I understand that the area under a non-negative function across an interval is defined as the definite integral on that interval. I have a question about how to apply this to finding the volume/area of a parallelepiped/parallelogram. Since they are similar, I'll just ask about the parallelogram.
Integrating gives area under graph $l_1\cdot l_2$. Here, I define the variable $y=f(x),\, 0\leq x\leq l_2$ as the length of the intersection of the line parallel to the base and the parallelogram at the point $x$.
Why is it that the first area under graph corresponds to the area of the parallelogram? (Let's just assume we don't know about the usual geometrical way of finding area of parallelogram, that is, cutting and pasting an extruding triangle to form a rectangle)