Question

please help?
Evaluate the definite integral (if it exists)

intergral (e^1/3)/-6x^2 dx from 1 to 2 ?

Answer 1

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

Answer 2

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

Answer 3

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

is that 1/3 or 1/x? need to verify

Answer 4

thats what it looks like properly :)

Answer 5

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

Answer 6

well, if this is going to be simple than with any luck it will come from something like:\[\Large \frac d{dx}e^{1/x}\]

Answer 7

so lets start by taking the derivative of that

Answer 8

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

Answer 9

exactly, @hartnn, this substution does the trick.

Answer 10

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

Answer 11

we should verify that if its going to be simple first and see if we can modifiy it with a useful form of "1"

Answer 12

the integral exists anyway! i know that much

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

oh ok :)

Answer 17

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

Answer 18

let u= 1/x

du/dx = -1/2x^2

du = -1/2x^2 dx

Answer 19

how do i get the derivative of 1/x?

Answer 20

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

Answer 21

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

Answer 22

what do you mean a bad 2? :)

Answer 23

x^-1 goes to -x^(-2)

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

Answer 24

oh ok i get ya now

Answer 25

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

Answer 26

wheres the 1/6?

Answer 27

.....

Answer 28

\[\int\frac{e^{1/x}}{-6x^2}dx\]
\[\int\frac16 \frac{e^{1/x}}{-x^2}dx\]
\[\frac16\int \frac{e^{1/x}}{-x^2}dx\]

Answer 29

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

Answer 30

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

Answer 31

you got it from here?

Answer 32

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

Answer 33

something like that right?

Answer 34

yes, but apply your limits

Answer 35

what do you mean?

Answer 36

e^u/6

Answer 37

not, e^(1/x) /6

Answer 38

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

Answer 39

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

Answer 40

but u=1/x?

Answer 41

if you do a usub, then you need to do hartnns route

otherwise its just e^(1/x)/6 applied at 1 and 2

Answer 42

so i put 1 and 2 in for x?

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

and then i got that^

Answer 47

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

Answer 48

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

Answer 49

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

Answer 50

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

Answer 51

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

Answer 52

when i worked it out it was -0.782 but if i leave it as the 1/6... it works out just the same :)

Answer 53

thanks for the help :)