IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

aquafina_1991

Compute the area bounded by Witch of Agnesia

,

and positive part of y axis. [ ] denotes the floor function and { } denotes the fractional part function. (â??Witch of Agnesiâ? curve was mentioned by lady mathematician Maria Agnesi in her book. It is said that the lady had the habit of walking in the night in sleep and solving problems while not being aware of it. She used to get surprised in the morning by seeing the solution in her own handwriting.)

punnima

xy2=1-x

y^2=-1+ 1/x ------(1)

we can make the foll obs about the curve

1.it is symm abt the x-axis

2. for y=0,

1/x -1=0

or x=1

i.e, the curve cuts the x axis at (1,0)

3.LHS is always positive

(1/x -1) >= 0

or x<= 1, i.e, the curve exists for x<1

4.the curve doesnot exist for x<0

5.thrfore domain of f(x) is 0<x<1

6. as x----> 0, y------>

punnima

y= [ (x/2) -( {x} /3] ------(2)

the fractional part function is defined for x>0

as {x}=x-[x]

1.... at x=1, {x}=0

y=[1/2] =0

2......when 0<x<1

{x} =x-[x] =x-0=x

y=[x/2 -x/3] = x/6

0<x<1 , 0<x/6<1/6

y=[x/6]=0

equation (2) represents the x axix in the interval 0<x<1

area =_m ₀^mf(y) dy

=_m ₀^m( y² +1 )^-1 dy

=_m tan ^-1(y)

=tan^-1(

)

thus the area is pi/2 sq. units

akhil_o

Punnima is spot on!

second function is always 0 in (0,1]

so we just find area of first function

x=1/(1+y²)

integrating from limits 0 to infinity

we get tan^-1y from 0 to infinity

so tan^-1 infinty -tan^-10

= pi/2 - 0

= pi/2

punnima

akhil_o: thanx for the backup....i am glad that u found my ans. rite..
i am usually weak in qns. related to [ ] function and { } function.
pls. tell me how to thorough myself in this area.

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