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e^1/3 looks like a variable, but its not; its just a constant

\[\int\limits\limits_{1}^{2} \frac{ e^\frac{ 1 }{ x } }{ -6x^2 } dx\]

thats what it looks like properly :)

\[\frac{ 1 }{ x }\]

so lets start by taking the derivative of that

yeah, just put u= 1/x
du=... ?

do we not have to do something with the derivavtive of -6x^2?

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?

.....

oh yes i see :) got confused for a minute!

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?

\[\frac{ e^u }{ 6 }\]

\[\int_{1}^2f(x)dx=F(2)-F(1)\]
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 :)