Is there a geometric intuition for integration by parts? [duplicate]

Is there a geometric intuition for integration by parts?

$$\int f(x)g'(x)\,dx = f(x)g(x) - \int g(x)f'(x)\,dx$$

This can, of course, be shown algebraically by product rule, but still where is geometric intuition? I have seen geometry of IBP using parametric equations but I don't get it.

Newest edit: few similar questions has been asked before, but they use parametric equations to show geometry behind IBP. I am interested if there is geometric intuition which uses functions in Cartesian plane or some other, maybe more natural, explanation.

Note. Edited because Adayah pointed out (correctly, and to my chagrin) that this answer was totally sloppy—sloppier even than I intended it to be. Let's hope it's better now.

When we use integration by parts on an integral

$$ \int u(x) \, \mathrm{d}v(x) = \int u(x) v'(x) \, \mathrm{d}x $$

we implicitly treat $u$ and $v$ as parametric functions of $x$. If we plot these functions against each other on the $u$-$v$ plane, we might obtain something like the below:

(Note that $v$ is on the horizontal axis, and $u$ on the vertical.) In this diagram, the purple region below the curve represents the definite integral

$$ \int_{v(x)=2}^3 u(x) \, \mathrm{d}v(x) = \int_{x=v^{-1}(2)}^{v^{-1}(3)} u(x) v'(x) \, \mathrm{d}x $$

Similarly, the blue region to the left of the curve represents the definite integral

$$ \int_{u(x)=1}^2 v(x) \, \mathrm{d}u(x) = \int_{x=u^{-1}(1)}^{u^{-1}(2)} v(x) u'(x) \, \mathrm{d}x $$

Note that we can set

• $x_1$ such that $u(x_1) = 1$ and $v(x_1) = 2$
• $x_2$ such that $u(x_2) = 2$ and $v(x_1) = 3$

and so we can relate those two integrals by

$$ \int_{x=x_1}^{x_2} u(x) v'(x) \, \mathrm{d}x = \left. u(x) v(x) \phantom\int\!\!\!\!\! \right]_{x=x_1}^{x_2} - \int_{x=x_1}^{x^2} v(x) u'(x) \, \mathrm{d}x $$

Obviously this simple visualization of integration by parts relies (at least to some degree) on $u(x)$ and $v(x)$ being one-to-one; otherwise, we have to use signed areas. However, the necessary rigor can be added. I'm making the assumption that rigor was not what was needed here. (ETA: Though more than I provided at first!)

In light of @Adayah's observations, I'll offer a different geometric intuition for $fdg=d(fg)-gdf$, which integrates to the desired result. Consider the special case $f,\,g,\,df,\,dg>0$, so we can draw an $f\times g$ rectangle inside an $(f+df)\times (g+dg)$ rectangle. Apart from a negligible $df\times dg$ corner piece, the trimming $d(fg)$ outside the slightly smaller rectangle is two rectangles of areas $fdg,\,gdf$.