let A_n = ∫_0 ^n (x−[x])^2 dx 1) calculate A_n 2) find lim_(n→+∞) A


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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
$A_n = ∫_0 ^n ( x^2 −2x[x] +[x]^2 )dx = ∫_0 ^n x^2 dx −2 ∫_0 ^n x[x]dx +∫_0 ^n [x]^2 dx but ∫_0 ^n x[x]dx=Σ_(k=0) ^(n−1) ∫_k ^(k+1) kx dx =Σ_(k=0) ^(n−1) k ( (((k+1)^2 )/2) −(k^2 /2)) =Σ_(k=0) ^(n−1) k((k^2 +2k +1−k^2 )/2) =Σ_(k=0) ^(n−1) k( k +(1/2))=Σ_(k=0) ^(n−1) k^2 +(1/2)Σ_(k=0) ^(n−1) k (((n−1)(n−1+1)(2(n−1)+1))/6) +(1/2)(((n−1)n)/2) =((n(n−1)(2n−1))/6) +((n(n−1))/4) =((n(n−1))/2){ ((2n−1)/3) +(1/2)} =((n(n−1))/2){ ((4n−2 +3)/6)} =((n(n−1)(4n+1))/(12)) also ∫_0 ^n x^2 dx =[(x^3 /3)]_0 ^n = (n^3 /3) ∫_0 ^n [x]^2 dx =Σ_(k=0) ^(n−1) ∫_k ^(k+1) k^2 dx =Σ_(k=0) ^(n−1) k^2 =((n(n−1)(2n−1))/6) ⇒ A_n = (n^3 /3) −2 ((n(n−1)(4n+1))/(12)) +((n(n−1)(4n+1))/6) A_n =(n^3 /3) 2) lim_(n→+∞) A_n =+∞$
$A_{n} = \int_{0}^{n} (x^{2} - 2 x [x] + {[x]}^{2}) d x$ $= \int_{0}^{n} x^{2} d x - 2 \int_{0}^{n} x [x] d x + \int_{0}^{n} {[x]}^{2} d x b u t$ $\int_{0}^{n} x [x] d x = \sum_{k = 0}^{n - 1} \int_{k}^{k + 1} k x d x$ $= \sum_{k = 0}^{n - 1} k (\frac{{(k + 1)}^{2}}{2} - \frac{k^{2}}{2})$ $= \sum_{k = 0}^{n - 1} k \frac{k^{2} + 2 k + 1 - k^{2}}{2}$ $= \sum_{k = 0}^{n - 1} k (k + \frac{1}{2}) = \sum_{k = 0}^{n - 1} k^{2} + \frac{1}{2} \sum_{k = 0}^{n - 1} k$ $\frac{(n - 1) (n - 1 + 1) (2 (n - 1) + 1)}{6} + \frac{1}{2} \frac{(n - 1) n}{2}$ $= \frac{n (n - 1) (2 n - 1)}{6} + \frac{n (n - 1)}{4}$ $= \frac{n (n - 1)}{2} {\frac{2 n - 1}{3} + \frac{1}{2}}$ $= \frac{n (n - 1)}{2} {\frac{4 n - 2 + 3}{6}}$ $= \frac{n (n - 1) (4 n + 1)}{12} a l s o$ $\int_{0}^{n} x^{2} d x = {[\frac{x^{3}}{3}]}_{0}^{n} = \frac{n^{3}}{3}$ $\int_{0}^{n} {[x]}^{2} d x = \sum_{k = 0}^{n - 1} \int_{k}^{k + 1} k^{2} d x$ $= \sum_{k = 0}^{n - 1} k^{2} = \frac{n (n - 1) (2 n - 1)}{6} \Rightarrow$ $A_{n} = \frac{n^{3}}{3} - 2 \frac{n (n - 1) (4 n + 1)}{12} + \frac{n (n - 1) (4 n + 1)}{6}$ $A_{n} = \frac{n^{3}}{3}$ $2) {l i m}_{n \to + \infty} A_{n} = + \infty$
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