how to integrate 2e^1-2x with limit of ln2 and 0?
回答 (2)
2016-08-03 7:04 am
✔ 最佳答案
Let u = e^(1-2x)Then, du = -2 e^(1-2x) dx and dx = -1/[2 e^(1 - 2x)] du = -(1/2) (1/u) du
F(n)
= ∫2 e^(1-2x) dx
= ∫2 u [-(1/2) (1/u) du
= -∫du
= - u + C
= - e^(1-2x)
∫[0→ln2] e^(1-2x) dx
= F(ln2) - F(0)
= - e^(1-2ln2) + e^(1-0)
= e - e^(1-ln4)
= e[1 - e^(-ln4)]
2016-08-03 6:39 am
Hint: Let u = 1-2x
收錄日期: 2021-04-20 16:26:20
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