let-A-n-0-n-x-x-2-dx-1-calculate-A-n-2-find-lim-n-A-n-

Question Number 38100 by maxmathsup by imad last updated on 21/Jun/18

l e t A_{n} = \int_{0}^{n} {(x - [x])}^{2} d x

1) c a l c u l a t e A_{n}

2) f i n d {l i m}_{n \to + \infty} A_{n}

Commented by prof Abdo imad last updated on 22/Jun/18

A_{n} = \int_{0}^{n} (x^{2} - 2 x [x] + {[x]}^{2}) d x

= \int_{0}^{n} x^{2} d x - 2 \int_{0}^{n} x [x] d x + \int_{0}^{n} {[x]}^{2} d x b u t

\int_{0}^{n} x [x] d x = \sum_{k = 0}^{n - 1} \int_{k}^{k + 1} k x d x

= \sum_{k = 0}^{n - 1} k (\frac{{(k + 1)}^{2}}{2} - \frac{k^{2}}{2})

= \sum_{k = 0}^{n - 1} k \frac{k^{2} + 2 k + 1 - k^{2}}{2}

= \sum_{k = 0}^{n - 1} k (k + \frac{1}{2}) = \sum_{k = 0}^{n - 1} k^{2} + \frac{1}{2} \sum_{k = 0}^{n - 1} k

\frac{(n - 1) (n - 1 + 1) (2 (n - 1) + 1)}{6} + \frac{1}{2} \frac{(n - 1) n}{2}

= \frac{n (n - 1) (2 n - 1)}{6} + \frac{n (n - 1)}{4}

= \frac{n (n - 1)}{2} {\frac{2 n - 1}{3} + \frac{1}{2}}

= \frac{n (n - 1)}{2} {\frac{4 n - 2 + 3}{6}}

= \frac{n (n - 1) (4 n + 1)}{12} a l s o

\int_{0}^{n} x^{2} d x = {[\frac{x^{3}}{3}]}_{0}^{n} = \frac{n^{3}}{3}

\int_{0}^{n} {[x]}^{2} d x = \sum_{k = 0}^{n - 1} \int_{k}^{k + 1} k^{2} d x

= \sum_{k = 0}^{n - 1} k^{2} = \frac{n (n - 1) (2 n - 1)}{6} \Rightarrow

A_{n} = \frac{n^{3}}{3} - 2 \frac{n (n - 1) (4 n + 1)}{12} + \frac{n (n - 1) (4 n + 1)}{6}

A_{n} = \frac{n^{3}}{3}

2) {l i m}_{n \to + \infty} A_{n} = + \infty