Integral of absolute value of X and area under the curve. Here's my question. We know that the absolute value of X looks like:

Clearly, we can see, since the absolute value of x is always greater than or equal to 0, the area under the curve is always positive. Why then does it integrate to the following? Where the integral function takes negative values? I do understand, though, that the derivative of the following function works out to be what one would expect: |x|
EDIT: I guess this question is a little bit stupid. I am confusing definite integral with the indefinite integral. I do notice that if I take any two points and take the difference between the values of the indefinite integral evaluated at these points, I get a positive value for the area. 

 A: Your intuition seems to be telling you that the antiderivative of an always-positive function should be always positive. But this is not correct. This is a counterexample. Integrating $x^2 + 1$ is another example: it's antiderivative is $\frac{x^3}{3} + x + C$, which is not always positive.
Instead, the correct property that we should expect is for the function to be always increasing. Starting with a positive function $f(x)$, we know that $\displaystyle \int_a^b f(x) dx > 0$. In particular, this should mean that $\displaystyle F(x) = \int_0^x f(t) dt$, which is the antiderivative, to be a strictly increasing function.
For instance, $\int_a^b f(x) dx > 0 \iff  F(b) - F(a) > 0$, so that we see that $F(x)$ must be strictly increasing.
In this case, $\frac{1}{2}x^2 \text{sgn}(x)$ is a strictly increasing function, so that it might be the antiderivative of a positive function (like it is).
A: I am answering my own stupid question for the sake of completeness. 
I am confusing definite integral with the indefinite integral. I do notice that if I take any two points and take the difference between the values of the indefinite integral evaluated at these points, I get a positive value for the area. At the same time, the derivative of the indefinite integral gives |x|
i.e. 
$$
\left(\frac{x^2 * sgn(x)}{2}\right)_{-1}^{0} = (0 - \frac{-1}{2}) = \frac{1}{2}
$$
which is the same as the area of the triangle as below:

