Question

用换元法计算定积分∫【0到4】[√(2t + 1)]dt

Answer 1

令y = √(2t + 1)
y^2 = 2t + 1，两边微分
2y dy = 2 dt
dt = y dy，这里就是dt和dy之间的转换，看好了
当t = 0时y = √[2(0) + 1] = 1
当t = 4时y = √[2(4) + 1] = 3
∫[0→4] √(2t + 1) dt
= ∫[1→3] y * [y dy]
= ∫[1→3] y^2 dy
= (1/3)y^3 |[1→3]
= (1/3)[3^3 - 1^3]
= (1/3)(27 - 1)
= 26/3

追问

2y dy = 2 dt不懂

追答

(y^2)' = 2y，微分形式就是2y dy

(2t + 1)' = 2，微分形式就是2 dt

Answer 2

令√(2t + 1)=x
2t＋1=x²
t=【x²-1】/2
dt=2xdx
原式=∫【1,3】2x²dx
=2x³/3\[1，3]
=18-2/3
=52/3

追问

dt=2xdx不明白

追答

t=【x²-1】/2

dt=xdx
原式=∫【1,3】x²dx
=x³/3\[1，3]
=9-1/3
=26/3

追问

原式=∫【1,3】x²dx中间为什么是x^2？