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!

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

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.

To evaluate the integral ∫ |x - 2| dx without a calculator, we need to split the integral into two separate parts based on the sign of (x - 2).

Let's consider two cases:
1. x ≥ 2: In this case, the absolute value |x - 2| is equal to (x - 2). Therefore, we can write the integral as ∫ (x - 2) dx.

2. x < 2: In this case, the absolute value |x - 2| becomes -(x - 2) because it is negative. So, the integral becomes ∫ -(x - 2) dx.

Now, let's evaluate each case:

Case 1: x ≥ 2
∫ (x - 2) dx
= (1/2)x^2 - 2x + C1, where C1 is the constant of integration.

Case 2: x < 2
∫ -(x - 2) dx
= -[(1/2)x^2 - 2x] + C2, where C2 is another constant of integration.

At x = 2, the absolute value function changes sign. Therefore, we need to account for this in our final result by including the signum function (sgn) you mentioned in your question.

The signum function sgn(x) evaluates to 1 for x > 0, -1 for x < 0, and 0 when x = 0. In our case, sgn(2 - x) evaluates to 1 for x < 2, and -1 for x > 2.

Taking this into account, we modify our result for case 2:
∫ -(x - 2) dx
= -[(1/2)x^2 - 2x] + sgn(2 - x) + C2

Now, we can combine the results for both cases to get the final answer:
∫ |x - 2| dx =
(1/2)x^2 - 2x + sgn(2 - x) + C

Where C is the constant of integration, combining C1 and C2.

Therefore, the correct expression for ∫ |x - 2| dx is:
∫ |x - 2| dx = (1/2)x^2 - 2x + sgn(2 - x) + C

I hope this explanation helps! Let me know if you have any further questions.