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Question Number 55237 by peter frank last updated on 19/Feb/19

$${\int }_0^3{\int }_1^2({\mathrm{x}}^3+{\mathrm{y}}^2)\mathrm{dxdy}$$

Commented by Abdo msup. last updated on 20/Feb/19

$$I={\int }_0^{3}A\left(y\right)dywithA\left(y\right)={\int }_1^2(x^3+y^2)dx$$$$={\int }_1^2{x}^{3}dx+y^2{\int }_1^{2}dx={\left[\frac{x^4}{4}\right]}_1^2+y^2$$$$=\frac{1}{4}(2^4-1)+y^2=\frac{15}{4}+y^2\Rightarrow$$$$I={\int }_0^3(\frac{15}{4}+y^2)dy=\frac{45}{4}+{\left[\frac{1}{3}{y}^3\right]}_0^3$$$$=\frac{45}{4}+9=\frac{45+36}{4}=\frac{81}{4}$$

Answered by kaivan.ahmadi last updated on 20/Feb/19

$${\int }_0^3{(\frac{x^4}{4}+y^{2}x]}_1^2)dy={\int }_0^3(4+2y^2-\frac{1}{4}-y^2)dy=$$$${{\int }_0^3(y^2+\frac{15}{4})dy=\frac{y^3}{3}+\frac{15}{4}y]}_0^3=9+\frac{45}{4}=\frac{81}{4}$$$$$$

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