# Integrating with Absolute Value

```Integrating with Absolute Value
• We will integrate these by using the
(sum/difference) rules to separate the function into
several different integrals.
• When integrating a linear function with substitution
it is often best to draw a rough sketch to help you
decide where to break up the integral. This is
especially important if we are finding the definite
integral.
• Remember what an absolute value function looks
like, it takes every part of the function that is below
the x-axis and reflects it back up above the x-axis.
f ( x)  x
f ( x) | x |
This part of the function
is the opposite of the
other part or –f(x).
f ( x)   x
This part of the function
f(x)=x is still the line y=x
f ( x)  x
So an absolute value function is simply a piecewise, it is the positive
part of f(x) and the other part is –f(x).
 x, when x  0
| x | 
 x, when x  0
The cutoff points for x and –x is critical
here, (0,0) we will always want to look
for where the x-intercepts are.
So we will integrate the following function by breaking it up into 2
parts. We need to break it up so we get rid of the absolute value bars.

4
4
0
4
| x | dx  4  xdx  0 xdx
Now simply integrate both new integrals.
Actually integrate both parts and evaluate.
0
4
4
0
   xdx   xdx
If you have this graph
could you also use the
area formula for a
triangle? Are these
two triangles the
same?
• What about when the absolute value function is
transformed? (moved away from (0,0)) We want to
keep the same process in mind. Lets first work with
a left/right transformation.

5
5
| x  3 | dx
• What does this graph actually look like? Make a
quick sketch
• When you add something inside the |bars| then it moves the figure
left, if you subtract something inside the bars it moves it right.
• This changes the x-intercept from (0,0) and in this case puts it at
(-3,0) which is critical for when we integrated the different parts.

3
5
5
( x  3)dx   ( x  3)dx
3
Find the definite integral

3
5
5
( x  3)dx   ( x  3)dx
3
• Now what if we want to move that function up or
down?
f ( x) | x  2 | 2
Find the area under the curve on the
interval of [-4,3]

3
4
(| x  2 | 2)dx
We will still break it up the
absolute value part into parts, but
also will deal with the +2 by itself.

3
4
3
| x  2 | dx   2dx
4

3
4
3
| x  2 | dx   2dx
4
• Now break up the absolute value part.

2
4
3
3
2
4
( x  2)dx   ( x  2)dx   2dx
Integrate and evaluate the definite integral
• Things get a little more tricky when the function becomes nonlinear.
f ( x)  x  3x  4
2
When we take the absolute value of this
function, everything that is below the x-axis
is reflected to the positive side.
f ( x) | x  3x  4 |
2
The key points that determine what region is reflected
over the x-axis are the x-intercepts, we will need to
find those x-intercepts, because we will use them as
our limits of integration for the above shaded part.
(note that the shaded part is the opposite (negative) of
what the original function was.
Steps
• Take the function and factor it to find the x-intercepts. In this case
the x-intercepts are x=-4 and x=1. Those will be the limits of
integration for the “bubbled part”. Break up the absolute value into
the 3 parts and integrate.
2

5
| ( x  3x  4) | dx
2

4
5
( x  3x  4)dx
2
+

1
4
( x 2  3x  4)dx
+

2
1
( x 2  3x  4)dx
Now integrate each part and add up

2
4
( x  3x  4)dx + 
4
5

5
| ( x2  3x  4) | dx
2
1
2
( x  3x  4)dx +  ( x 2  3x  4)dx
2
1
```