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 214917 by Spillover last updated on 23/Dec/24

Answered by cemoosky last updated on 23/Dec/24

$${\int }_0^1{\int }_0^1\cdot \cdot \cdot {\int }_0^1(x_1^3+x_2^3+\cdot \cdot \cdot +x_n^3){dx}_1{dx}_2\cdot \cdot \cdot {dx}_n$$$$forn=2018$$$$n{\int }_0^1{\int }_0^1\cdot \cdot \cdot {\int }_0^1\left(x_n^3\right){dx}_1{dx}_2\cdot \cdot \cdot {dx}_n=n{\int }_0^1\left(x_n^3\right){dx}_n$$$$=n{\left[\frac{x^4}{4}\right]}_0^1=n[\frac{{\left(1\right)}^4}{4}-\frac{{\left(0\right)}^4}{4}]=n\left[\frac{1}{4}\right]$$$$=2018\left[\frac{1}{4}\right]=\frac{2018}{4}=504.5$$

Commented by Spillover last updated on 24/Dec/24

$$great$$

Answered by Spillover last updated on 24/Dec/24

Terms of Service

Privacy Policy

Contact: info@tinkutara.com