![](https://iknow-base.cdn.bcebos.com/lxb/notice.png)
微积分求解答
展开全部
∫D∫e^[-(x^2+y^2)]dσ
=4∫(0,π/2)dθ∫(2,3)e^(-r^2)rdr
=-2∫(0,π/2)dθ∫(2,3)e^(-r^2)d(-r^2)
=-2∫(0,π/2)[e^(-r^2)](2,3)dθ
=-2[e^(-3^2)-e^(-2^2)]∫(0,π/2)dθ
=-2(1/e^9-1/e^4)[θ](0,π/2)
=2(e^9-e^4)/e^13(π/2-0)
=π(e^9-e^4)/e^13
=4∫(0,π/2)dθ∫(2,3)e^(-r^2)rdr
=-2∫(0,π/2)dθ∫(2,3)e^(-r^2)d(-r^2)
=-2∫(0,π/2)[e^(-r^2)](2,3)dθ
=-2[e^(-3^2)-e^(-2^2)]∫(0,π/2)dθ
=-2(1/e^9-1/e^4)[θ](0,π/2)
=2(e^9-e^4)/e^13(π/2-0)
=π(e^9-e^4)/e^13
![](https://ecmc.bdimg.com/public03/b4cb859ca634443212c22993b0c87088.png)
2024-07-12 广告
当您光临我们的门店或参与活动时,只需轻松打开微信,对准我们提供的二维码一扫,即可参与激动人心的抽奖环节。奖品丰富多样,从精美小礼品到超值优惠券,应有尽有。我们致力于为您带来便捷、有趣的互动体验,让每一次扫码都成为一次惊喜的开启。感谢您的支持...
点击进入详情页
本回答由哎呦互动提供
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询