> 数学 >
∫x*e^(-k^2*x^2)dx 扩号里是e的次方 从零到正无穷的定积分
人气:220 ℃ 时间:2020-03-17 01:44:18
解答
∫xe^(-k²x²) dx
= (1/2)∫e^(-k²x²) d(x²)
= (1/2)(1/-k²)∫e^(-k²x²) d(-k²x²)
= -1/(2k²)*e^(-k²x²)
= -1/(2k²)*[0-1/e^(0)]
= -1/(2k²)*(-1)
= 1/(2k²)
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版