On-line Math 21

5.2  The Fundamental Theorem of Calculus

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Example 4

ó õ 2 -1  |x|dx.

Solution

Here you have to split it up into two pieces:

ó õ 2 -1  |x|dx = ó õ 0 -1  |x|dx+ ó õ 2 0  |x|dx.

The first integral has x always negative, and the second has x always positive, so we can remove the absolute value signs:

ó õ 2 -1  |x|dx = ó õ 0 -1  |x|dx+ ó õ 2 0  |x|dx = ó õ 0 -1  -xdx+ ó õ 2 0  xdx = -x2/2| -10+ x2/2| 02 = (0-(-1/2))+(4/2-0) = 5/2.

The only ``antidifferentiation'' trick we use here is that

æ ç è 1 2 x2 ö ÷ ø ¢ = x,

which certainly is true.

Copyright (c) 2000 by David L. Johnson.