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mags093

  • 2 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
    • 2 years ago
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    e^1/3 looks like a variable, but its not; its just a constant

  2. mags093
    • 2 years ago
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    \[\int\limits\limits_{1}^{2} \frac{ e^\frac{ 1 }{ x } }{ -6x^2 } dx\]

  3. amistre64
    • 2 years ago
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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

  4. mags093
    • 2 years ago
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    thats what it looks like properly :)

  5. mags093
    • 2 years ago
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    \[\frac{ 1 }{ x }\]

  6. amistre64
    • 2 years ago
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    well, if this is going to be simple than with any luck it will come from something like:\[\Large \frac d{dx}e^{1/x}\]

  7. amistre64
    • 2 years ago
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    so lets start by taking the derivative of that

  8. hartnn
    • 2 years ago
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    yeah, just put u= 1/x du=... ?

  9. Spacelimbus
    • 2 years ago
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    exactly, @hartnn, this substution does the trick.

  10. mags093
    • 2 years ago
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    do we not have to do something with the derivavtive of -6x^2?

  11. amistre64
    • 2 years ago
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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"

  12. mags093
    • 2 years ago
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    the integral exists anyway! i know that much

  13. amistre64
    • 2 years ago
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    \[\frac d{dx}e^{1/x}=\frac{e^{1/x}}{-2x^2}\]agree?

  14. mags093
    • 2 years ago
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    where did the -2x^2 come from? is the derivative of -6x^2 not 12x?

  15. amistre64
    • 2 years ago
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    your confusing your rules .... its best to see these things on a more holistic level :)

  16. mags093
    • 2 years ago
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    oh ok :)

  17. amistre64
    • 2 years ago
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    \[\frac d{dx}e^{u(x)}=\frac{d}{dx}u(x)~e^{u(x)}=u' ~e^u\]

  18. amistre64
    • 2 years ago
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    let u= 1/x du/dx = -1/2x^2 du = -1/2x^2 dx

  19. mags093
    • 2 years ago
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    how do i get the derivative of 1/x?

  20. amistre64
    • 2 years ago
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    1/x = x^-1 which is then just the power rule .... which means i got a bad 2 in there :)

  21. amistre64
    • 2 years ago
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    im thinking of a sqrt function ..... silly me

  22. mags093
    • 2 years ago
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    what do you mean a bad 2? :)

  23. amistre64
    • 2 years ago
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    x^-1 goes to -x^(-2) 1/x goes to -1/x^2 NOT -1/2x^2

  24. mags093
    • 2 years ago
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    oh ok i get ya now

  25. amistre64
    • 2 years ago
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    so just pull out the 1/6 and integrate the easy way

  26. mags093
    • 2 years ago
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    wheres the 1/6?

  27. amistre64
    • 2 years ago
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    .....

  28. amistre64
    • 2 years ago
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    \[\int\frac{e^{1/x}}{-6x^2}dx\] \[\int\frac16 \frac{e^{1/x}}{-x^2}dx\] \[\frac16\int \frac{e^{1/x}}{-x^2}dx\]

  29. mags093
    • 2 years ago
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    oh yes i see :) got confused for a minute!

  30. amistre64
    • 2 years ago
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    just becasue you were doing vector calculus the other day doesnt mean you have to forget the fundamentals ;)

  31. amistre64
    • 2 years ago
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    you got it from here?

  32. mags093
    • 2 years ago
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    \[\frac{ e^\frac{ 1 }{ x } }{ 6 }\]

  33. mags093
    • 2 years ago
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    something like that right?

  34. amistre64
    • 2 years ago
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    yes, but apply your limits

  35. mags093
    • 2 years ago
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    what do you mean?

  36. hartnn
    • 2 years ago
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    e^u/6

  37. hartnn
    • 2 years ago
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    not, e^(1/x) /6

  38. amistre64
    • 2 years ago
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    i mean ....\[\int_{a}^{b}f(x)~dx=F(b)-F(a)\]

  39. hartnn
    • 2 years ago
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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

  40. mags093
    • 2 years ago
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    but u=1/x?

  41. amistre64
    • 2 years ago
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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

  42. mags093
    • 2 years ago
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    so i put 1 and 2 in for x?

  43. amistre64
    • 2 years ago
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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

  44. mags093
    • 2 years ago
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    \[\frac{ e^u }{ 6 }\]

  45. amistre64
    • 2 years ago
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    \[\int_{1}^2f(x)dx=F(2)-F(1)\] and we determined that \[F(x)=\frac16e^{1/x}\]

  46. mags093
    • 2 years ago
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    and then i got that^

  47. mags093
    • 2 years ago
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    so i sub in 2 and 1 for x then subtract them? is that it?

  48. amistre64
    • 2 years ago
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    yes\[\frac16(e^{1/2}-e^{1/1})\] yes\[\frac16(e^{1/2}-e)\]

  49. mags093
    • 2 years ago
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    if i work that out it gives me something like -0.782?

  50. mags093
    • 2 years ago
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    if i just leave it as \[\frac{ 1 }{ 6 } (e^\frac{ 1 }{ 2 } - e\]

  51. hartnn
    • 2 years ago
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    did u mean -0.1782 ? and you can keep your answer as 1/6(e^(1/2)-e)

  52. mags093
    • 2 years ago
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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 :)

  53. mags093
    • 2 years ago
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    thanks for the help :)

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