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ksaimouli
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1353714182799:dw

ksaimouli
 2 years ago
Best ResponseYou've already chosen the best response.0@zepdrix integrals not introduced

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1not introduced? what does that mean? :o

ksaimouli
 2 years ago
Best ResponseYou've already chosen the best response.0i dont have any idea of integrals

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Powers 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 antidifferentiate 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
 2 years ago
Best ResponseYou've already chosen the best response.1How can we apply this to the first term?

ksaimouli
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1353714722655:dw

ksaimouli
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1353714766187:dw

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Integration 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 antidifferentiation for now. We're trying to UNDO the process that we're familiar with.

ksaimouli
 2 years ago
Best ResponseYou've already chosen the best response.0so e^x remain same why?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1\[\huge y=e^{2x}\]\[\huge y'=2e^{2x}\]So we're familiar with this process right? Let's see if we understand what's happening in reverse.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1\[\huge y'=e^{2x}\]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
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1353715382636:dw

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1\[\huge y'=e^{x}\]\[\huge y=\frac{1}{1}e^{x}\]Hmm it looks like you might have too many negatives in there.

ksaimouli
 2 years ago
Best ResponseYou've already chosen the best response.0how? when i cancel we do get e^x

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1We 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 :)

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1The only last little bit to remember is to add a CONSTANT to the end of your equation when you antidifferentiate. 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 antiderivative of 2x, what would you say??

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1You 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 antidifferentiate, you throw on an unknown constant "C" to take care of all of those values.
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