#Need help with this please

164 messages · Page 1 of 1 (latest)

burnt roostBOT
pure vine
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What have u tried?

rain thistle
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hi

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Thanks for messaging back

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so i know its an improper integral becasue the function e^1/x/x^2 becomes undefined at x=0

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and then i rewrote the integral as a limit

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which i got lim a-->0 - ∫^a -4 e1/2/x^2 dx

pure vine
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let me tex that up for ya
$\lim_{a\to0^{-}}\int_{-4}^a \frac{e^{1/x}}{x^2},dx$

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right?

rain thistle
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yes correct

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sorry

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i donnt know how to do that

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LOL

pure vine
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not worries at all, its only for clarity really

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well, so after this, what did u try then?

placid daggerBOT
pure vine
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sorry missed the a lol

rain thistle
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i checked near x= 0 , that 1/x^2 grows very largely

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and e ^1/x approaches 0

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but e^1/2/x^2 still goes up and then i concluded with that the function diverges as x--> 0 ^-

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so i concluded that the integral is divergent

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but I got it wrong sadly...

pure vine
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I see, well as a start, i rarely recommend checking if the integrand converges so to speak; as theres many scenarioes when the integrand can very well diverge in some sense but still have a convergent integral

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but it just so happens actually in this case that e^1/x / x^2 does converge when x approches 0 from the left

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so consider x < 0, then e^(1/x) is the same as 1/e^(-1/x), where -1/x is positive

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agree so far?

rain thistle
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yes

rain thistle
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ok

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am i on the right track at least?

pure vine
pure vine
placid daggerBOT
pure vine
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Actually

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i think it might be more comfortable if we rewrite this into a standard limit, which hopefully is standard

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Are you familiar with your standard limits?

rain thistle
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yes i am actually

rain thistle
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so the answer is not divergent 100%?

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because i thought it was divergent but i guess its not

pure vine
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the above reasoning is not really about the convergence of the intergral

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but more so about the argument you had made

rain thistle
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ooh ok

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so whats next

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because i really want to get this question right

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and i dont want to get it wrong

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im thinking about this question the whole day

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LOL

pure vine
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i guess it depends, do you want to check before computing if its convergent or later by simply computing the integral as you would with the limit above?

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in either case you would have to compute it

rain thistle
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ok sure

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how woudl i go about doing that

pure vine
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to answer it in your thingy

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well what have u tried, did you manage to find an anti derivative or?

rain thistle
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no not yet, im having trouble with that

pure vine
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notice what happens if u make the substitution u = 1/x

rain thistle
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um

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doesnt it just transfor the integrall into a simpler version

pure vine
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yes!

rain thistle
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ok perfect

pure vine
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hopefully youre able to compute it by then

rain thistle
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sorry im lost

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i know it transform the integral into a simpler version where it can be easily evaluated but now what?

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i compute it with the antiderivative

pure vine
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yes and then take the limit!

rain thistle
pure vine
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yes

pure vine
rain thistle
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the first one

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u sent me

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okay

pure vine
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yes

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that "middle picture" is not really about the integral

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so you can ignore that

rain thistle
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okay solving right now

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is it -e^-1/4?

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please be right

pure vine
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Let's check by checking ur work!

rain thistle
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okay

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do you know the answer yourself or no

pure vine
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Yes, but i want you to understand how to find it

rain thistle
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okay

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should i tell you my steos?

pure vine
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Yes please

rain thistle
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okok

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so i let u = 1/x then x=1/u

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and dx= -1/u^2 du

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then the integral become ∫ a above and -4 below e^1/x/x^2 dx= ∫- infinty above and -1/4 below e^u times by u^2 times (-1/u^2)du

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and with that

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i simplified

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which got be ∫a above and -4 below e^1/x/x^2 dx = - ∫- infinity above and -1/4 below e^u du

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and then the antiderivative of e^u is e^u

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so pretty much

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i did - ∫- infintity above and -1/4 below e^u du= -[e^u] ^-1/4 - infinity

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and with that

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i evaluated the limits and subsituted them into the expression -[e^u] -1/4 - infinity = - (e^-1/4-0) \

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simplified that - (e^-1/4 - 0) which finally got me the answer

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-e^-1/4

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so if everything checks out

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this answer should be right correct?

pure vine
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Almost!

rain thistle
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shoot

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what mistake

pure vine
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The order in which you subtract the endpoints

pure vine
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do you see that you got the order mixed up?

rain thistle
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oooooo

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i see it

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your right

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good catch

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okay

pure vine
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Yes apart from that youre correct, i would personally make some changes to how you present the answer

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and theres some "shortcuts" too

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would you like too see how i'd present it if that seems helpful?

rain thistle
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yes please

pure vine
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I'll present it as how i would write it instead of explaning it so to speak if that makes sense if thats cool?

rain thistle
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sure yea

pure vine
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(thought id indirectly be explaning it of course since thats the point!)

placid daggerBOT
pure vine
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So thats that!

rain thistle
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ooohh

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okay

pure vine
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But yeah you already got it nailed down

rain thistle
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so i forgot to add the -0

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right

pure vine
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No

rain thistle
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its just positive e

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instead of negative

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cuz i put -e

pure vine
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You forgot to write out the a of the integral, thats the cruical thing imo

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as infinty doesnt really mean anything when u use it like a number like that

rain thistle
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i see

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why is your e positive?

pure vine
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Which e?

rain thistle
pure vine
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The order in which you subtract them matters

rain thistle
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okayyy makes sense now

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so the final answer would be this

image.png
pure vine
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i only added -0 for clarity, but yes

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e^(-1/4)

rain thistle
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woww

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okay

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thank you so much for your help

pure vine
pure vine
rain thistle
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i understand it well now thanks for all your help

rain thistle
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ur a genious

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LOL

pure vine
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lmao

rain thistle
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can you help me out with one more question please?

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if possible?

pure vine
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Maybe, if its short!

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If i dont have time i recommend opening up a new channel so not to confuse future helpers

rain thistle
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okok

pure vine
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It's namley 2:30 am over here

rain thistle
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ill just do that

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okok

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i will do that