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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
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
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