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...](https://www.tinkutara.com/question/Q79517.png)
$${I}\:=\sum_{{k}=\mathrm{2}} ^{\mathrm{999}} \:\int_{{k}} ^{{k}+\mathrm{1}} {kx}\:{dx}\:=\sum_{{k}=\mathrm{2}} ^{\mathrm{999}} {k}\:\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{{k}} ^{{k}+\mathrm{1}} \\ $$$$=\sum_{{k}=\mathrm{2}} ^{\mathrm{999}} \frac{{k}}{\mathrm{2}}\left\{\:\left({k}+\mathrm{1}\right)^{\mathrm{2}} −{k}^{\mathrm{2}} \right\} \\ $$$$=\sum_{{k}=\mathrm{2}} ^{\mathrm{999}} \:\frac{{k}}{\mathrm{2}}\left\{\:{k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{1}−{k}^{\mathrm{2}} \right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{2}} ^{\mathrm{999}} {k}\left\{\mathrm{2}{k}+\mathrm{1}\right\}\:=\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{2}} ^{\mathrm{999}} \left(\mathrm{2}{k}^{\mathrm{2}} \:+{k}\right) \\ $$$$=\sum_{{k}=\mathrm{2}} ^{\mathrm{999}} \:{k}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{2}} ^{\mathrm{999}} {k} \\ $$$$=\sum_{{k}=\mathrm{1}} ^{\mathrm{999}} \:{k}^{\mathrm{2}} \:−\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{1}} ^{\mathrm{999}} {k}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${we}\:{know}\:\sum_{{k}=\mathrm{1}} ^{{n}} {k}^{\mathrm{2}} \:=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}\:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{999}\left(\mathrm{1000}\right)\left(\mathrm{2}×\mathrm{999}\:+\mathrm{1}\right)}{\mathrm{6}}\:+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{999}\left(\mathrm{1000}\right)}{\mathrm{2}}\:−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${rest}\:{finishing}\:{the}\:{calculus}… \\ $$
Commented by Pratah last updated on 25/Jan/20
$$\mathrm{thanks} \\ $$
Commented by mathmax by abdo last updated on 25/Jan/20
$${you}\:{are}\:{welcome}. \\ $$