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Here's my question. We know that the absolute value of X looks like: Absolute Value of x

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.

Integral of Absolute Value of x

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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).

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

enter image description here

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