求0到1e^(√x+1)dx的定积分
来源:学生作业帮 编辑:百度作业网作业帮 分类:数学作业 时间:2024/07/31 20:42:01
求0到1e^(√x+1)dx的定积分
![求0到1e^(√x+1)dx的定积分](/uploads/image/z/310234-58-4.jpg?t=%E6%B1%820%E5%88%B01e%5E%28%E2%88%9Ax%2B1%29dx%E7%9A%84%E5%AE%9A%E7%A7%AF%E5%88%86)
∫ (0->1) e^(√x+1) dx
let
y =√x+1
dy = dx/(2√x)
dx = 2(y-1) dy
x=0,y=1
x=1 ,y=2
∫ (0->1) e^(√x+1) dx
= 2∫ (1->2) (y-1)e^y dy
=2∫ (1->2) y d(e^y) - 2∫ (1->2) e^y dy
=2 [ye^y](1->2) - 4∫ (1->2) e^y dy
= 2(2e^2 - e) - 4(e^2 - e)
=2e
再问:
��ô�ʹ�2�����4
再答: 2�� (1->2) y d(e^y) - 2�� (1->2) e^y dy =2 [ye^y](1->2) - 2�� (1->2) e^y dy - 2�� (1->2) e^y dy
let
y =√x+1
dy = dx/(2√x)
dx = 2(y-1) dy
x=0,y=1
x=1 ,y=2
∫ (0->1) e^(√x+1) dx
= 2∫ (1->2) (y-1)e^y dy
=2∫ (1->2) y d(e^y) - 2∫ (1->2) e^y dy
=2 [ye^y](1->2) - 4∫ (1->2) e^y dy
= 2(2e^2 - e) - 4(e^2 - e)
=2e
再问:
![](http://img.wesiedu.com/upload/f/52/f528e9d52f7877a7a2a8dd349a10c78b.jpg)
再答: 2�� (1->2) y d(e^y) - 2�� (1->2) e^y dy =2 [ye^y](1->2) - 2�� (1->2) e^y dy - 2�� (1->2) e^y dy