Let ' x ' be any real number, then x=[x]+{x}, where [x] is integer part of x and {x} is fractional part of x. Then, x−[x]={x}, Also period of {x}=1 Now,
100
∑
n=1
n
∫
n−1
ex−[x]dx=
100
∑
n=1
n
∫
n−1
e[x]dx [Difference between upper and lower limit is 1 unit] =