Calculus question... density and cumulative distribution function?

After measuring the duration of many telephone calls, the telephone company found their data was well approximated by the density function: p(x) = 0.4e^(-0.4x), where x is the duration of a call, in minutes.

(a) What percentage of calls last between 1 and 2 minutes?
(b) What percentage of calls last 1 minutes or less?
(c) What percentage of calls last 3 minutes or more?
(d) Find the cumulative distribution function.

I know what the answers to each of these questions are, because they are in the back of my book. But I can't figure out HOW to answer...

The density function p(x) indicates the number of calls at a certain duration: the fraction of calls that had duration between x and x + dx is:
fraction(x, x+dx) = p(x)*dx

The fraction between two values a and b (in minutes) is:
fraction(a, b) = integral(x=a,b) [p(x)dx]

In particular, if ask, "What is the fraction that had duration between 0 and infinity?", the answer had better be 1. Checking it out:
fraction(0, infinity) = integral(0, infinity)[0.4*e^(-0.4*x) dx]
= integral(0, infinity)[e^(-y) dy]
= 1
So it checks out.
a) fraction(1,2) = integral(1,2)[0.4*e^(-0.4*x) dx]
= integral(0.4, 0.8)[e(-y) dy]
= e^(-0.4) - e^(-0.8)
(Multiply by 100 to get the %-age.)

b) fraction(0,1) = integral(0,1)[0.4*e^(-0.4*x) dx]
= integral(0, 0.4)[e(-y) dy]
= e^(0) - e^(-0.4)
= 1 - e^(-0.4)
(Multiply by 100 to get the %-age.)

c) fraction(3, infinity) = integral(3,infinity)[0.4*e^(-0.4*x) dx]
= integral(1.2, infinity)[e(-y) dy]
= e^(-1.2) - e^(-infinity)
= e^(-1.2)
(Multiply by 100 to get the %-age.)

d) The cumulative distribution function is defined as the fraction that have duration UP TO a given value. In other words:
cdf(x) = integral(0,x)[p(x)dx]
= integral(0, x)[0.4*e^(-0.4*x) dx]
= integral(0, 0.4*x)[e^(-y) dy]
= 1 - e^(-0.4*x)

I hope this clears the matter up.

The proper way to represent your pdf is to use f(x), not p(x).

f(x) = .4e^(-.4x), 0 < x < infinity

If f(x) is a pdf, then p(a <= x <= b) is the area under the curve f(x) between x=a and x=b.

So, p(a <= x <= b) is going to be the integral of f(x) evaluated from a to b.

integral of f(x) = -e^(-.4x)

(a) evaluate the integral from 2 to 1. You get -.44 - (-.67) = 23 percent.

(b) evaluate from 1 to 0. You get 33 percent

(c) I think you have the idea at this point.

(d) F(x) = P(X <= x) = 1 - e^(-.4x)

Note: You can verify that f(x) is a pdf by making sure that the area under the curve f(x) for the entire range of x is 1. If you calculate the integral of f(x) from 0 to infinity, you get 1.