integral (abs(x-2)) dx
The issue here is that the abs ()changes sign when the argument is negative, so you have to counter act that with the signum function
ok so apparently there's no way to express
integral |f(x)|dx
in standard mathematical functions... which I don't exactly buy...
but ya the issue came up when I was trying to evaluate
integral |x - 2|dx
and apparently this is correct
integral |x-2|dx = 2 - [(x-2)^2sgn(2-x)]/2
now I'm not saying that it's wrong or anything I'm just carious as to why it's correct and if somebody could show me how one would get to that without a calculator and done by hand somehow... any help would be great... if you don't know the signum function, sgn(x), is defined as sgn(x) = x/|x| = e^(i arg(x)) = x/SQRT(x^2)
were arg(x) is the complex argument function
THANKS!
4 answers
http://en.wikipedia.org/wiki/Sign_function
ya i realized that but I still have no idea how you get 2 - [(x-2)^2sgn(2-x)]/2 by hand...
This has to be a definite integral, right?
otherswise,how did a constant get into the integration result?
I integrate it as (x-2)^2SGN(x-2)/2+C
if the limits are zero to x, then
int=(x-2)^2SGN(x-2)/2 -(-2)^2(SGN2)/2=
= 2-(x-2)^2SGN(2-x)/2
I hope this helps.
otherswise,how did a constant get into the integration result?
I integrate it as (x-2)^2SGN(x-2)/2+C
if the limits are zero to x, then
int=(x-2)^2SGN(x-2)/2 -(-2)^2(SGN2)/2=
= 2-(x-2)^2SGN(2-x)/2
I hope this helps.
3 answers