mags093
please help?
Evaluate the definite integral (if it exists)
intergral (e^1/3)/-6x^2 dx from 1 to 2 ?
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amistre64
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e^1/3 looks like a variable, but its not; its just a constant
mags093
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\[\int\limits\limits_{1}^{2} \frac{ e^\frac{ 1 }{ x } }{ -6x^2 } dx\]
amistre64
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\[\huge \int\limits\limits_{1}^{2} \frac{ e^\frac{ 1 }{ x } }{ -6x^2 } dx\]
is that 1/3 or 1/x? need to verify
mags093
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thats what it looks like properly :)
mags093
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\[\frac{ 1 }{ x }\]
amistre64
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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
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so lets start by taking the derivative of that
hartnn
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yeah, just put u= 1/x
du=... ?
Spacelimbus
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exactly, @hartnn, this substution does the trick.
mags093
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do we not have to do something with the derivavtive of -6x^2?
amistre64
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we should verify that if its going to be simple first and see if we can modifiy it with a useful form of "1"
mags093
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the integral exists anyway! i know that much
amistre64
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\[\frac d{dx}e^{1/x}=\frac{e^{1/x}}{-2x^2}\]agree?
mags093
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where did the -2x^2 come from? is the derivative of -6x^2 not 12x?
amistre64
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your confusing your rules .... its best to see these things on a more holistic level :)
mags093
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oh ok :)
amistre64
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\[\frac d{dx}e^{u(x)}=\frac{d}{dx}u(x)~e^{u(x)}=u' ~e^u\]
amistre64
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let u= 1/x
du/dx = -1/2x^2
du = -1/2x^2 dx
mags093
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how do i get the derivative of 1/x?
amistre64
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1/x = x^-1 which is then just the power rule .... which means i got a bad 2 in there :)
amistre64
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im thinking of a sqrt function ..... silly me
mags093
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what do you mean a bad 2? :)
amistre64
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x^-1 goes to -x^(-2)
1/x goes to -1/x^2 NOT -1/2x^2
mags093
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oh ok i get ya now
amistre64
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so just pull out the 1/6 and integrate the easy way
mags093
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wheres the 1/6?
amistre64
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.....
amistre64
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\[\int\frac{e^{1/x}}{-6x^2}dx\]
\[\int\frac16 \frac{e^{1/x}}{-x^2}dx\]
\[\frac16\int \frac{e^{1/x}}{-x^2}dx\]
mags093
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oh yes i see :) got confused for a minute!
amistre64
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just becasue you were doing vector calculus the other day doesnt mean you have to forget the fundamentals ;)
amistre64
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you got it from here?
mags093
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\[\frac{ e^\frac{ 1 }{ x } }{ 6 }\]
mags093
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something like that right?
amistre64
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yes, but apply your limits
mags093
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what do you mean?
hartnn
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e^u/6
hartnn
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not, e^(1/x) /6
amistre64
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i mean ....\[\int_{a}^{b}f(x)~dx=F(b)-F(a)\]
hartnn
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plus you need to change limits beforehand,
u=1/x
when x=1,u=1
when x=2, u=1/2
mags093
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but u=1/x?
amistre64
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if you do a usub, then you need to do hartnns route
otherwise its just e^(1/x)/6 applied at 1 and 2
mags093
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so i put 1 and 2 in for x?
amistre64
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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
mags093
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\[\frac{ e^u }{ 6 }\]
amistre64
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\[\int_{1}^2f(x)dx=F(2)-F(1)\]
and we determined that \[F(x)=\frac16e^{1/x}\]
mags093
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and then i got that^
mags093
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so i sub in 2 and 1 for x then subtract them? is that it?
amistre64
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yes\[\frac16(e^{1/2}-e^{1/1})\]
yes\[\frac16(e^{1/2}-e)\]
mags093
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if i work that out it gives me something like -0.782?
mags093
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if i just leave it as \[\frac{ 1 }{ 6 } (e^\frac{ 1 }{ 2 } - e\]
hartnn
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did u mean -0.1782 ?
and you can keep your answer as 1/6(e^(1/2)-e)
mags093
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when i worked it out it was -0.782 but if i leave it as the 1/6... it works out just the same :)
mags093
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thanks for the help :)