> 数学 >
求∫∫yx²e^xy ( 0≤x≤1 0≤y≤3)
人气:249 ℃ 时间:2020-05-08 20:48:13
解答
∫[0,1]∫[0,3]yx²e^xydxdy
=∫[0,1]x^2e^xdx*∫[0,3]ye^ydy
=[∫[0,1]x^2de^x]*[∫[0,3]yde^y]
=[x^2e^x[0,1]-∫[0,1]2xde^x]*[ye^y[0,3]-∫[0,3]de^y]
=[e-2xe^x[0,1]+2∫[0,1]de^x]*[3e^3-e^y[0,3]]
=[e-2e+2e^x[0,1]]*[3e^3-e^3]
=e*2e^3
=2e^4不对吧~e^xy不等于e^x*e^y吧,e^x*e^y等于e^(x+y)噢 ∫[0,1]∫[0,3]yx²e^xydxdy =∫[0,1]∫[0,3]yxe^xydxydx =∫[0,1]∫[0,3]yxde^xydx =∫[0,1][xye^(xy)[0,3]-∫[0,3]e^xyd(xy)]dx =∫[0,1][3xe^(3x)-∫[0,3]e^xyd(xy)]dx =∫[0,1][3xe^(3x)-e^(xy)[0,3]]dx =∫[0,1][3xe^(3x)-e^(3x)+1]dx =∫[0,1]3xe^(3x)dx-∫[0,1]e^(3x)dx+∫[0,1]dx =xe^(3x)[0,1]-∫[0,1]e^(3x)dx-∫[0,1]e^(3x)dx+1 =e^3+1-2∫[0,1]e^(3x)dx =e^3+1-2/3e^(3x)[0,1] =e^3+1-2/3e^3+2/3 =5/3+e^3/3
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版