Question Number 197500 by yaslm last updated on 19/Sep/23
Commented by SANOGO last updated on 19/Sep/23
$${meeci}\:{bien} \\ $$
Answered by Mathspace last updated on 20/Sep/23
![∫_0 ^1 ∫_(−1) ^0 e^(2x−3y+1) dxdy =∫_0 ^1 (∫_(−1) ^0 e^(−3y+1) dy)e^(2x) dx =e∫_0 ^1 ([−(1/3)e^(−3y) ]_(−1) ^o )e^(2x) dx =−(e/3)∫_0 ^1 (1−e^3 )e^(2x) dx =((e^4 −e)/3)∫_0 ^1 e^(2x) dx =((e^4 −e)/3)[(1/2)e^(2x) ]_0 ^1 =((e^4 −e)/6)(e^2 −1)](https://www.tinkutara.com/question/Q197532.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{−\mathrm{1}} ^{\mathrm{0}} \:{e}^{\mathrm{2}{x}−\mathrm{3}{y}+\mathrm{1}} {dxdy} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\int_{−\mathrm{1}} ^{\mathrm{0}} {e}^{−\mathrm{3}{y}+\mathrm{1}} {dy}\right){e}^{\mathrm{2}{x}} {dx} \\ $$$$={e}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\left[−\frac{\mathrm{1}}{\mathrm{3}}{e}^{−\mathrm{3}{y}} \right]_{−\mathrm{1}} ^{{o}} \right){e}^{\mathrm{2}{x}} {dx} \\ $$$$=−\frac{{e}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{e}^{\mathrm{3}} \right){e}^{\mathrm{2}{x}} {dx} \\ $$$$=\frac{{e}^{\mathrm{4}} −{e}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{\mathrm{2}{x}} {dx} \\ $$$$=\frac{{e}^{\mathrm{4}} −{e}}{\mathrm{3}}\left[\frac{\mathrm{1}}{\mathrm{2}}{e}^{\mathrm{2}{x}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{{e}^{\mathrm{4}} −{e}}{\mathrm{6}}\left({e}^{\mathrm{2}} −\mathrm{1}\right) \\ $$