Find: 𝛀(n) =∫_( 1) ^( n) ([x]^2 ∙{x} + [x]∙{x}^2 )dx n∈N ; [∗]-GIF ; {x}=x-[x]


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Question Number 157021 by MathSh last updated on 18/Oct/21
$Find: 𝛀(n) =∫_( 1) ^( n) ([x]^2 ∙{x} + [x]∙{x}^2 )dx n∈N ; [∗]-GIF ; {x}=x-[x]$
$Find :$ $Ω (n) = \underset{1}{\int^{n}} ({[x]}^{2} \cdot {x} + [x] \cdot {x}^{2}) dx$ $n \in N; [*] - GIF; {x} = x - [x]$
	Answered by ghimisi last updated on 18/Oct/21
	$Ω(n)=∫_1 ^n [x]{x}([x]+{x})dx=∫_1 ^n x[x](x−[x])dx= Σ_(k=1) ^(n−1) ∫_k ^(k+1) (x^2 k−xk^2 )dx=Σ_(k=1) ^(n−1) (k∙(((k+1)^3 −k^3 )/3)−k^2 ∙(((k+1)^2 −k^2 )/2))= Σ_(k=1) ^(n−1) (((3k^3 +3k^2 +k)/3)−((2k^3 +k^2 )/2))=Σ_(k=1) ^(n−1) ((3k^2 +2k)/6)= (1/6)(3∙(((n−1)n(2n−1))/6)+2∙(((n−1)n)/2))=((n(n−1)(2n+1))/(12))$
	$Ω (n) = \underset{1}{\int^{n}} [x] {x} ([x] + {x}) d x = \underset{1}{\int^{n}} x [x] (x - [x]) d x =$ $\underset{k = 1}{\sum^{n - 1}} \underset{k}{\int^{k + 1}} (x^{2} k - {x k}^{2}) d x = \underset{k = 1}{\sum^{n - 1}} (k \cdot \frac{{(k + 1)}^{3} - k^{3}}{3} - k^{2} \cdot \frac{{(k + 1)}^{2} - k^{2}}{2}) =$ $\underset{k = 1}{\sum^{n - 1}} (\frac{3 k^{3} + 3 k^{2} + k}{3} - \frac{2 k^{3} + k^{2}}{2}) = \underset{k = 1}{\sum^{n - 1}} \frac{3 k^{2} + 2 k}{6} =$ $\frac{1}{6} (3 \cdot \frac{(n - 1) n (2 n - 1)}{6} + 2 \cdot \frac{(n - 1) n}{2}) = \frac{n (n - 1) (2 n + 1)}{12}$
	Commented by MathSh last updated on 18/Oct/21

	$Perfect dear Ser, thank you$
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