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 190487 by stvnmaxi last updated on 03/Apr/23

Answered by aleks041103 last updated on 04/Apr/23

$$A={\int }_0^1{\int }_0^1{x}^2{y}^{2}dxdy=\left({\int }_0^1{x}^{2}dx\right)\left({\int }_0^1{y}^{2}dy\right)=$$$$=\frac{1}{3}\frac{1}{3}=\frac{1}{9}=A$$$$B={\int }_0^1{\int }_0^1(x^2+y^2)dxdy=$$$$={\int }_0^1{\int }_0^1{x}^{2}dxdy+{\int }_0^1{\int }_0^1{y}^{2}dxdy=$$$$=\left({\int }_0^{1}dy\right)\left({\int }_0^1{x}^{2}dx\right)+\left({\int }_0^{1}dx\right)\left({\int }_0^1{y}^{2}dy\right)=$$$$=1\times \frac{1}{3}+1\times \frac{1}{3}=\frac{2}{3}=B$$$$C={\int }_1^{\sqrt{3}}{\int }_0^1\frac{dxdy}{x^2+y^2}=$$$$={\int }_1^{\sqrt{3}}\left({\int }_0^1\frac{dx}{x^2+y^2}\right)dy$$$${\int }_0^1\frac{dx}{x^2+y^2}=\frac{1}{y}{\int }_0^{1/y}\frac{d\left(x/y\right)}{{\left(x/y\right)}^2+1}=\frac{1}{y}arctan\left(\frac{1}{y^2}\right)$$$$\Rightarrow C={\int }_1^{\sqrt{3}}arctan\left(\frac{1}{y^2}\right)\frac{dy}{y}=$$$$=\frac{1}{2}{\int }_1^{\sqrt{3}}arctan\left(1/y^2\right)\frac{d\left(y^2\right)}{y^2}=$$$$=\frac{1}{2}{\int }_1^3\frac{arctan\left(1/z\right)}{z}dz$$$$thisintegralisnotsolvableinelementary$$$$functions$$

Commented by stvnmaxi last updated on 05/Apr/23

$$\mathrm{there}\mathrm{is}\mathrm{not}\mathrm{another}\mathrm{way}\mathrm{to}\mathrm{calculer}\mathrm{the}\mathrm{last}\mathrm{question}\left(\mathrm{C}\right)$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com