Question Number 207509 by 073 last updated on 17/May/24
Answered by mr W last updated on 17/May/24
![∫_k ^(k+1) [x]x⌈x⌉dx =k(k+1)∫_k ^(k+1) xdx =k(k+1)(((k+1)^2 −k^2 )/2) =((k(k+1)(2k+1))/2) =((2k^3 +3k^2 +k)/2) ∫_0 ^5 [x]x⌈x⌉dx =Σ_(k=0) ^4 ∫_k ^(k+1) [x]x⌈x⌉dx =Σ_(k=0) ^4 ((2k^3 +3k^2 +k)/2) =((4^2 ×(4+1)^2 )/4)+((4×(4+1)×(2×4+1))/4)+((4×(4+1))/4) =150 ✓](https://www.tinkutara.com/question/Q207510.png)
$$\int_{{k}} ^{{k}+\mathrm{1}} \left[{x}\right]{x}\lceil{x}\rceil{dx} \\ $$$$={k}\left({k}+\mathrm{1}\right)\int_{{k}} ^{{k}+\mathrm{1}} {xdx} \\ $$$$={k}\left({k}+\mathrm{1}\right)\frac{\left({k}+\mathrm{1}\right)^{\mathrm{2}} −{k}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$=\frac{{k}\left({k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$=\frac{\mathrm{2}{k}^{\mathrm{3}} +\mathrm{3}{k}^{\mathrm{2}} +{k}}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{5}} \left[{x}\right]{x}\lceil{x}\rceil{dx} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\int_{{k}} ^{{k}+\mathrm{1}} \left[{x}\right]{x}\lceil{x}\rceil{dx} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\frac{\mathrm{2}{k}^{\mathrm{3}} +\mathrm{3}{k}^{\mathrm{2}} +{k}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{4}^{\mathrm{2}} ×\left(\mathrm{4}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{4}×\left(\mathrm{4}+\mathrm{1}\right)×\left(\mathrm{2}×\mathrm{4}+\mathrm{1}\right)}{\mathrm{4}}+\frac{\mathrm{4}×\left(\mathrm{4}+\mathrm{1}\right)}{\mathrm{4}} \\ $$$$=\mathrm{150}\:\checkmark \\ $$
Commented by mr W last updated on 21/May/24
![when k≤x<k+1: [x]=k what is wrong? if my answer is wrong, please show your right answer.](https://www.tinkutara.com/question/Q207627.png)
$${when}\:{k}\leqslant{x}<{k}+\mathrm{1}: \\ $$$$\left[{x}\right]={k} \\ $$$${what}\:{is}\:{wrong}? \\ $$$${if}\:{my}\:{answer}\:{is}\:{wrong},\:{please}\:{show} \\ $$$${your}\:{right}\:{answer}. \\ $$