How comes that ( after Desmos graphing calculator) the " area function" of f(x)=x² gives back negative values when x<0? I'm currently trying to understand basic calculus concepts. 

I asked to Desmos graphing calculator to draw the graph 

(1) of f(x) = x²

(2) of the " area function" or " accumulation function" of function f, that I defined as the integral from 0 to x of f(t)dt. 

What astonishes me is that the curve of the area function is below zero for x&lt;0, while the curve of the original function is always above the X-axis. 

If my " area function " represents the area under the curve of f and the X axis ( limited by x=0 and  the vertical line passsing through x), how comes that the values of the area function are negative for x &lt; 0? 

In order the " area function" to correspond to its informal concept , shouldn't we define it as 

A(x)= absolute value of ( integral from 0 to x of f(t)dt ) ?
Integrals explicitly are signed areas. Your original function is always positive, and if you're integrating left-to-right, so $x_1&gt;x_0$:
$$\int_{x_0}^{x_1}f(x) dx &gt;0$$
but if you're integrating right-to-left, so $x_1&lt;x_0$:
$$\int_{x_0}^{x_1}f(x) dx &lt;0$$
The area under your original function between, say $x_0=-2$ and $x_1=0$, is $F(0)-F(-2)$, which you'll find is positive.

What you're plotting is a primitive function $F$, which has the useful property that the signed area under $f$ between $x_0$ and $x_1$ is $F(x_1)-F(x_0)$. Taking absolute values as you propose destroys this useful property.
It's normal. You have negative values for x&lt;0
Intégration of f(x)=x^2 from a to b is always positif only if a &lt; b