double integral of e^y/x dy dx with outer limits as 0 and 2 and inner limits as 0 and x^2 ???
@TuringTest why do we dont change order in this one...ans, is 1/2
\[\int\limits_{0}^{1}\int\limits_{0}^{x^2} e^ y/x dy dx\]
@ kainui then why we change order here let me tag you in one...
I'm sorry, I'm either really tired or confused. It seems to me that this integral can only be done by changing the bounds. Is that what you are saying @chakshu ? You are asking why we have to change the bonds?
m asking that why this question is not solved by changing order of integeration its just solved simply to give ans. as 1/2
i will repeat turing's word. in last Q, it was difficult to integrate w.r.t y after x was integrated, thats why bounds were changed. in this case, its easy to integrate without changing bounds
http://openstudy.com/users/chakshu#/updates/5080305ee4b0b56960054f2d this is another questn that involves changing order i just wanna knoe the theoritical differnce that when do we have to change order to integerate ????hope this is simple to understnd
It might help to sketch a picture of the graph first to better explain this.
i may be totally wrong (probably am) but isn't \[\int_0^{x^2}\frac{e^y}{x}dy=\frac{e^{x^2}-1}{x}\]
@hartnn so ur sayin since in previous questn we had difficult limts so we changed order and in this one we have easy limits so we dont??
then second job would be to compute \[\int_0^1\frac{e^{x^2}-1}{x}dx\]
ohhhhhhhh my bad frnds its e^y/x sorrrrrrrrryyyyyy for that mistake
its not about limits, its about what u get after integrating w.r.t one of the variables, sometimes the resulting function is very difficult to integrate w.r.t other variable...
In general: \[\int\limits \int\limits f(x,y)dA = \int\limits_{a}^{b} \int\limits_{g_1(x)}^{g_2(x)}f(x,y)dydx \]
Use parentheses. e^(y/x) or (e^y)/x?
@chakshu no one can help you until you answer this last question I just asked you lol.
I answered his question through facebook everyone, my connect here sucks it was e^(y/x)
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