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AravindG
 one year ago
Best ResponseYou've already chosen the best response.0y=mod x has no derivative

AravindG
 one year ago
Best ResponseYou've already chosen the best response.0then why ask for anti derivative?

Babyslapmafro
 one year ago
Best ResponseYou've already chosen the best response.0I'm trying to find the area between the following two curves. \[y=\frac{ 2 }{ 1+x^2 }, y=x\]

Babyslapmafro
 one year ago
Best ResponseYou've already chosen the best response.0Between the points, x=1,1

Babyslapmafro
 one year ago
Best ResponseYou've already chosen the best response.0Right but how do I integrate with an absolute value of x?

chethana_bhaskara
 one year ago
Best ResponseYou've already chosen the best response.1integral(1 to 0) (x)dx+ integral(0 to 1)xdx

Babyslapmafro
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{1}^{1}(\frac{ 2 }{ 1+x^2 }x)dx\]

Babyslapmafro
 one year ago
Best ResponseYou've already chosen the best response.0oh i see, i have to find two definite integrals

chethana_bhaskara
 one year ago
Best ResponseYou've already chosen the best response.1u get the value of that integral of x as 1

chethana_bhaskara
 one year ago
Best ResponseYou've already chosen the best response.1yea because, x can either be +x or x ...so u integrate x from 1 to 0 and +x from 0 to 1

Babyslapmafro
 one year ago
Best ResponseYou've already chosen the best response.0Then you just add the two areas together?

Babyslapmafro
 one year ago
Best ResponseYou've already chosen the best response.0Ok could you just explain again why we use x when find the definite integral between 1 and 0?

Babyslapmafro
 one year ago
Best ResponseYou've already chosen the best response.0I got the answer right, I just want to be clear as to why that is done.

chethana_bhaskara
 one year ago
Best ResponseYou've already chosen the best response.1because x is negative and you so u consider the negative values of x .....I actually don't have complete knowledge on this topic :/

Babyslapmafro
 one year ago
Best ResponseYou've already chosen the best response.0Ok well I'll ask when I get to class thanks for the help
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