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Question Number 98621 by abony1303 last updated on 15/Jun/20

Commented by abony1303 last updated on 15/Jun/20

$$\mathrm{pls}\mathrm{help}$$

Answered by mr W last updated on 15/Jun/20

$$letx=n+twith0⩽t<1$$$${\int }_1^{10}x(x-\left[x\right])dx$$$$=\underset{n=1}{\sum ^9}{\int }_n^{n+1}x(x-\left[x\right])dx$$$$=\underset{n=1}{\sum ^9}{\int }_0^1(n+t)tdt$$$$=\underset{n=1}{\sum ^9}(\frac{n}{2}+\frac{1}{3})$$$$=\frac{1}{2}\times \frac{9\times 10}{2}+\frac{9}{3}$$$$=\frac{51}{2}$$

Answered by mathmax by abdo last updated on 15/Jun/20

$$\mathrm{I}={\int }_1^{10}\mathrm{x}(\mathrm{x}-\left[\mathrm{x}\right])\mathrm{dx}\Rightarrow \mathrm{I}={\int }_1^{10}{\mathrm{x}}^2\mathrm{dx}-{\int }_1^{10}\mathrm{x}\left[\mathrm{x}\right]\mathrm{dx}$$$${\int }_1^{10}{\mathrm{x}}^2\mathrm{dx}={\left[\frac{{\mathrm{x}}^3}{3}\right]}_1^{10}=\frac{1}{3}({10}^3-1)=\frac{999}{3}=333$$$${\int }_1^{10}\mathrm{x}\left[\mathrm{x}\right]=\sum _{\mathrm{k}=1}^9{\int }_{\mathrm{k}}^{\mathrm{k}+1}\mathrm{kx}\mathrm{dx}=\sum _{\mathrm{k}=1}^9\mathrm{k}(\frac{{(\mathrm{k}+1)}^2}{2}-\frac{{\mathrm{k}}^2}{2})$$$$=\frac{1}{2}\sum _{\mathrm{k}=1}^9\mathrm{k}(2\mathrm{k}+1)=\sum _{\mathrm{k}=1}^9{\mathrm{k}}^2+\frac{1}{2}\sum _{\mathrm{k}=1}^9\mathrm{k}$$$$=\frac{9(9+1)(2.9+1)}{6}+\frac{1}{2}\frac{9(9+1)}{2}=\frac{9\times 10\times 19}{6}+\frac{1}{4}\times 9\times 10$$$$=3\times 5\times 19+\frac{45}{2}=15\times 19+\frac{45}{2}=285+\frac{45}{2}\Rightarrow$$$$\mathrm{I}=333-285-\frac{45}{2}=48-\frac{45}{2}=\frac{96-45}{2}=\frac{51}{2}$$$$$$

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