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 82871 by abdomathmax last updated on 25/Feb/20

$$find\int {\int }_{{[0,1]}^2}\mid x^2-y^2\mid dxdy$$

Commented by mathmax by abdo last updated on 25/Feb/20

$$I=\int {\int }_{{[0,1]}^2}\mid x+y\mid \mid x-y\mid dxdy=\int {\int }_{{[0,1]}^2}(x+y)\mid x-y\mid dxdy$$$$={\int }_0^1({\int }_0^x(x+y)\mid x-y\mid dy+{\int }_x^1(x+y)\mid x-y\mid dy)dx$$$$={\int }_0^1({\int }_0^x(x+y)(x-y)dy+{\int }_x^1(x+y)(y-x)dy)dx$$$$={\int }_0^1\left({\int }_0^x(x+y)(x-y)dy\right)dx+{\int }_0^1\left({\int }_x^1(y+x)(y-x)dy\right)dx$$$$wehave{\int }_0^x(x+y)(x-y)dy={\int }_0^x(x^2-y^2)dy$$$$=x^3-{\left[\frac{1}{3}{y}^3\right]}_0^x=x^3-\frac{x^3}{3}=\frac{2}{3}{x}^3\Rightarrow {\int }_0^1(....dy)dx=\frac{2}{3}{\int }_0^1{x}^{3}dx$$$$=\frac{2}{3}\times \frac{1}{4}=\frac{1}{6}$$$${\int }_x^1(y+x)(y-x)dy={\int }_x^1(y^2-x^2)dy={\left[\frac{y^3}{3}\right]}_x^1-x^2(1-x)$$$$=\frac{1}{3}-\frac{x^3}{3}-x^2+x^3=\frac{1}{3}-x^2+\frac{2}{3}{x}^3\Rightarrow$$$${\int }_0^1(....dy)dx={\int }_0^1(\frac{1}{3}-x^2+\frac{2}{3}{x}^3)dx={[\frac{x}{3}-\frac{1}{3}{x}^3+\frac{2}{12}{x}^4]}_0^1$$$$=\frac{1}{6}\Rightarrow I=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com