Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 155619 by mathdanisur last updated on 02/Oct/21

$$\mathbf{\Omega }=\underset{0}{\stackrel{1}{\int }}\frac{{\mathrm{x}}^{49}}{1+\mathrm{x}+{\mathrm{x}}^2+{\mathrm{x}}^3...{\mathrm{x}}^{100}}\mathrm{dx}=?$$

Answered by mindispower last updated on 04/Oct/21

$$={\int }_0^1\frac{x^{49}(1-x)}{1-x^{50}}dx$$$$u=x^{50}\Rightarrow x^{49}dx=\frac{du}{50}$$$$={\int }_0^1\frac{(1-u^{\frac{1}{50}})}{1-u}\frac{du}{50}$$$$=\frac{1}{50}{\int }_0^1\frac{1-u^{\frac{1}{50}}}{1-u}du$$$$\mathrm{\Psi }(1+x)=-\gamma +{\int }_0^1\frac{1-t^x}{1-t}dx$$$$\mathrm{\Omega }=\frac{1}{50}(\mathrm{\Psi }\left(\frac{51}{50}\right)+\gamma ).$$

Commented by puissant last updated on 04/Oct/21

$$\int \frac{x^{49}}{1+x+...+x^{100}}dx=\int \frac{x^{49}(1-x)}{(1-x^{101})}dx\ne \int \frac{x^{49}(1-x)}{(1-x^{50})}dx$$

Commented by mindispower last updated on 04/Oct/21

$$changedquationitwas1+x...+x^{49}$$

Commented by puissant last updated on 04/Oct/21

����

Terms of Service

Privacy Policy

Contact: info@tinkutara.com