Let a thing rod of length a have a density distribution function p(x)+10e^(-.1x), where x is measured in cm and p in grams per centimeter.

Question

A) if the mass of the rod is 30 g, what is the value of a?
B) For the 30g rod, will the center of mass lie at its midpoint, to the left, or to the right? Why?
C) For the 30g rod, find the center of mass and compare to your prediction in (B).
D) At what value of x should the 30g rod be cut in order to form two pieces of equal mass?

Answer 1

I suspect you meant

p(x) = 10e^(-.1x)

So, the mass will be

∫[0,a] 10e^(-0.1x) dx
= -100e^(-.1x) [0,a]
= 100 - 100e^(-a/10)
= 100(1-e^(-a/10))

If that is 30, then a = 10 ln(10/7)

So, the density decreases with increasing distance, meaning more of the mass is closer to x=0. So, the center of mass will be to the left of the midpoint.

The center of mass is is thus
∫[0,a] 10xe^(-0.1x) dx
---------------------------
30

So, just evaluate that new integral, plug in your a, and you're done.

For (D), split the integral
and find k where

∫[0,k] 10e^(-0.1x) dx = ∫[k,a] 10e^(-0.1x) dx