作业帮反常积分问题计算-4∫(0 +∞)(y^2)×[e^(-4y)-e^(-2y)]dy=8∫(0 +∞)y[(-1/4)(e^- 4y)+(1/2)(e^-2y)]dy =-8∫(0 +∞)[(1/16)(e^-4y)-(1/4)(e^-2y)]dy=7/8请问这两步是怎么得到的?
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/29 23:44:26
![反常积分问题计算-4∫(0 +∞)(y^2)×[e^(-4y)-e^(-2y)]dy=8∫(0 +∞)y[(-1/4)(e^- 4y)+(1/2)(e^-2y)]dy =-8∫(0 +∞)[(1/16)(e^-4y)-(1/4)(e^-2y)]dy=7/8请问这两步是怎么得到的?](/uploads/image/z/13339160-8-0.jpg?t=%E5%8F%8D%E5%B8%B8%E7%A7%AF%E5%88%86%E9%97%AE%E9%A2%98%E8%AE%A1%E7%AE%97-4%E2%88%AB%280+%2B%E2%88%9E%29%28y%5E2%29%C3%97%5Be%5E%28-4y%29-e%5E%28-2y%29%5Ddy%3D8%E2%88%AB%280+%2B%E2%88%9E%29y%5B%28-1%2F4%29%28e%5E-+4y%29%2B%281%2F2%29%28e%5E-2y%29%5Ddy+%3D-8%E2%88%AB%280+%2B%E2%88%9E%29%5B%281%2F16%29%28e%5E-4y%29-%281%2F4%29%28e%5E-2y%29%5Ddy%3D7%2F8%E8%AF%B7%E9%97%AE%E8%BF%99%E4%B8%A4%E6%AD%A5%E6%98%AF%E6%80%8E%E4%B9%88%E5%BE%97%E5%88%B0%E7%9A%84%3F)
反常积分问题计算-4∫(0 +∞)(y^2)×[e^(-4y)-e^(-2y)]dy=8∫(0 +∞)y[(-1/4)(e^- 4y)+(1/2)(e^-2y)]dy =-8∫(0 +∞)[(1/16)(e^-4y)-(1/4)(e^-2y)]dy=7/8请问这两步是怎么得到的?
反常积分问题计算
-4∫(0 +∞)(y^2)×[e^(-4y)-e^(-2y)]dy=8∫(0 +∞)y[(-1/4)(e^- 4y)+(1/2)(e^-2y)]dy =-8∫(0 +∞)[(1/16)(e^-4y)-(1/4)(e^-2y)]dy=7/8请问这两步是怎么得到的?
反常积分问题计算-4∫(0 +∞)(y^2)×[e^(-4y)-e^(-2y)]dy=8∫(0 +∞)y[(-1/4)(e^- 4y)+(1/2)(e^-2y)]dy =-8∫(0 +∞)[(1/16)(e^-4y)-(1/4)(e^-2y)]dy=7/8请问这两步是怎么得到的?
用的是分部积分
-4∫[0→+∞] y²[e^(-4y)-e^(-2y)] dy
=-4∫[0→+∞] y²e^(-4y) dy + 4∫[0→+∞] y²e^(-2y) dy
=∫[0→+∞] y² de^(-4y) - 2∫[0→+∞] y² de^(-2y)
=y²e^(-4y) - 2∫[0→+∞] ye^(-4y) dy - 2y²e^(-2y) + 4∫[0→+∞] ye^(-2y)dy |[0→+∞]
=-2∫[0→+∞] ye^(-4y) dy + 4∫[0→+∞] ye^(-2y)dy
=(1/2)∫[0→+∞] y de^(-4y) - 2∫[0→+∞] yde^(-2y)
=(1/2)ye^(-4y) - (1/2)∫[0→+∞] e^(-4y) dy - 2ye^(-2y) + 2∫[0→+∞] e^(-2y)dy |[0→+∞]
=-(1/2)∫[0→+∞] e^(-4y) dy + 2∫[0→+∞] e^(-2y)dy
=(1/8)e^(-4y) - e^(-2y) |[0→+∞]
=-1/8 + 1
=7/8
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