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ksaimouli

  • 2 years ago

find all possible function with given derivative

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  1. ksaimouli
    • 2 years ago
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    |dw:1353714182799:dw|

  2. ksaimouli
    • 2 years ago
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    @zepdrix integrals not introduced

  3. zepdrix
    • 2 years ago
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    not introduced? what does that mean? :o

  4. ksaimouli
    • 2 years ago
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    i dont have any idea of integrals

  5. ksaimouli
    • 2 years ago
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    i have no idea

  6. zepdrix
    • 2 years ago
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    Powers of x aren't too bad, we simply do the REVERSE that we did in differentiation, but we also do it in the REVERSE order. The power rule for integration is, we increase the power by 1, then we divide by the NEW power. Example:\[\huge x^n\]When we anti-differentiate this we get,\[\huge \frac{x^{n+1}}{n+1}\]You'll sometimes see the division instead written as a fraction in front, I prefer this at least, hehe.\[\huge \frac{1}{n+1} x^{n+1}\]

  7. zepdrix
    • 2 years ago
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    How can we apply this to the first term?

  8. ksaimouli
    • 2 years ago
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    |dw:1353714722655:dw|

  9. ksaimouli
    • 2 years ago
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    |dw:1353714766187:dw|

  10. zepdrix
    • 2 years ago
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    Yah looks good :D

  11. zepdrix
    • 2 years ago
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    Integration can be a little daunting at first, with the big fancy S bar, and trying to understand what the dx is there for. So we'll ignore all of that for right now :D Just dip your feet into the water, don't jump in yet. We'll just call it anti-differentiation for now. We're trying to UNDO the process that we're familiar with.

  12. ksaimouli
    • 2 years ago
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    so e^-x remain same why?

  13. zepdrix
    • 2 years ago
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    \[\huge y=e^{2x}\]\[\huge y'=2e^{2x}\]So we're familiar with this process right? Let's see if we understand what's happening in reverse.

  14. zepdrix
    • 2 years ago
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    \[\huge y'=e^{2x}\]Hmmm we don't have that 2 there anymore, so we won't get back e^2x, does that make sense? We're going to be off by a factor of 2. So to fix that, we'll divide by 2.\[\huge y=\frac{1}{2}e^{2x}\]

  15. ksaimouli
    • 2 years ago
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    |dw:1353715382636:dw|

  16. zepdrix
    • 2 years ago
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    \[\huge y'=e^{-x}\]\[\huge y=\frac{1}{-1}e^{-x}\]Hmm it looks like you might have too many negatives in there.

  17. ksaimouli
    • 2 years ago
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    how? when i cancel we do get -e^-x

  18. zepdrix
    • 2 years ago
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    Wut? :o

  19. zepdrix
    • 2 years ago
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    We are starting with e^-x. Maybe your just a little confused remembering the DERIVATIVE of this term. \[\huge (e^{-x})' \neq e^{-x}\] \[\huge (e^{-x})' = -e^{-x}\]Right? Let's make sure we have that much straightened out :)

  20. ksaimouli
    • 2 years ago
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    thx i got it now

  21. zepdrix
    • 2 years ago
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    The only last little bit to remember is to add a CONSTANT to the end of your equation when you anti-differentiate. Here is why:\[\large y=x^2+2\]\[\large y'=2x\] Separately, let's examine another function.\[\large y=x^2+7\]\[\large y'=2x\]See how they both gave us the same result?? So if I asked you, what is the anti-derivative of 2x, what would you say??

  22. zepdrix
    • 2 years ago
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    You just need to remember that it could have been any number, any constant attached to the end, that turned into a 0 when we differentiated. So when you anti-differentiate, you throw on an unknown constant "C" to take care of all of those values.

  23. ksaimouli
    • 2 years ago
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    2x+c

  24. zepdrix
    • 2 years ago
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    yay ksaim c:

  25. ksaimouli
    • 2 years ago
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    ;-) you r awesome

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