## ksaimouli Group Title find all possible function with given derivative one year ago one year ago

1. ksaimouli Group Title

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2. ksaimouli Group Title

@zepdrix integrals not introduced

3. zepdrix Group Title

not introduced? what does that mean? :o

4. ksaimouli Group Title

i dont have any idea of integrals

5. ksaimouli Group Title

i have no idea

6. zepdrix Group Title

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 Group Title

How can we apply this to the first term?

8. ksaimouli Group Title

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9. ksaimouli Group Title

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10. zepdrix Group Title

Yah looks good :D

11. zepdrix Group Title

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 Group Title

so e^-x remain same why?

13. zepdrix Group Title

$\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 Group Title

$\huge y'=e^{2x}$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 Group Title

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16. zepdrix Group Title

$\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 Group Title

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

18. zepdrix Group Title

Wut? :o

19. zepdrix Group Title

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 Group Title

thx i got it now

21. zepdrix Group Title

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 Group Title

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 Group Title

2x+c

24. zepdrix Group Title

yay ksaim c:

25. ksaimouli Group Title

;-) you r awesome