Question Number 67946 by Cmr 237 last updated on 02/Sep/19
$$\mathrm{use}\:\boldsymbol{\mathrm{Green}}−\boldsymbol{\mathrm{Riemann}}\:\boldsymbol{\mathrm{formuler}} \\ $$$$\mathrm{to}\:\mathrm{determined}: \\ $$$$\boldsymbol{\mathrm{I}}=\int\int_{\boldsymbol{\mathrm{D}}} \boldsymbol{\mathrm{xy}}\mathrm{dxdy} \\ $$$$\boldsymbol{\mathrm{D}}=\left\{\left(\mathrm{x},\mathrm{y}\right)\in\mathbb{R}^{\mathrm{2}} \mid\mathrm{x}\geqslant\mathrm{0};\mathrm{y}\geqslant;\mathrm{x}+{y}\leqslant\mathrm{1}\right\} \\ $$
Commented by mathmax by abdo last updated on 02/Sep/19
![I =∫_0 ^1 (∫_0 ^(1−y) xydx)dy =∫_0 ^1 (∫_0 ^(1−y) xdx)ydy =∫_0 ^1 ((((1−y)^2 )/2))ydy =(1/2) ∫_0 ^1 y(y^2 −2y+1)dy =(1/2)∫_0 ^1 (y^3 −2y^2 +y)dy =(1/2)[(y^4 /4)−(2/3)y^3 +(y^2 /2)]_0 ^1 =(1/2){(1/4)−(2/3)+(1/2)} =(1/2){(3/4)−(2/3)} =(3/8)−(1/3) =(1/(24)) .](https://www.tinkutara.com/question/Q67976.png)
$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\int_{\mathrm{0}} ^{\mathrm{1}−{y}} {xydx}\right){dy}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\int_{\mathrm{0}} ^{\mathrm{1}−{y}} {xdx}\right){ydy} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\left(\mathrm{1}−{y}\right)^{\mathrm{2}} }{\mathrm{2}}\right){ydy}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} {y}\left({y}^{\mathrm{2}} −\mathrm{2}{y}+\mathrm{1}\right){dy} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \left({y}^{\mathrm{3}} −\mathrm{2}{y}^{\mathrm{2}} \:+{y}\right){dy}\:=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{y}^{\mathrm{4}} }{\mathrm{4}}−\frac{\mathrm{2}}{\mathrm{3}}{y}^{\mathrm{3}} \:+\frac{{y}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}\right\}\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{3}}{\mathrm{4}}−\frac{\mathrm{2}}{\mathrm{3}}\right\}\:=\frac{\mathrm{3}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{3}}\:=\frac{\mathrm{1}}{\mathrm{24}}\:. \\ $$
Commented by Cmr 237 last updated on 05/Sep/19
$$\mathrm{thk}\:\mathrm{sir} \\ $$
Commented by mathmax by abdo last updated on 10/Sep/19
$${you}\:{are}\:{welcome}. \\ $$