Consider the integral,
$$ \int_{2}^{0} x dx$$
Evaluating, we get a negative area but why is when add up rectangles starting at x=2 and go to x=0, is the area accumulated negative?
I could easily prove this by a change of bounds substitution $ x= -t$, but I am looking for a more geometric perspective.
If you approximate an integral with rectangles, you are using signed lengths. So for example when you integrate $-x$, the height of the rectangle is $x$, but when you measure from the horizontal axis in a downward direction you assign a negative sign to it. So the area of the rectangle will be $$dA=-|x||dx|$$ Similarly, if your lower limit is greater than the upper limit, you go in the negative direction, so $dx$ will transform into $-|dx|$: $$\int_2^0 x dx=\sum_i |x_i|(-|\Delta x|)=-\sum_i|x_i||\Delta x|$$
