Question

For a real number x let [x] denote the largest integer less than or equal to x and {x} = x - [x] . The smallest possible integer value of n for which ${\int }_1^n\left[x\right]\left\{x\right\}dx$ exceeds 2013 is

• A. 63
• B. 64
• C. 90
• D. 91

Answer 1

Answer: (D)

91

${\int }_1^n\left[x\right]\left\{x\right\}dx$

$={\int }_1^{2}1(x-1)dx+{\int }_2^{3}2(x-2)dx+...+{\int }_{n-1}^n(n-1)(x-n+1)dx$

This can be generalized as $\sum _{t=1}^{n-1}{\int }_t^{t+1}t(x-t)dx$

$=\sum _{t=1}^{n-1}{[\frac{t{x}^2}{2}-t^{2}x]}_t^{t+1}$

$=\sum _{t=1}^{n-1}[\frac{t(t^2+2t+1-t^2)}{2}-t^2(t+1-t\left)\right]$

$=\sum _{t=1}^{n-1}[t^2+\frac{t}{2}-t^2]$

$=\frac{1}{2}\times \frac{(n-1)n}{2}$

$=\frac{(n-1)n}{4}$

We need the smallest n so that $(n-1)n>2013\times 4$

i.e. $(n-1)n>8052$

The smallest square number greater than $8052$ is $8100$, whose square root is $90$.

If $n=90,90\ast 89<8052$

$\therefore n=91$