How to calculate this? #int_0^1{nx}^2dx#; #ninNN#*

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Answer 1 · 2017-05-18T20:53:07

# int_0^1{nx}^2dx = n^2/3 #

# int_0^1{nx}^2dx = int_0^1n^2x^2dx #
# " " = n^2int_0^1x^2dx #
# " " = n^2[x^3/3]_0^1 #
# " " = n^2(1/3-0) #
# " " = n^2/3 #

Answer 2 · 2017-05-18T23:19:08

See below.

Assuming #{x}=x-floor(x)# we have

#int_0^1 {nx}^2 dx = int_0^1 (nx-floor(nx))^2 dx# and making #y = nx# we have

# int_0^1 (nx-floor(nx))^2 dx = 1/nint_0^n(y-floor(y))^2dy =#

now #y-floor(y) = xi in [0,1)#

so

#1/nint_0^n(y-floor(y))^2dy = 1/n(sum_(k=0)^(n-1)int_0^1xi^2d xi)#

#=1/n(n 1/3) = 1/3#

Answer 3 · 2017-05-19T00:11:35

#int_0^1 {nx}^2dx = 1/3#

independently of #n#.

If #{nx}^2# is the fractional part #(nx-[nx])^2# then we have to consider that the function is continuous in the intervals:

#0 < nx < 1#

#1 < nx < 2#

etc... that is in the intervals:

#0 < x < 1/n#

#1/n < x < 2/n#

etc...

For every #n# consider then #j=0,1,..., n-1#. In the interval #x in [j/n,(j+1)/n)# the function is continuous and its value is:

#{nx}^2 = (nx-j)^2 =n^2(x-j/n)^2#

We can then express the integral as:

#int_0^1 {nx}^2dx = n^2 sum_(j=0)^(n-1) int_(j/n)^((j+1)/n) (x-j/n)^2dx#

Substitute now in the #j#-th term #y=x-j/n#:

#int_(j/n)^((j+1)/n) (x-j/n)^2dx = int_0^(1/n) y^2dy =[y^3/3]_0^(1/n) = 1/(3n^3)#

So the terms are all equal and their sum is:

#sum_(j=0)^(n-1) int_(j/n)^((j+1)/n) (x-j/n)^2dx = sum_(j=0)^(n-1) 1/(3n^3) = n xx 1/(3n^3) = 1/(3n^2)#

Finally:

#int_0^1 {nx}^2dx = n^2 xx 1/(3n^2) = 1/3 #