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please help? Evaluate the definite integral (if it exists) intergral (e^1/3)/-6x^2 dx from 1 to 2 ?

Mathematics
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e^1/3 looks like a variable, but its not; its just a constant
\[\int\limits\limits_{1}^{2} \frac{ e^\frac{ 1 }{ x } }{ -6x^2 } dx\]
\[\huge \int\limits\limits_{1}^{2} \frac{ e^\frac{ 1 }{ x } }{ -6x^2 } dx\] is that 1/3 or 1/x? need to verify

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Other answers:

thats what it looks like properly :)
\[\frac{ 1 }{ x }\]
well, if this is going to be simple than with any luck it will come from something like:\[\Large \frac d{dx}e^{1/x}\]
so lets start by taking the derivative of that
yeah, just put u= 1/x du=... ?
exactly, @hartnn, this substution does the trick.
do we not have to do something with the derivavtive of -6x^2?
we should verify that if its going to be simple first and see if we can modifiy it with a useful form of "1"
the integral exists anyway! i know that much
\[\frac d{dx}e^{1/x}=\frac{e^{1/x}}{-2x^2}\]agree?
where did the -2x^2 come from? is the derivative of -6x^2 not 12x?
your confusing your rules .... its best to see these things on a more holistic level :)
oh ok :)
\[\frac d{dx}e^{u(x)}=\frac{d}{dx}u(x)~e^{u(x)}=u' ~e^u\]
let u= 1/x du/dx = -1/2x^2 du = -1/2x^2 dx
how do i get the derivative of 1/x?
1/x = x^-1 which is then just the power rule .... which means i got a bad 2 in there :)
im thinking of a sqrt function ..... silly me
what do you mean a bad 2? :)
x^-1 goes to -x^(-2) 1/x goes to -1/x^2 NOT -1/2x^2
oh ok i get ya now
so just pull out the 1/6 and integrate the easy way
wheres the 1/6?
.....
\[\int\frac{e^{1/x}}{-6x^2}dx\] \[\int\frac16 \frac{e^{1/x}}{-x^2}dx\] \[\frac16\int \frac{e^{1/x}}{-x^2}dx\]
oh yes i see :) got confused for a minute!
just becasue you were doing vector calculus the other day doesnt mean you have to forget the fundamentals ;)
you got it from here?
\[\frac{ e^\frac{ 1 }{ x } }{ 6 }\]
something like that right?
yes, but apply your limits
what do you mean?
e^u/6
not, e^(1/x) /6
i mean ....\[\int_{a}^{b}f(x)~dx=F(b)-F(a)\]
plus you need to change limits beforehand, u=1/x when x=1,u=1 when x=2, u=1/2
but u=1/x?
if you do a usub, then you need to do hartnns route otherwise its just e^(1/x)/6 applied at 1 and 2
so i put 1 and 2 in for x?
3 methods that im aware of; complete usub changes limits to us partial usub where you undo the u back to x and just working thru a normal integration were nothing is changed about.... which is what i did
\[\frac{ e^u }{ 6 }\]
\[\int_{1}^2f(x)dx=F(2)-F(1)\] and we determined that \[F(x)=\frac16e^{1/x}\]
and then i got that^
so i sub in 2 and 1 for x then subtract them? is that it?
yes\[\frac16(e^{1/2}-e^{1/1})\] yes\[\frac16(e^{1/2}-e)\]
if i work that out it gives me something like -0.782?
if i just leave it as \[\frac{ 1 }{ 6 } (e^\frac{ 1 }{ 2 } - e\]
did u mean -0.1782 ? and you can keep your answer as 1/6(e^(1/2)-e)
when i worked it out it was -0.782 but if i leave it as the 1/6... it works out just the same :)
thanks for the help :)

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