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### Calculus of Piecewise Functions

```Date: 06/07/2003 at 21:09:28
From: Matt
Subject: Calculus of the modulus function

The modulus function (absolute value function) is a piecewise function
that is customarily written as f(x)=|x|. Functions such as the floor
and ceiling functions are also piecewise, but they are written
concisely with such notations as g(x)=[x].

Most of the time, with some exceptions, the integrals and derivatives
of piecewise functions are piecewise themselves.

Take for example INT(|x|)dx. Its integral is piecewise, i.e. x^2/2 + C
for x=>0 and -x^2/2 + C for x<0.  This can be written |x^2/2| + C for
all x.

But what if the expression within the absolute value signs is more
complicated than just x?  What if the integral is INT(|u|)u'dx?  Then
the arbitrary C's are different for each case of the modulus, and
there would seem to be no way to consolidate the two cases into one
expression using case notation.

I guess my real question is, can I take the integral or derivative of
a piecewise function like the floor function [u] or the absolute value
function |u| and still somehow notate it in concise form, such as |U|
or [U]?
```

```
Date: 06/07/2003 at 23:17:42
From: Doctor Peterson
Subject: Re: Calculus of the modulus function

Hi, Matt.

for INT(|x| dx) is wrong. As you say, it is

x^2/2 + C for x >= 0

-x^2/2 + C for x < 0

But that is not the same as |x^2| + C, since this is always positive,
and the correct antiderivative is not!

So let's introduce the function that helps us deal with this more
effectively, the signum, or sign, function

sgn(x) = 1  if x > 0
0  if x = 0
-1  if x < 0

Using this, we can write your antiderivative as

sgn(x) * x^2/2 + C

Further, what did you do with the two different C's in your
antiderivative? It looks as if you ignored the distinction, but it
worked out okay. What you have to do is to find the connection
between them, which is based on the fact that the antiderivative has
to be continuous. You find the C in one piece that makes it match up
with the one next to it. In our example, we really have

x^2/2 + C1  for x >= 0

-x^2/2 + C2  for x < 0

and to be continuous, we need

0^2/2 + C1 = -0^2/2 + C2

In this case, this means that C1=C2 and we can just call it C. In
harder cases, you might find that, say, C2 = C1+1, and you would have
to use C+1 in the second piece.

Now I'll demonstrate a way to handle absolute values in integrals that
can be a little risky sometimes if you aren't careful, but can get you

Let's write the absolute value function in terms of signum:

|x| = sgn(x) * x

Do you see why this is true? And do you see that, since signum is
constant within each piece of its definition, its derivative is zero
and it commutes with the integral sign? Then

INT(|x| dx) = INT(sgn(x) * x dx)

= sgn(x) INT(x dx)

= sgn(x) x^2/2 + C

Note, by the way, that this approach also makes it obvious that
d/dx |x| = sgn(x), if you ignore x=0.

I can't guarantee that this will always work correctly; we're being
a little loose with theorems about integration here, and the
discontinuity of signum at x=0 in particular makes this potentially
illegal. In particular, we are saved by the fact that x^2 is zero
when sgn(x) is zero, which makes the result continuous. You have to
check things like that before blindly accepting the result, so the
best thing to do is to use this method to find a neat answer, use the
piecewise approach to get the right answer, and then compare them.

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Calculus
High School Calculus
High School Functions

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