IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

ananth_patri

plz find the value of ₀

^{[ x]} 2^x/2^[x] dx.......plz reply fast to get rates..

ananth_patri

plz reply,...

akhil_o

[x]/log 2?

ananth_patri

excellent but proceure plz....

akhil_o

see takin denominator in power we get
2^(x-[x])
=2^{x}
{x} is periodic with period 1

so 2^{x} also repaets after every 1 unit
also integrating from 0 to [x]
we have integral no of such units
area of 1 unit
=integral (from 0 to 1) 2^x
=2^x/log 2 from 0 to 1
=(2-1)/log 2
=1/log 2
and there are [x] such units

so total=[x]/log 2

ananth_patri

y did u consider one unit??and solved??well so in case of periodic func can we cosider in one uints and solve??/ consider sinx now in this case if we consider one unit and solve we wont get correct answer....well plz explain...

akhil_o

yup for all periodic functions we can write
where f(x) has period=a

_[0]

^{[ na]} f(x)dx=n*_{[ 0]}

^{[ a]} f(x)dx

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