Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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...

$${I}\:=\sum_{{k}=\mathrm{2}} ^{\mathrm{999}} \:\int_{{k}} ^{{k}+\mathrm{1}} {kx}\:{dx}\:=\sum_{{k}=\mathrm{2}} ^{\mathrm{999}} {k}\:\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{{k}} ^{{k}+\mathrm{1}} \\ $$$$=\sum_{{k}=\mathrm{2}} ^{\mathrm{999}} \frac{{k}}{\mathrm{2}}\left\{\:\left({k}+\mathrm{1}\right)^{\mathrm{2}} −{k}^{\mathrm{2}} \right\} \\ $$$$=\sum_{{k}=\mathrm{2}} ^{\mathrm{999}} \:\frac{{k}}{\mathrm{2}}\left\{\:{k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{1}−{k}^{\mathrm{2}} \right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{2}} ^{\mathrm{999}} {k}\left\{\mathrm{2}{k}+\mathrm{1}\right\}\:=\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{2}} ^{\mathrm{999}} \left(\mathrm{2}{k}^{\mathrm{2}} \:+{k}\right) \\ $$$$=\sum_{{k}=\mathrm{2}} ^{\mathrm{999}} \:{k}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{2}} ^{\mathrm{999}} {k} \\ $$$$=\sum_{{k}=\mathrm{1}} ^{\mathrm{999}} \:{k}^{\mathrm{2}} \:−\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{1}} ^{\mathrm{999}} {k}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${we}\:{know}\:\sum_{{k}=\mathrm{1}} ^{{n}} {k}^{\mathrm{2}} \:=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}\:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{999}\left(\mathrm{1000}\right)\left(\mathrm{2}×\mathrm{999}\:+\mathrm{1}\right)}{\mathrm{6}}\:+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{999}\left(\mathrm{1000}\right)}{\mathrm{2}}\:−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${rest}\:{finishing}\:{the}\:{calculus}... \\ $$

Commented by Pratah last updated on 25/Jan/20

thanks

$$\mathrm{thanks} \\ $$

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

you are welcome.

$${you}\:{are}\:{welcome}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com