Integrating an Absolute Value

[Pages:1]Integrating an Absolute Value

4

|x3 - 5x2 + 6x| dx

0

There is no anti-derivative for an absolute value; however, we know it's definition.

|x| =

x -x

if x 0 elsewise

Thus we can split up our integral depending on where x3 - 5x2 + 6x is non-negative.

x3 - 5x2 + 6x 0. x(x2 - 5x + 6) 0.

x(x - 2)(x - 3) 0.

After testing the intervals (-, 0), (0, 2), (2, 3), and (3, ) we discover x3 - 5x2 + 6x 0 when x (0, 2) (3, ). Now we can integrate.

4

|x3 - 5x2 + 6x| dx = applying the definition of absolute value

0

2

x3 - 5x2 + 6x dx +

0

3

-(x3 - 5x2 + 6x) dx +

2

4

x3 - 5x2 + 6x dx = using anti-derivative

3

1 x4 - 5 x3 + 3x2 2 +

43

0

- 1 x4 + 5 x3 - 3x2 3 +

43

2

1 x4 - 5 x3 + 3x2 4

=

8 5 37 ++

43

3

3 12 12

74 =.

12

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