Why is the definite integral from $a$ to $b$ is negative of integral from $b$ to $a$ graphically? I was studying properties of definite Integrals, and I came across this property which was easy to prove.
$\int_{b}^{a}f(x) \,\mathrm{d}x = -\int_{a}^{b}f(x) \,\mathrm{d}x$
However, I had a problem with the graphical approach. If we look at the graph of $x^3-x$

Now the definite integral represents algebraic area, but in $[-1,0]$, the graph is above $x-axis$, So I thought the area should be positive either going from $a$ to $b$ or from $b$ to $a$.
But obviously, this is not true, and I guessed that it is because of the $dx$ in multiplication but I couldn't find out the proper explaination anywhere, so posting it here.
 A: Suppose $f$ has antiderivative $F$. Then
$$\int_a^b f(x)\mathrm{d}x=F(b)-F(a)$$
And,
$$\int_b^a f(x)\mathrm{d}x=F(a)-F(b)=-(F(b)-F(a))=-\int_a^b f(x)\mathrm{d}x$$
How do we think about this graphically? If we have some function $f$, the graph of $f(-x)$ will be the usual graph of $f(x)$ flipped about the $y$ axis. That is, the point at $a$ will move to $-a$ and $b$ will move to $-b$. Since we assume from the beginning $a<b$, then we can say $-a>-b$. So graphically, it is easy to justify that
$$\int_a^b f(x)\mathrm{d}x=\int_{-b}^{-a}f(-x)\mathrm{d}x$$
This can be visualized here https://www.desmos.com/calculator/ivgeeuba0k
The above equation can be verified using the coordinate transformation $x'=-x$:
$$\int_{-b}^{-a}f(-x)\mathrm{d}x=\int_{b}^{a}f(x')(-\mathrm{d}x')=-\int_{b}^{a}f(x')\mathrm{d}x'=-\int_{b}^{a}f(x)\mathrm{d}x.$$
A: The claim: The area should be positive either going from a to b or from b to a
$\!>$ is incorrect
Consider the conversion of addition to integration. In this, one assumes an element $\delta x=x_{final}-x_{initial}$.
So while going from $b$ towards $a$, the result will surely be negative of what it would have been for $a$ to $b$.
This was formulated by Rienmann as well, and called Riemann Sums.

In the above graph, $\xi_i$ are sample points in region bounded by $x_{i-1}<x<x_i$. The integration can now be simplified into summation of rectangular regions. Given by : $${A = \lim\limits_{\left| P \right| \to 0} \sum\limits_{i = 1}^n {f\left( {{\xi _i}} \right)\Delta {x_i}} }$$
where ${\left\| P \right\| }={ \max \left\{ {\Delta {x_1},\Delta {x_2}, \ldots ,\Delta {x_n}} \right\}}$ and ${\Delta {x_i} = {x_i} – {x_{i – 1}}}$
