IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

bubunbhatt

_{[ 0]}

^{[ 100 ]} (tan~1x) dx

Here (.) is greatest integer function (box func.)

and tan~1x is tan inverse x

Option :

(a) tan1-100 (b)

/2- tan1 (c) 100 - tan1 (d) none of these

Please post full explanation..........

rishikesh_anshu

breaking limit from 0 to tan1 and from tan1 to 100
so in limit 0 to tan1 tan^-1x<1 hence [tan^-1x]=0
and in limit tan1 to 100 <1tan^-1x<2 hence [tan^-1x]=1

hence

_{[ 0]}

^{[tan1 ]} 0dx+_{[tan1 ]}

^{[ 100]} 1dx
=100-tan1

bubunbhatt

Please
explain why you broke the limits from 0 to tan1
and tan1 to 100
Anyways THANKS A LoT

t_c511

here we have to break the limts because for evaluating integration involving greatest integer function, limits are to be adjusted such that we get values of function 0 for 0 to 1, 1 for 1 to 2.......etc

bubunbhatt

But how does the value of tan1 to 100 lie between 2??
coz see putting x=100 we get tan~1 (100)=89.4275
Please help i am not getting this
Bubun

bubunbhatt

PLEASE HELP SOMEONE

cbg2007

dear bubunbhatt,
integral tan^-1(x) dx . 1=
x tan^-1 x - integral x/1+x^2 dx

x tan^-1 x - 1/2 log(1+x^2) +c
apply limits to the above

gokul07

lt 0 to tan1 tan^-1x<1 hence [tan^-1x]=0
and in lt tan1 to 100 <1tan^-1x<2 hence [tan^-1x]=1

hence

_{[ 0]}

^{[tan1 ]} 0dx+_{[tan1 ]}

^{[ 100]} 1dx
=100-tan1

bubunbhatt

Guys please explan how tan1 to 100 is less than 2??????????????

KAB

When x lies between 0 to tan1 [tan(^-1)x] will be zero.
When x lies between tan1 to 100 [tan(^-1)x] will be 1 because tan(^-1)100=1.56 radians So [1.56]=1

bubunbhatt

Thanks Kab
You have cleared my dout.
plz accept my salute
Thanks

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