求c,使lim（x趋近正无穷）（x+c/x-c）^x=∫（负无穷,c）te^2tdt这是定积分题目 括号里是表示范围
2019-06-02
求c,使lim（x趋近正无穷）（x+c/x-c）^x=∫（负无穷,c）te^2tdt
这是定积分题目 括号里是表示范围
优质解答
左边
=（x+c/x-c）^x
=[1+2c/(x-c)]^x
=[1+2c/(x-c)]^[(x-c)/2c]*2cx/(x-c)
x趋于正无穷时 原式 = e^[2cx/(x-c)] = e^2c
右边
=(1/2)∫tde^2t
=(1/2)[te^2t|(负无穷到C) - ∫（负无穷到C）e^2tdt]
=(C/2 - 1/4)e^2c
所以C/2 - 1/4 = 1
C = 5/2
左边
=（x+c/x-c）^x
=[1+2c/(x-c)]^x
=[1+2c/(x-c)]^[(x-c)/2c]*2cx/(x-c)
x趋于正无穷时 原式 = e^[2cx/(x-c)] = e^2c
右边
=(1/2)∫tde^2t
=(1/2)[te^2t|(负无穷到C) - ∫（负无穷到C）e^2tdt]
=(C/2 - 1/4)e^2c
所以C/2 - 1/4 = 1
C = 5/2