# Definite integral and area/volume

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.

This is the "right" way to integrate, which gives area under graph $$l_1\cdot h$$

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)

Your parallelogram is described by two curves. Let's call the lower left corner $$A$$, then $$B$$ is on the lower right, $$C$$ is upper right, and $$D$$ is upper left. For simplicity, $$x$$ axis is along $$AB$$. Then the first function gets the $$y$$ coordinate along $$ADC$$ and the second function gets the $$y$$ coordinate along $$ABC$$. Then the area of the parallelogram is \begin{align}\int_{x_A}^{x_C}[f_1(x)-f_2(x)] dx&=\int_{x_A}^{x_C}f_1(x)dx-\int_{x_A}^{x_C}f_2(x)] dx\\&=\int_{x_A}^{x_D}f_1(x)dx +\int_{x_D}^{x_C}f_1(x)dx-\int_{x_A}^{x_B}f_2(x)dx-\int_{x_B}^{x_C}f_2(x)dx\\&=\int_{x_A}^{x_D}\left[\frac{y_D-y_A}{x_D-x_A}(x-x_A)+y_A\right]dx+\int_{x_D}^{x_C}y_Ddx\\&-\int_{x_A}^{x_B}y_Adx-\int_{x_B}^{x_C}\left[\frac{y_C-y_B}{x_C-x_B}(x-x_B)+y_B\right]dx\end{align} Now for the parallelogram $$y_A=y_B=0$$, $$y_C=y_D=h$$, $$x_C-x_D=x_B-x_A=l_1$$, $$x_A=0$$, $$x_B=l_1$$. If you also notice that $$x_C-x_B=x_D-x_A$$, then the formula for the area can be simplified a lot. You should be able to cancel the first and last integrals. This is equivalent of "cutting and pasting" the triangles: you cut out the triangle with hypotenuse $$BC$$ and paste it as a triangle with hypotenuse $$AD$$.