Question Number 217356 by SciMaths last updated on 11/Mar/25
Answered by profcedricjunior last updated on 11/Mar/25
![i=∫_1 ^2 ∫_y ^y^2 ∫_0 ^(ln(y+z)) e^x dxdydz =∫_1 ^2 ∫_y ^y^2 [e^x ]_0 ^(ln(y+z)) dydz =∫_1 ^2 ∫_y ^y^2 (y+z−1)dydz=∫_1 ^2 [yz+(z^2 /2)−z]_y ^y^2 dy =[(y^4 /4)+(y^4 /8)−(y^3 /3)−(y^3 /3)−(y^3 /6)+(y^2 /2)]_1 ^2 =[4+2−(8/3)−(8/3)−(8/6)+2−(1/4)−(1/8)+(1/3)+(1/3)+(1/6)−(1/2) =8−((20)/3)−(3/8)+(5/6)−(1/2)=8−((35)/6)−(7/8) =8−((280+42)/(48))=8−((322)/(48))=8−((161)/(24))](https://www.tinkutara.com/question/Q217357.png)
$$\boldsymbol{{i}}=\int_{\mathrm{1}} ^{\mathrm{2}} \int_{\boldsymbol{{y}}} ^{\boldsymbol{{y}}^{\mathrm{2}} } \int_{\mathrm{0}} ^{\boldsymbol{{ln}}\left(\boldsymbol{{y}}+\boldsymbol{{z}}\right)} \boldsymbol{{e}}^{\boldsymbol{{x}}} \boldsymbol{{dxdydz}} \\ $$$$\:\:=\int_{\mathrm{1}} ^{\mathrm{2}} \int_{\boldsymbol{{y}}} ^{\boldsymbol{{y}}^{\mathrm{2}} } \left[\boldsymbol{{e}}^{\boldsymbol{{x}}} \right]_{\mathrm{0}} ^{\boldsymbol{{ln}}\left(\boldsymbol{{y}}+\boldsymbol{{z}}\right)} \boldsymbol{{dydz}} \\ $$$$\:\:=\int_{\mathrm{1}} ^{\mathrm{2}} \int_{\boldsymbol{{y}}} ^{\boldsymbol{{y}}^{\mathrm{2}} } \left(\boldsymbol{{y}}+\boldsymbol{{z}}−\mathrm{1}\right)\boldsymbol{{dydz}}=\int_{\mathrm{1}} ^{\mathrm{2}} \left[\boldsymbol{{yz}}+\frac{\boldsymbol{{z}}^{\mathrm{2}} }{\mathrm{2}}−\boldsymbol{{z}}\right]_{\boldsymbol{{y}}} ^{\boldsymbol{{y}}^{\mathrm{2}} } \boldsymbol{{dy}} \\ $$$$\:\:=\left[\frac{\boldsymbol{{y}}^{\mathrm{4}} }{\mathrm{4}}+\frac{\boldsymbol{{y}}^{\mathrm{4}} }{\mathrm{8}}−\frac{\boldsymbol{{y}}^{\mathrm{3}} }{\mathrm{3}}−\frac{\boldsymbol{{y}}^{\mathrm{3}} }{\mathrm{3}}−\frac{\boldsymbol{{y}}^{\mathrm{3}} }{\mathrm{6}}+\frac{\boldsymbol{{y}}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{1}} ^{\mathrm{2}} \: \\ $$$$\:\:=\left[\mathrm{4}+\mathrm{2}−\frac{\mathrm{8}}{\mathrm{3}}−\frac{\mathrm{8}}{\mathrm{3}}−\frac{\mathrm{8}}{\mathrm{6}}+\mathrm{2}−\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{2}}\right. \\ $$$$\:\:\:=\mathrm{8}−\frac{\mathrm{20}}{\mathrm{3}}−\frac{\mathrm{3}}{\mathrm{8}}+\frac{\mathrm{5}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{8}−\frac{\mathrm{35}}{\mathrm{6}}−\frac{\mathrm{7}}{\mathrm{8}} \\ $$$$\:\:\:=\mathrm{8}−\frac{\mathrm{280}+\mathrm{42}}{\mathrm{48}}=\mathrm{8}−\frac{\mathrm{322}}{\mathrm{48}}=\mathrm{8}−\frac{\mathrm{161}}{\mathrm{24}} \\ $$