Question Number 100074 by kungmikami last updated on 24/Jun/20
Answered by smridha last updated on 25/Jun/20
![∫_0 ^1 [∫_0 ^(√y) (x^2 y+xy^2 )dx]dy =∫_0 ^1 dy [(1/3)x^3 y+(1/2)x^2 y^2 ]_0 ^(√y) =∫_0 ^1 [(1/3)y^(1+(3/2)) +(1/2)y^3 ]dy =[(1/3).(y^(2+(3/2)) /(2+(3/2)))+(1/8)y^4 ]_0 ^1 =(2/(21))+(1/8)=((37)/(168))(unit)^2](https://www.tinkutara.com/question/Q100139.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \left[\int_{\mathrm{0}} ^{\sqrt{\boldsymbol{{y}}}} \left(\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{y}}+\boldsymbol{{xy}}^{\mathrm{2}} \right)\boldsymbol{{dx}}\right]\boldsymbol{{dy}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \boldsymbol{{dy}}\:\left[\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{x}}^{\mathrm{3}} \boldsymbol{{y}}+\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{y}}^{\mathrm{2}} \right]_{\mathrm{0}} ^{\sqrt{{y}}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left[\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{y}}^{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}} +\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{y}}^{\mathrm{3}} \right]\boldsymbol{{dy}} \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{3}}.\frac{\boldsymbol{{y}}^{\mathrm{2}+\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{2}+\frac{\mathrm{3}}{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{8}}\boldsymbol{{y}}^{\mathrm{4}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{21}}+\frac{\mathrm{1}}{\mathrm{8}}=\frac{\mathrm{37}}{\mathrm{168}}\left(\boldsymbol{{unit}}\right)^{\mathrm{2}} \\ $$