What most adequately sums up the 90's?

Question

anonymous · Answer

Is this a legit question?

anonymous · Answer

Really?

kainui · Answer

@ParthKohli says 

90+91+92+93+...+99 but I'm not sure that's good enough.

anonymous · Answer

It is good enough baby.

anonymous · Answer

dude... you must be 1 20s boy

parthkohli · Answer

$$\int_{90}^{100} xdx$$Memories, memories. ;-;

parthkohli · Answer

If you define 90 to be in the 90s, then add 90 to that.

shamil98 · Answer

$$\Large \sum_{i = 90}^{99} i$$
same thing rite idk i suck at math

kainui · Answer

Yes but wait are these truly all the 90s? I mean not just the 90s that are integers are included which is good. But what about -92?

parthkohli · Answer

Then the sum should be zero.

shamil98 · Answer

if you included the negatives from -90 to -99 it would just be zero..

kainui · Answer

Yeah, so what is it? Also, I just realized something possibly important. $$\Large \pi ^2*10=98.696...$$ Could this mean something we didn't consider?

anonymous · Answer

@ParthKohli  I don't understand why I can get the same answer from your methods
$$\int_{90}^{100} xdx =950$$ while 90+91+....+99=945
why?

texaschic101 · Answer

lol...I thought you were talking about the 90's like 1990 .....memories

shamil98 · Answer

he's going from 90 to 100 
we went from 90 to 99

parthkohli · Answer

Because integrating is not summing up stepwise.

shamil98 · Answer

or something like that

kainui · Answer

Well in fact there are an infinite number of 90s between 90 and 100, so we multiply each one by an infinitesimal amount and then add it up. So we're getting kind of like a "relative" sum or some such. Not sure. But the answer might be infinity or zero.

parthkohli · Answer

Yeah, you would think that the 99 to 100 should be giving me a good advantage, but it doesn't.  Know why?|dw:1403835763477:dw|