JEE Main 26 Feb 2021 Shift 1 Solved Paper

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]
‌‌=

1

∫

0

e[x]dx+

2

∫

1

e{x⟩dx+...+

100

∫

99

e[x]dx
‌=ex]01+e(x−1)]12+...+e(x−99)]99100
‌‌=(e−1)+(e−1)+...+(e−1)=100(e−1)