Antiderivative of an Absolute Value?

Question

When integrating, is there anything special about integrating the absolute value of x ( |x| )? Does it follow a different set of rules than just x by itself? Is its antiderivative something other than x^2/2?

Thank you!

Answer 1

it can be a little different than just x^2/2. it will depend upon the limits of integration.

Recall:

|x| = x if x >0
|x| = -x if x<0

suppose we have the integral from a to b of |x| where a is less than zero.

we can write it as two integrals:

integral from a to 0 of -x plus integral from 0 to b of x.

when you are done and evaluate at the endpoints you get:

b^2/2 + a^2/2

for example:

∫│x│dx from x = -2 to x = 5

∫-xdx from -2 to 0 + ∫xdx from 0 to 5

-x^2/2 from -2 to 0 = 0 --(-2)^2/ = 2
plus
x^2/2 from 0 to 5 = (5^2)/2 - 0 = 12.5

for a grand total of 14.5

if however the lower limit is greater than zero, then it will be just like x^2/2

Answer 2

Yes:

|x| = {x, x>=0; -x, x<= 0

So ∫ |x| = x^2/2 for x > 0, -x^2/2 for x < 0. This can be expressed as x |x| / 2.

For instance, ∫(-3 to -1) |x| dx = (-1)(1)/2 - (-3)(3)/2 = 4, whereas x^2/2 would give us -4.

Also, ∫ (-1 to 1) |x| dx = 1(1)/2 - (-1)(1)/2 = 1, whereas x^2/2 would give us 0.

Answer 3

Well, I'm stumped. Seems to me the antiderivative is x * abs(x), where abs(x) is the absolute value of x, but you're saying absolute value signs can't be used. You could also define a piecewise function, but you say that's not allowed, either. Hmm...

Answer 4

No! not that simple: ∫|x|dx = |x|*x/2;