求0到1e^(√x+1)dx的定积分

网友回答

∫ (0->1) e^(√x+1) dx
lety =√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