Question Number 100362 by Dara last updated on 26/Jun/20
$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {e}^{\mathrm{2}{x}+{y}} {dydx} \\ $$
Answered by smridha last updated on 26/Jun/20
![∫_0 ^1 dx.[e^(2x+y) ]_0 ^1 =∫_0 ^1 [e^(2x+1) −e^(2x) ]dx =(e−1)[(e^(2x) /2)]_(0 ) ^1 =(1/2)(e−1)(e^2 −1)=(1/2)(e^3 −e^2 −e+1)](https://www.tinkutara.com/question/Q100364.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \boldsymbol{{dx}}.\left[\boldsymbol{{e}}^{\mathrm{2}\boldsymbol{{x}}+\boldsymbol{{y}}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left[\boldsymbol{{e}}^{\mathrm{2}\boldsymbol{{x}}+\mathrm{1}} −\boldsymbol{{e}}^{\mathrm{2}\boldsymbol{{x}}} \right]\boldsymbol{{dx}} \\ $$$$=\left(\boldsymbol{{e}}−\mathrm{1}\right)\left[\frac{\boldsymbol{{e}}^{\mathrm{2}\boldsymbol{{x}}} }{\mathrm{2}}\right]_{\mathrm{0}\:} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\boldsymbol{{e}}−\mathrm{1}\right)\left(\boldsymbol{{e}}^{\mathrm{2}} −\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\boldsymbol{{e}}^{\mathrm{3}} −\boldsymbol{{e}}^{\mathrm{2}} −\boldsymbol{{e}}+\mathrm{1}\right) \\ $$
Answered by Ar Brandon last updated on 26/Jun/20
![I=∫_0 ^1 ∫_0 ^1 e^(2x+y) dxdy=[∫_0 ^1 e^(2x) dx][∫_0 ^1 e^y dy]=[(e^(2x) /2)]_0 ^1 ∙[(e^y /1)]_0 ^1 ⇒I=(((e^2 −1)/2))(((e−1)/1))=(1/2)(e+1)(e−1)^2](https://www.tinkutara.com/question/Q100367.png)
$$\mathcal{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{e}^{\mathrm{2x}+\mathrm{y}} \mathrm{dxdy}=\left[\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{e}^{\mathrm{2x}} \mathrm{dx}\right]\left[\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{e}^{\mathrm{y}} \mathrm{dy}\right]=\left[\frac{\mathrm{e}^{\mathrm{2x}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} \centerdot\left[\frac{\mathrm{e}^{\mathrm{y}} }{\mathrm{1}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\Rightarrow\mathcal{I}=\left(\frac{\mathrm{e}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\mathrm{e}−\mathrm{1}}{\mathrm{1}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{e}+\mathrm{1}\right)\left(\mathrm{e}−\mathrm{1}\right)^{\mathrm{2}} \\ $$
Answered by mathmax by abdo last updated on 26/Jun/20
![I =∫_0 ^1 ∫_0 ^1 e^(2x+y) dy dx ⇒ I =∫_0 ^1 e^(2x) dx.∫_0 ^1 e^y dy =[(1/2)e^(2x) ]_0 ^1 .[e^y ]_0 ^1 =(1/2)(e^2 −1)(e−1) =(1/2)(e^3 −e^2 −e+1)](https://www.tinkutara.com/question/Q100392.png)
$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{e}^{\mathrm{2x}+\mathrm{y}} \:\mathrm{dy}\:\mathrm{dx}\:\Rightarrow\:\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{e}^{\mathrm{2x}} \:\mathrm{dx}.\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{e}^{\mathrm{y}} \:\mathrm{dy} \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e}^{\mathrm{2x}} \right]_{\mathrm{0}} ^{\mathrm{1}} .\left[\mathrm{e}^{\mathrm{y}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{e}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{e}−\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{e}^{\mathrm{3}} −\mathrm{e}^{\mathrm{2}} −\mathrm{e}+\mathrm{1}\right) \\ $$