Greatest Integral

Calculus Level 4

Υ = ∫ 0 n [ x ] dx \large \Upsilon = \int_0^n [ x ] \text{ dx}

Choose the correct option:

A) If n is an integer then Υ = n ( n + 1 ) 2 \Upsilon = \dfrac{n(n + 1)}{2}

B) If n = x then Υ = x [ x ] − [ x ] 2 ( [ x ] + 1 ) \Upsilon = x[x] - \dfrac{[x]}{2}( [x] +1)

C) If n = x then Υ = x [ x ] + [ x ] 2 ( [ x ] + 1 ) \Upsilon = x[x] + \dfrac{[x]}{2}( [x] +1)

D)If n is an integer then Υ = ( n − 1 ) ( n + 1 ) 2 \Upsilon = \dfrac{(n - 1)(n + 1)}{2}

Details :

  • n > 0 n > 0
  • [ . ] represents Greatest Integer function
More than one option correct None of these D B A C

Akhil Bansal by Akhil Bansal , 22, India

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1 solution

If n n is an integer, then Υ = ∫ 0 n [ x ] d x = ∑ k = 0 n − 1 k = n ( n − 1 ) 2 . \Upsilon = \int_0^n [x]dx = \sum_{k = 0}^{n-1} k = \frac{n(n-1)}2. We can also write this as ⋯ = n 2 − n 2 = 2 n 2 − ( n 2 + n ) 2 = n 2 − n 2 ( n + 1 ) . \dots = \frac{n^2-n}2 = \frac{2n^2 - (n^2 + n)}2 = n^2 - \frac n2 (n+1). Now we have Υ = ∫ 0 [ n ] [ x ] d x + ∫ [ n ] n [ x ] d x = [ n ] 2 − [ n ] 2 ( [ n ] + 1 ) + [ n ] ( n − [ n ] ) , \Upsilon = \int_0^{[n]} [x]dx + \int_{[n]}^n [x]dx = [n]^2-\frac {[n]}2 ([n]+1) + [n](n- [n]), because the second integrand is constant and equal to [ n ] [n] on the interval from [ n ] [n] to n n . Canceling the first term [ n ] 2 [n]^2 against the last term [ n ] ( ⋯ − [ n ] ) [n](\dots -[n]) , we find Υ = ∫ 0 [ n ] [ x ] d x + ∫ [ n ] n [ x ] d x = [ n ] n − [ n ] 2 ( [ n ] + 1 ) . \Upsilon = \int_0^{[n]} [x]dx + \int_{[n]}^n [x]dx = [n]n-\frac {[n]}2 ([n]+1).

Thus the solution is B.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

I also wrote that if n = x... and my name is Akhil not Aknil..

Akhil Bansal - 5 years, 8 months ago

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Apologies-- I changed the name :) The use of x x is not technically incorrect, but it is confusing, since you also use x x as a integration variable.

Arjen Vreugdenhil - 5 years, 8 months ago

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