anonymous
  • anonymous
please help? Evaluate the definite integral (if it exists) intergral (e^1/3)/-6x^2 dx from 1 to 2 ?
Mathematics
katieb
  • katieb
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amistre64
  • amistre64
e^1/3 looks like a variable, but its not; its just a constant
anonymous
  • anonymous
\[\int\limits\limits_{1}^{2} \frac{ e^\frac{ 1 }{ x } }{ -6x^2 } dx\]
amistre64
  • amistre64
\[\huge \int\limits\limits_{1}^{2} \frac{ e^\frac{ 1 }{ x } }{ -6x^2 } dx\] is that 1/3 or 1/x? need to verify

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anonymous
  • anonymous
thats what it looks like properly :)
anonymous
  • anonymous
\[\frac{ 1 }{ x }\]
amistre64
  • amistre64
well, if this is going to be simple than with any luck it will come from something like:\[\Large \frac d{dx}e^{1/x}\]
amistre64
  • amistre64
so lets start by taking the derivative of that
hartnn
  • hartnn
yeah, just put u= 1/x du=... ?
anonymous
  • anonymous
exactly, @hartnn, this substution does the trick.
anonymous
  • anonymous
do we not have to do something with the derivavtive of -6x^2?
amistre64
  • amistre64
we should verify that if its going to be simple first and see if we can modifiy it with a useful form of "1"
anonymous
  • anonymous
the integral exists anyway! i know that much
amistre64
  • amistre64
\[\frac d{dx}e^{1/x}=\frac{e^{1/x}}{-2x^2}\]agree?
anonymous
  • anonymous
where did the -2x^2 come from? is the derivative of -6x^2 not 12x?
amistre64
  • amistre64
your confusing your rules .... its best to see these things on a more holistic level :)
anonymous
  • anonymous
oh ok :)
amistre64
  • amistre64
\[\frac d{dx}e^{u(x)}=\frac{d}{dx}u(x)~e^{u(x)}=u' ~e^u\]
amistre64
  • amistre64
let u= 1/x du/dx = -1/2x^2 du = -1/2x^2 dx
anonymous
  • anonymous
how do i get the derivative of 1/x?
amistre64
  • amistre64
1/x = x^-1 which is then just the power rule .... which means i got a bad 2 in there :)
amistre64
  • amistre64
im thinking of a sqrt function ..... silly me
anonymous
  • anonymous
what do you mean a bad 2? :)
amistre64
  • amistre64
x^-1 goes to -x^(-2) 1/x goes to -1/x^2 NOT -1/2x^2
anonymous
  • anonymous
oh ok i get ya now
amistre64
  • amistre64
so just pull out the 1/6 and integrate the easy way
anonymous
  • anonymous
wheres the 1/6?
amistre64
  • amistre64
.....
amistre64
  • amistre64
\[\int\frac{e^{1/x}}{-6x^2}dx\] \[\int\frac16 \frac{e^{1/x}}{-x^2}dx\] \[\frac16\int \frac{e^{1/x}}{-x^2}dx\]
anonymous
  • anonymous
oh yes i see :) got confused for a minute!
amistre64
  • amistre64
just becasue you were doing vector calculus the other day doesnt mean you have to forget the fundamentals ;)
amistre64
  • amistre64
you got it from here?
anonymous
  • anonymous
\[\frac{ e^\frac{ 1 }{ x } }{ 6 }\]
anonymous
  • anonymous
something like that right?
amistre64
  • amistre64
yes, but apply your limits
anonymous
  • anonymous
what do you mean?
hartnn
  • hartnn
e^u/6
hartnn
  • hartnn
not, e^(1/x) /6
amistre64
  • amistre64
i mean ....\[\int_{a}^{b}f(x)~dx=F(b)-F(a)\]
hartnn
  • hartnn
plus you need to change limits beforehand, u=1/x when x=1,u=1 when x=2, u=1/2
anonymous
  • anonymous
but u=1/x?
amistre64
  • amistre64
if you do a usub, then you need to do hartnns route otherwise its just e^(1/x)/6 applied at 1 and 2
anonymous
  • anonymous
so i put 1 and 2 in for x?
amistre64
  • amistre64
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
anonymous
  • anonymous
\[\frac{ e^u }{ 6 }\]
amistre64
  • amistre64
\[\int_{1}^2f(x)dx=F(2)-F(1)\] and we determined that \[F(x)=\frac16e^{1/x}\]
anonymous
  • anonymous
and then i got that^
anonymous
  • anonymous
so i sub in 2 and 1 for x then subtract them? is that it?
amistre64
  • amistre64
yes\[\frac16(e^{1/2}-e^{1/1})\] yes\[\frac16(e^{1/2}-e)\]
anonymous
  • anonymous
if i work that out it gives me something like -0.782?
anonymous
  • anonymous
if i just leave it as \[\frac{ 1 }{ 6 } (e^\frac{ 1 }{ 2 } - e\]
hartnn
  • hartnn
did u mean -0.1782 ? and you can keep your answer as 1/6(e^(1/2)-e)
anonymous
  • anonymous
when i worked it out it was -0.782 but if i leave it as the 1/6... it works out just the same :)
anonymous
  • anonymous
thanks for the help :)

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