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## anonymous 3 years 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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1. amistre64

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

2. anonymous

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

3. 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

4. anonymous

thats what it looks like properly :)

5. anonymous

$\frac{ 1 }{ x }$

6. amistre64

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

7. amistre64

so lets start by taking the derivative of that

8. hartnn

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

9. anonymous

exactly, @hartnn, this substution does the trick.

10. anonymous

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

11. 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"

12. anonymous

the integral exists anyway! i know that much

13. amistre64

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

14. anonymous

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

15. amistre64

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

16. anonymous

oh ok :)

17. amistre64

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

18. amistre64

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

19. anonymous

how do i get the derivative of 1/x?

20. amistre64

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

21. amistre64

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

22. anonymous

what do you mean a bad 2? :)

23. amistre64

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

24. anonymous

oh ok i get ya now

25. amistre64

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

26. anonymous

wheres the 1/6?

27. amistre64

.....

28. amistre64

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

29. anonymous

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

30. amistre64

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

31. amistre64

you got it from here?

32. anonymous

$\frac{ e^\frac{ 1 }{ x } }{ 6 }$

33. anonymous

something like that right?

34. amistre64

yes, but apply your limits

35. anonymous

what do you mean?

36. hartnn

e^u/6

37. hartnn

not, e^(1/x) /6

38. amistre64

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

39. hartnn

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

40. anonymous

but u=1/x?

41. 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

42. anonymous

so i put 1 and 2 in for x?

43. 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

44. anonymous

$\frac{ e^u }{ 6 }$

45. amistre64

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

46. anonymous

and then i got that^

47. anonymous

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

48. amistre64

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

49. anonymous

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

50. anonymous

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

51. hartnn

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

52. 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 :)

53. anonymous

thanks for the help :)

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