Question-79516

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

Commented by abdomathmax last updated on 25/Jan/20

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 \dots

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 .