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Question Number 207509 by 073 last updated on 17/May/24

Answered by mr W last updated on 17/May/24

$${\int }_k^{k+1}\left[x\right]x\lceil x\rceil dx$$$$=k(k+1){\int }_k^{k+1}xdx$$$$=k(k+1)\frac{{(k+1)}^2-k^2}{2}$$$$=\frac{k(k+1)(2k+1)}{2}$$$$=\frac{2k^3+3k^2+k}{2}$$$${\int }_0^5\left[x\right]x\lceil x\rceil dx$$$$=\underset{k=0}{\sum ^4}{\int }_k^{k+1}\left[x\right]x\lceil x\rceil dx$$$$=\underset{k=0}{\sum ^4}\frac{2k^3+3k^2+k}{2}$$$$=\frac{4^2\times {(4+1)}^2}{4}+\frac{4\times (4+1)\times (2\times 4+1)}{4}+\frac{4\times (4+1)}{4}$$$$=150✓$$

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