DEFINITE INTEGRALS WITH ABSOLUTE VALUE
This note addresses the calculation of definite integrals of the form
ˆ b
|f (x)| dx
a
when f (x) is a linear or quadratic polynomial. The absolute value function
(
x,
0x
(1)
|x| =
x, x < 0
has the familiar “V”-shaped graph. Integrands featuring absolute value must be split up
according to this piecewise definition. Following are two examples of this process.
´ 10
Example 1. Find 0 |x 5| dx.
Solution. Replacing x with x
5 in equation (1) gives
(
x 5, 5  x
|x 5| =
5 x, x < 5
where we’ve rewritten 0  x 5 and x 5 < 0 as 5  x and x < 5. Since 5 is in the interval
of integration [0, 10], we have to use both parts of this definition. (If we integrate over [6, 10]
or [0, 4], for example, then we only need to use the first line of the definition.) Therefore
ˆ 10
ˆ 5
ˆ 10
|x 5| dx =
|x 5| dx +
|x 5| dx
0
0
5
ˆ 5
ˆ 10
=
5 x dx +
x 5 dx
0
5


10
1 25
1
= 5x
x
+ x2 5x
2 0
2
5
✓
◆ ✓
◆
✓
◆ ✓
◆
1 2
1 2
1
1 2
2
=
5(5)
(5)
5(0)
(0)
+
(10)
5(10)
(5)
5(5)
2
2
2
2
= 25
⇤
using FTC2.
The crucial steps in example 1 are to determine the values of x where the definition of
|x 5| changes, determine whether they are in the interval of integration, then split the
integral into pieces accordingly. The same steps apply in general.
(1) Determine the values of x when the definition of |f (x)| changes. Say they are a <
x1 < x2 < · · · < xn < b.
(2) Check which (if any) of these values lie in the interval of integration [a, b].
1
(3) Split the integral of |f (x)| over [a, b] into several pieces
ˆ b
ˆ x1
ˆ x2
ˆ
(2)
|f (x)| dx =
|f (x)| dx +
|f (x)| dx + · · · +
a
a
x1
b
xn
|f (x)| dx.
Replace each |f (x)| by either f (x) or f (x), depending on the details of your particular problem.
This turns the problem of finding the integral of |f (x)| into the problems of finding integrals
of f (x), f (x).
In example 1 we had only x1 = 5, and so we ended up with two integrals.
Example 2. Set up the equation (2) for the absolute value integral
ˆ 5
| t2 + 6t 8| dt.
2
Solution. For ease of reference call that function v(t). To evaluate this integral we must
understand the sign of v(t) = (t 4)(t 2) on the interval [ 2, 5]. Because the roots of
v(t) are t = 2, 4 and the leading coefficient of v(t) is negative, it follows that v(t) is negative
on [ 2, 2], positive on [2, 4], and negative again on [4, 5]. You can graph this to see for
yourself. This means the definition of |v(t)| changes at t1 = 2, t2 = 4, t3 = 5. Accordingly
we must split up
[ 2, 5] = [ 2, 2] [ [2, 4] [ [2, 5]
and use the definitions according to equation (1)
8
>
2t2
< v(t),
|v(t)| = v(t),
2t4
>
: v(t), 4  t  5
Therefore
ˆ
5
2
|v(t)| dt =
ˆ
=
ˆ
2
v(t) dt +
2
2
4
v(t) dt +
2
t
2
ˆ
2
6t + 8 dt +
ˆ
5
v(t) dt
4
ˆ
2
4
2
t + 6t
8 dt +
ˆ
5
t2
6t + 8 dt
4
We leave this unevaluated for the sake of neatness, but the reader should be able to use
FTC2 and linearity/power rule to calculate these three definite integrals. The final answer
is 40.
⇤
2