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Question Number 217356 by SciMaths last updated on 11/Mar/25

Answered by profcedricjunior last updated on 11/Mar/25

$$\mathit{i}={\int }_1^2{\int }_{\mathit{y}}^{{\mathit{y}}^2}{\int }_0^{\mathit{l}\mathit{n}(\mathit{y}+\mathit{z})}{\mathit{e}}^{\mathit{x}}\mathit{d}\mathit{x}\mathit{d}\mathit{y}\mathit{d}\mathit{z}$$$$={\int }_1^2{\int }_{\mathit{y}}^{{\mathit{y}}^2}{\left[{\mathit{e}}^{\mathit{x}}\right]}_0^{\mathit{l}\mathit{n}(\mathit{y}+\mathit{z})}\mathit{d}\mathit{y}\mathit{d}\mathit{z}$$$$={\int }_1^2{\int }_{\mathit{y}}^{{\mathit{y}}^2}(\mathit{y}+\mathit{z}-1)\mathit{d}\mathit{y}\mathit{d}\mathit{z}={\int }_1^2{[\mathit{y}\mathit{z}+\frac{{\mathit{z}}^2}{2}-\mathit{z}]}_{\mathit{y}}^{{\mathit{y}}^2}\mathit{d}\mathit{y}$$$$={[\frac{{\mathit{y}}^4}{4}+\frac{{\mathit{y}}^4}{8}-\frac{{\mathit{y}}^3}{3}-\frac{{\mathit{y}}^3}{3}-\frac{{\mathit{y}}^3}{6}+\frac{{\mathit{y}}^2}{2}]}_1^2$$$$=[4+2-\frac{8}{3}-\frac{8}{3}-\frac{8}{6}+2-\frac{1}{4}-\frac{1}{8}+\frac{1}{3}+\frac{1}{3}+\frac{1}{6}-\frac{1}{2}$$$$=8-\frac{20}{3}-\frac{3}{8}+\frac{5}{6}-\frac{1}{2}=8-\frac{35}{6}-\frac{7}{8}$$$$=8-\frac{280+42}{48}=8-\frac{322}{48}=8-\frac{161}{24}$$

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