# Integration question by Saket

**Question:**

What is the integral of !!2^{{y}}!! with respect to !!y!!. Limits of integration are !!0!! to !![5.8]!!. Where !!{x}!! is fractional part of !!x!! and !![x]!! is g.i.f

**Solution:**

So we are finding out the integral of the question given by

!!int_0^[5.8] 2^{{y}} dy!!

!!=int_0^5 2^{{y}} dy!!

!!=int_0^1 2^{{y}} dy+int_1^2 2^{{y}} dy+int_2^3 2^{{y}} dy+int_3^4 2^{{y}} dy+int_4^5 2^{{y}} dy!!

Since !!{y}!! is the fractional part of !!y!! so !!{y}={y-k}!! for any integer !!k!!.

So we get !!int_k^(k+1) 2^{y} dy=int_0^1 2^{y} dy!!

Hence we have !!int_0^[5.8] 2^{{y}} dy=5int_0^1 2^{{y}} dy!!

!!=5int_0^1 2^y dy!!

!!=5/ln2(2^1-2^0)=5/ln2!!