几道关于定积分的问题
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/07/31 04:07:40
几道关于定积分的问题
![](http://img.wesiedu.com/upload/d/7f/d7f1fd7847e50f1ff56e2bdbc0dffe63.jpg)
![](http://img.wesiedu.com/upload/d/7f/d7f1fd7847e50f1ff56e2bdbc0dffe63.jpg)
![几道关于定积分的问题](/uploads/image/z/7163037-45-7.jpg?t=%E5%87%A0%E9%81%93%E5%85%B3%E4%BA%8E%E5%AE%9A%E7%A7%AF%E5%88%86%E7%9A%84%E9%97%AE%E9%A2%98)
∫(1-e^-2x)^(1/2)dx
=∫(t^2-1)^(1/2)/t^2dt (t=e^x)
(t^2-1)^(1/2)/t^2=(t^2-1)^(-1/2)-1/t^2(t^2-1)^(1/2)
∫(t^2-1)^(1/2)/t^2dt
=In|t+(t^2-1)^(1/2)|-∫dt/t^2(t^2-1)^(1/2)
=In|t+(t^2-1)^(1/2)|+sgn(u)∫udu/(1-u^2)^(1/2) (u=1/t)
=In|t+(t^2-1)^(1/2)|-(t^2-1)^(1/2)/t
∴原定积分下上限为1,2
原定积分=In(2+3^(1/2))-3^(1/2)/2
=∫(t^2-1)^(1/2)/t^2dt (t=e^x)
(t^2-1)^(1/2)/t^2=(t^2-1)^(-1/2)-1/t^2(t^2-1)^(1/2)
∫(t^2-1)^(1/2)/t^2dt
=In|t+(t^2-1)^(1/2)|-∫dt/t^2(t^2-1)^(1/2)
=In|t+(t^2-1)^(1/2)|+sgn(u)∫udu/(1-u^2)^(1/2) (u=1/t)
=In|t+(t^2-1)^(1/2)|-(t^2-1)^(1/2)/t
∴原定积分下上限为1,2
原定积分=In(2+3^(1/2))-3^(1/2)/2