find all possible function with given derivative

- ksaimouli

find all possible function with given derivative

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- ksaimouli

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- ksaimouli

@zepdrix integrals not introduced

- zepdrix

not introduced? what does that mean? :o

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## More answers

- ksaimouli

i dont have any idea of integrals

- ksaimouli

i have no idea

- zepdrix

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.\[\huge \frac{1}{n+1} x^{n+1}\]

- zepdrix

How can we apply this to the first term?

- ksaimouli

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- ksaimouli

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- zepdrix

Yah looks good :D

- zepdrix

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.

- ksaimouli

so e^-x remain same why?

- zepdrix

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

- zepdrix

\[\huge y'=e^{2x}\]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.\[\huge y=\frac{1}{2}e^{2x}\]

- ksaimouli

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- zepdrix

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

- ksaimouli

how? when i cancel we do get -e^-x

- zepdrix

Wut? :o

- zepdrix

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 :)

- ksaimouli

thx i got it now

- zepdrix

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??

- zepdrix

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.

- ksaimouli

2x+c

- zepdrix

yay ksaim c:

- ksaimouli

;-) you r awesome

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