Is there a geometric intuition for integration by parts?

Question

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.

Answer 1

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.

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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 $v(x_2) = 3$

and so we can relate those two integrals by

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

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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!)

Answer 2

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 $fdg,\,gdf$.