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Question Number 79516 by Pratah last updated on 25/Jan/20

Commented by abdomathmax last updated on 25/Jan/20
$I =Σ_(k=2) ^(999) ∫_k ^(k+1) kx dx =Σ_(k=2) ^(999) k [(x^2 /2)]_k ^(k+1) =Σ_(k=2) ^(999) (k/2){ (k+1)^2 −k^2 } =Σ_(k=2) ^(999) (k/2){ k^2 +2k+1−k^2 } =(1/2)Σ_(k=2) ^(999) k{2k+1} =(1/2)Σ_(k=2) ^(999) (2k^2 +k) =Σ_(k=2) ^(999) k^2 +(1/2)Σ_(k=2) ^(999) k =Σ_(k=1) ^(999) k^2 −1 +(1/2)Σ_(k=1) ^(999) k−(1/2) we know Σ_(k=1) ^n k^2 =((n(n+1)(2n+1))/6) ⇒ I =((999(1000)(2×999 +1))/6) +(1/2)×((999(1000))/2) −(3/2) rest finishing the calculus...$
$I = \sum_{k = 2}^{999} \int_{k}^{k + 1} k x d x = \sum_{k = 2}^{999} k {[\frac{x^{2}}{2}]}_{k}^{k + 1}$ $= \sum_{k = 2}^{999} \frac{k}{2} {{(k + 1)}^{2} - k^{2}}$ $= \sum_{k = 2}^{999} \frac{k}{2} {k^{2} + 2 k + 1 - k^{2}}$ $= \frac{1}{2} \sum_{k = 2}^{999} k {2 k + 1} = \frac{1}{2} \sum_{k = 2}^{999} (2 k^{2} + k)$ $= \sum_{k = 2}^{999} k^{2} + \frac{1}{2} \sum_{k = 2}^{999} k$ $= \sum_{k = 1}^{999} k^{2} - 1 + \frac{1}{2} \sum_{k = 1}^{999} k - \frac{1}{2}$ $w e k n o w \sum_{k = 1}^{n} k^{2} = \frac{n (n + 1) (2 n + 1)}{6} \Rightarrow$ $I = \frac{999 (1000) (2 \times 999 + 1)}{6} + \frac{1}{2} \times \frac{999 (1000)}{2} - \frac{3}{2}$ $r e s t f i n i s h i n g t h e c a l c u l u s . . .$
Commented by Pratah last updated on 25/Jan/20

$thanks$
Commented by mathmax by abdo last updated on 25/Jan/20

$y o u a r e w e l c o m e .$
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