
in (1, infty) because the left side has a bigger denominator and so the value of the overall fraction would be less

Answer 13

graphing it gives you more clarity, thats all...

Answer 14

k thank you. so integral of e^-x is... -e^-x?

Answer 15

xD (i know silly question)

Answer 16

yes, are you trying to prove convergence ?
then u need to look for a bigger function right ?

Answer 17

Here is the theorem :

1)
if the bigger known integral converges,
then the smaller unknown integral also converges

2)
if the smaller known integral diverges,
then the bigger unknown integral also diverges

Answer 18

yeah so it turns out 1/e^x is convergent so that checks out

Answer 19

i dont think so..

Answer 20

your original question is to test convergence of \(\large \int\limits_1^{\infty} \dfrac{dx}{x+e^{2x}}\) right ?