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

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

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

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

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

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

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

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

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

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

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

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

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

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

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