i need help with finding chebyshev inequality, please

The average value of a function is the integral divided by the range it's integrated over.

From your description I have a couple of problems. If the function is

f(x)=(3/2)e^-|3x| over the range from -∞ < x < ∞

Why are you integrating

f(x)=(3/2)xe^-|3x|dx where did the x come from?

second look at the absolute value

isn't |-3x|=|3x|

So isn't the integral of f(x)=(3/2)e^-|3x| over the range from -∞ < x < ∞
equal to twice the integral of f(x)=(3/2)e^-3x over the range from 0 < x < ∞

Hope that's some help.

Rap

You made no mistakes , friend . f(x) is the p.d.f. (I in fact verified it is a p.d.f.) so mean is the integration of xf(x) over domain of definition of x
which in this case is from - infinity t0 + infinity ; now xf(x)=(3/2)xe^ -l3xl is an odd function so the integration must yield o , and it is the mean . It is not unusual to have o as the mean for a p.d.f. , as you might know the standard-normal distribution , the standard Cauchy distribution etc. have mean 0