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vikrantg4 Group TitleBest ResponseYou've already chosen the best response.1
Don't get it :S
 2 years ago

Vaidehi09 Group TitleBest ResponseYou've already chosen the best response.0
me neither...how can u find the average of just one quantity?
 2 years ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.6
You need \(<\sin^2\theta>\) right?
 2 years ago

shruti Group TitleBest ResponseYou've already chosen the best response.0
not a quantity.
 2 years ago

Vaidehi09 Group TitleBest ResponseYou've already chosen the best response.0
*just one term
 2 years ago

shruti Group TitleBest ResponseYou've already chosen the best response.0
yes @apoorvk
 2 years ago

vikrantg4 Group TitleBest ResponseYou've already chosen the best response.1
average must be 1/2 ? :P
 2 years ago

shruti Group TitleBest ResponseYou've already chosen the best response.0
how you can say that?
 2 years ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.6
@Vaidehi09 basically that means average value of \(\sin^2\theta\) in its period  which is \(\pi\). (the graph for rekindling old memories:)dw:1343221614291:dw
 2 years ago

shruti Group TitleBest ResponseYou've already chosen the best response.0
is this graph applicable for that one? @apoorvk
 2 years ago

vikrantg4 Group TitleBest ResponseYou've already chosen the best response.1
You can try integration
 2 years ago

shruti Group TitleBest ResponseYou've already chosen the best response.0
please contineu how to do so.. ? need to integrate it from zero o pi. but in which period or how?
 2 years ago

vikrantg4 Group TitleBest ResponseYou've already chosen the best response.1
\[1\div(ba)\int\limits_{a}^{b}\sin^2\theta d \theta \]
 2 years ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.6
For this we need to integrate \(\sin^2\theta\) from 0 to \(\pi\),
 2 years ago

vikrantg4 Group TitleBest ResponseYou've already chosen the best response.1
take a = 0 and b = 2pi
 2 years ago

vikrantg4 Group TitleBest ResponseYou've already chosen the best response.1
no bro @apoorvk it should be 2pi
 2 years ago

Vaidehi09 Group TitleBest ResponseYou've already chosen the best response.0
@apoorvk i cant understand why the period is pi.
 2 years ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.6
\[\int\limits_0^{\pi}\sin^2\theta.d\theta\] and then I divide this by the difference of the limit, that is \(\pi 0=\pi\). Basically like the sum of the values by the 'domain covered'.
 2 years ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.6
@Vaidehi09, do you understand that the period sinx of 2pi?
 2 years ago

shruti Group TitleBest ResponseYou've already chosen the best response.0
okay. thanks to all.
 2 years ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.6
now, @Vaidehi09 when I square the function  two changes happen to curve  1. The curve's shape changes. 2. The negative part of the graph is flipped up in the positive axis. So, whereas the period was 2 pi for sinx, the \(\sin^2x\) graph repeats itself after every \(\pi\)distance. Hence the period is pi.
 2 years ago

vikrantg4 Group TitleBest ResponseYou've already chosen the best response.1
OmG ! Expert ! xD Keep it up brother.
 2 years ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.6
Not really :P
 2 years ago

Vaidehi09 Group TitleBest ResponseYou've already chosen the best response.0
@apoorvk ah yes..of course! i was getting confused with the ve part and what kind of change will happen to the shape of the graph....but i get it now. thanks a lot! :))
 2 years ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.6
No worries :]
 2 years ago

vikrantg4 Group TitleBest ResponseYou've already chosen the best response.1
I am new to calc.. so I am not sure about my replies :P
 2 years ago

sami21 Group TitleBest ResponseYou've already chosen the best response.0
@apoorvk good Explaination.
 2 years ago
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