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

- anonymous

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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- amistre64

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

- anonymous

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

- 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

thats what it looks like properly :)

- anonymous

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

- 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

so lets start by taking the derivative of that

- hartnn

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

- anonymous

exactly, @hartnn, this substution does the trick.

- anonymous

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

- 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

the integral exists anyway! i know that much

- amistre64

\[\frac d{dx}e^{1/x}=\frac{e^{1/x}}{-2x^2}\]agree?

- anonymous

where did the -2x^2 come from? is the derivative of -6x^2 not 12x?

- amistre64

your confusing your rules .... its best to see these things on a more holistic level :)

- anonymous

oh ok :)

- amistre64

\[\frac d{dx}e^{u(x)}=\frac{d}{dx}u(x)~e^{u(x)}=u' ~e^u\]

- amistre64

let u= 1/x
du/dx = -1/2x^2
du = -1/2x^2 dx

- anonymous

how do i get the derivative of 1/x?

- amistre64

1/x = x^-1 which is then just the power rule .... which means i got a bad 2 in there :)

- amistre64

im thinking of a sqrt function ..... silly me

- anonymous

what do you mean a bad 2? :)

- amistre64

x^-1 goes to -x^(-2)
1/x goes to -1/x^2 NOT -1/2x^2

- anonymous

oh ok i get ya now

- amistre64

so just pull out the 1/6 and integrate the easy way

- anonymous

wheres the 1/6?

- 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

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

- amistre64

just becasue you were doing vector calculus the other day doesnt mean you have to forget the fundamentals ;)

- amistre64

you got it from here?

- anonymous

\[\frac{ e^\frac{ 1 }{ x } }{ 6 }\]

- anonymous

something like that right?

- amistre64

yes, but apply your limits

- anonymous

what do you mean?

- hartnn

e^u/6

- hartnn

not, e^(1/x) /6

- amistre64

i mean ....\[\int_{a}^{b}f(x)~dx=F(b)-F(a)\]

- hartnn

plus you need to change limits beforehand,
u=1/x
when x=1,u=1
when x=2, u=1/2

- anonymous

but u=1/x?

- 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

so i put 1 and 2 in for x?

- 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

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

- amistre64

\[\int_{1}^2f(x)dx=F(2)-F(1)\]
and we determined that \[F(x)=\frac16e^{1/x}\]

- anonymous

and then i got that^

- anonymous

so i sub in 2 and 1 for x then subtract them? is that it?

- amistre64

yes\[\frac16(e^{1/2}-e^{1/1})\]
yes\[\frac16(e^{1/2}-e)\]

- anonymous

if i work that out it gives me something like -0.782?

- anonymous

if i just leave it as \[\frac{ 1 }{ 6 } (e^\frac{ 1 }{ 2 } - e\]

- hartnn

did u mean -0.1782 ?
and you can keep your answer as 1/6(e^(1/2)-e)

- 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

thanks for the help :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.