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

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

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

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

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

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

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

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

chethana_bhaskara
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0Then you just add the two areas together?

Babyslapmafro
 2 years 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
 2 years 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
 2 years 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
 2 years 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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