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 one year ago
please help?
Evaluate the definite integral (if it exists)
intergral (e^1/3)/6x^2 dx from 1 to 2 ?
 one year ago
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
 one year ago
Best ResponseYou've already chosen the best response.2e^1/3 looks like a variable, but its not; its just a constant

mags093
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits\limits_{1}^{2} \frac{ e^\frac{ 1 }{ x } }{ 6x^2 } dx\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2\[\huge \int\limits\limits_{1}^{2} \frac{ e^\frac{ 1 }{ x } }{ 6x^2 } dx\] is that 1/3 or 1/x? need to verify

mags093
 one year ago
Best ResponseYou've already chosen the best response.0thats what it looks like properly :)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2well, 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
 one year ago
Best ResponseYou've already chosen the best response.2so lets start by taking the derivative of that

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2yeah, just put u= 1/x du=... ?

Spacelimbus
 one year ago
Best ResponseYou've already chosen the best response.0exactly, @hartnn, this substution does the trick.

mags093
 one year ago
Best ResponseYou've already chosen the best response.0do we not have to do something with the derivavtive of 6x^2?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2we should verify that if its going to be simple first and see if we can modifiy it with a useful form of "1"

mags093
 one year ago
Best ResponseYou've already chosen the best response.0the integral exists anyway! i know that much

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac d{dx}e^{1/x}=\frac{e^{1/x}}{2x^2}\]agree?

mags093
 one year ago
Best ResponseYou've already chosen the best response.0where did the 2x^2 come from? is the derivative of 6x^2 not 12x?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2your confusing your rules .... its best to see these things on a more holistic level :)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac d{dx}e^{u(x)}=\frac{d}{dx}u(x)~e^{u(x)}=u' ~e^u\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2let u= 1/x du/dx = 1/2x^2 du = 1/2x^2 dx

mags093
 one year ago
Best ResponseYou've already chosen the best response.0how do i get the derivative of 1/x?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.21/x = x^1 which is then just the power rule .... which means i got a bad 2 in there :)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2im thinking of a sqrt function ..... silly me

mags093
 one year ago
Best ResponseYou've already chosen the best response.0what do you mean a bad 2? :)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2x^1 goes to x^(2) 1/x goes to 1/x^2 NOT 1/2x^2

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2so just pull out the 1/6 and integrate the easy way

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2\[\int\frac{e^{1/x}}{6x^2}dx\] \[\int\frac16 \frac{e^{1/x}}{x^2}dx\] \[\frac16\int \frac{e^{1/x}}{x^2}dx\]

mags093
 one year ago
Best ResponseYou've already chosen the best response.0oh yes i see :) got confused for a minute!

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2just becasue you were doing vector calculus the other day doesnt mean you have to forget the fundamentals ;)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2you got it from here?

mags093
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ e^\frac{ 1 }{ x } }{ 6 }\]

mags093
 one year ago
Best ResponseYou've already chosen the best response.0something like that right?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2yes, but apply your limits

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2i mean ....\[\int_{a}^{b}f(x)~dx=F(b)F(a)\]

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2plus you need to change limits beforehand, u=1/x when x=1,u=1 when x=2, u=1/2

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2if you do a usub, then you need to do hartnns route otherwise its just e^(1/x)/6 applied at 1 and 2

mags093
 one year ago
Best ResponseYou've already chosen the best response.0so i put 1 and 2 in for x?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.23 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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2\[\int_{1}^2f(x)dx=F(2)F(1)\] and we determined that \[F(x)=\frac16e^{1/x}\]

mags093
 one year ago
Best ResponseYou've already chosen the best response.0so i sub in 2 and 1 for x then subtract them? is that it?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2yes\[\frac16(e^{1/2}e^{1/1})\] yes\[\frac16(e^{1/2}e)\]

mags093
 one year ago
Best ResponseYou've already chosen the best response.0if i work that out it gives me something like 0.782?

mags093
 one year ago
Best ResponseYou've already chosen the best response.0if i just leave it as \[\frac{ 1 }{ 6 } (e^\frac{ 1 }{ 2 }  e\]

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2did u mean 0.1782 ? and you can keep your answer as 1/6(e^(1/2)e)

mags093
 one year ago
Best ResponseYou've already chosen the best response.0when i worked it out it was 0.782 but if i leave it as the 1/6... it works out just the same :)

mags093
 one year ago
Best ResponseYou've already chosen the best response.0thanks for the help :)
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