Find the area of the region enclosed by the curve y =[1/(1+x^2)] and it’s asymptote.

Intersection point of the curve with the axis can be obtained by puting x=0=> y=1 i.e. (0,1)

eq of curve can be written as y+yx^2=1

Asymptotes parralel to the axis is given by equating to zero the coeff. of highest power of x

i.e. y=0

Required area is 2integral(lim x=0 to infinity) y dx

=2integral [1/(1+x^2)]dx under lim 0 to infinity

=2 [arctan(x)] lim 0 to infinity

=2[arctan(infinity)-arctan(0)]

=2.pi/2

=pi

Thanks & Regards

Rinkoo Gupta

AskIITians Faculty