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Question: The value of \(\displaystyle \int_0^x \lfloor t+1\rfloor^3 \,dt\) (where \(\lfloor\cdot\rfloor\) den...

The value of ∫0x⌊t+1⌋3 dt\displaystyle \int_0^x \lfloor t+1\rfloor^3 \,dt (where ⌊⋅⌋\lfloor\cdot\rfloor denotes the greatest integer function of xx) is equal to

A

([x]([x]+1)2)2+([x]+1)3{x}\bigl(\tfrac{[x]\bigl([x]+1\bigr)}{2}\bigr)^2 + \bigl([x]+1\bigr)^3\{x\}

B

([x]([x]+1)2)3+([x]+1)3{x}\bigl(\tfrac{[x]\bigl([x]+1\bigr)}{2}\bigr)^3 + \bigl([x]+1\bigr)^3\{x\}

C

([x]([x]+1)2)3+([x]+1)2{x}\bigl(\tfrac{[x]\bigl([x]+1\bigr)}{2}\bigr)^3 + \bigl([x]+1\bigr)^2\{x\}

D

None of these

Answer

([x]([x]+1)2)2+([x]+1)3{x}\bigl(\tfrac{[x]([x]+1)}{2}\bigr)^2 + ([x]+1)^3\{x\}

Explanation

Solution

Solution Outline

  1. Partition the integral from t=0t=0 to t=xt=x at integer points.
  2. For integer kk from 1 to ⌊x⌋\lfloor x\rfloor: ∫k−1k⌊t+1⌋3 dt=∫k−1kk3 dt=k3. \int_{k-1}^{k} \lfloor t+1\rfloor^3\,dt = \int_{k-1}^{k} k^3\,dt = k^3.
  3. For the remaining part t∈[⌊x⌋,x]t\in[\lfloor x\rfloor,x]: ∫⌊x⌋x⌊t+1⌋3 dt=∫⌊x⌋x( ⌊x⌋+1 )3 dt=( ⌊x⌋+1 )3{x}. \int_{\lfloor x\rfloor}^{x} \lfloor t+1\rfloor^3\,dt = \int_{\lfloor x\rfloor}^{x} (\,\lfloor x\rfloor+1\,)^3\,dt = (\,\lfloor x\rfloor+1\,)^3\{x\}.
  4. Sum of cubes formula: ∑k=1nk3=(n(n+1)2)2,n=⌊x⌋. \sum_{k=1}^{n} k^3 = \Bigl(\frac{n(n+1)}{2}\Bigr)^{2},\quad n=\lfloor x\rfloor.
  5. Combine to obtain ∫0x⌊t+1⌋3 dt=([x]([x]+1)2)2+([x]+1)3{x}. \int_{0}^{x}\lfloor t+1\rfloor^3\,dt = \Bigl(\frac{[x]([x]+1)}{2}\Bigr)^{2} + ([x]+1)^{3}\{x\}.