Question Number 55237 by peter frank last updated on 19/Feb/19
$$\int_{\mathrm{0}} ^{\mathrm{3}} \int_{\mathrm{1}} ^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{2}} \right)\mathrm{dxdy} \\ $$
Commented by Abdo msup. last updated on 20/Feb/19
$${I}\:=\int_{\mathrm{0}} ^{\mathrm{3}} \:{A}\left({y}\right){dy}\:{with}\:{A}\left({y}\right)=\int_{\mathrm{1}} ^{\mathrm{2}} \left({x}^{\mathrm{3}} \:+{y}^{\mathrm{2}} \right){dx} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{2}} \:{x}^{\mathrm{3}} {dx}\:+{y}^{\mathrm{2}} \:\int_{\mathrm{1}} ^{\mathrm{2}} {dx}\:=\left[\frac{{x}^{\mathrm{4}} }{\mathrm{4}}\right]_{\mathrm{1}} ^{\mathrm{2}} \:+{y}^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\:\mathrm{2}^{\mathrm{4}} −\mathrm{1}\right)\:+{y}^{\mathrm{2}} \:=\frac{\mathrm{15}}{\mathrm{4}}\:+{y}^{\mathrm{2}} \:\Rightarrow \\ $$$${I}\:=\int_{\mathrm{0}} ^{\mathrm{3}} \:\left(\frac{\mathrm{15}}{\mathrm{4}}+{y}^{\mathrm{2}} \right){dy}\:=\frac{\mathrm{45}}{\mathrm{4}}\:+\left[\frac{\mathrm{1}}{\mathrm{3}}{y}^{\mathrm{3}} \right]_{\mathrm{0}} ^{\mathrm{3}} \\ $$$$=\frac{\mathrm{45}}{\mathrm{4}}\:+\mathrm{9}\:=\frac{\mathrm{45}+\mathrm{36}}{\mathrm{4}}\:=\frac{\mathrm{81}}{\mathrm{4}} \\ $$
Answered by kaivan.ahmadi last updated on 20/Feb/19
$$\left.\int_{\mathrm{0}} ^{\mathrm{3}} \left(\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+{y}^{\mathrm{2}} {x}\right]_{\mathrm{1}} ^{\mathrm{2}} \right){dy}=\int_{\mathrm{0}} ^{\mathrm{3}} \left(\mathrm{4}+\mathrm{2}{y}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}−{y}^{\mathrm{2}} \right){dy}= \\ $$$$\left.\int_{\mathrm{0}} ^{\mathrm{3}} \left({y}^{\mathrm{2}} +\frac{\mathrm{15}}{\mathrm{4}}\right){dy}=\frac{{y}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{15}}{\mathrm{4}}{y}\right]_{\mathrm{0}} ^{\mathrm{3}} =\mathrm{9}+\frac{\mathrm{45}}{\mathrm{4}}=\frac{\mathrm{81}}{\mathrm{4}} \\ $$$$ \\ $$