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. 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).

• think it should be $\frac{x^3}{3}+x$ – The Poor Jew Aug 14 at 0:41
• Yes, you're right. Thank you. – davidlowryduda Aug 14 at 16:03

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: 