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:

enter image description here

(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!)

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  • $\begingroup$ A sketch of how we address cases where it's not one-to-one: split the integration range at turning points, then handle each such range separately, multiplying $v$ by $-1$ on some if necessary to ensure it's increasing; this doesn't change whether $\int_a^buv^\prime dx=[uv]_a^b-\int_a^bvu^\prime dx$. $\endgroup$ – J.G. Jul 11 at 10:23
  • $\begingroup$ That prompts a question then: let $g(x)=x$, explain the identity $$\int_a^b xf'(x) dx = \int_{f(a)}^{f(b)} f^{-1}(y) dy$$ $\endgroup$ – Ben Jul 11 at 14:23
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    $\begingroup$ @BrianTung I appreciate your professional response. After the correction I think this is a great answer which I can whole-heartedly upvote. $\endgroup$ – Adayah Jul 11 at 19:19
  • $\begingroup$ This is interesting. In the limit, the diagram just illustrates the chain rule, but working with integrals instead allows it to be exact without any infinitesimals. $\endgroup$ – Carsten S Jul 12 at 15:09

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 $fdg,\,gdf$.

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  • $\begingroup$ This is not exactly what I was looking for. This is nice intuition for the product rule, so can you tranform this into IBP? $\endgroup$ – 1b3b Jul 12 at 11:34
  • $\begingroup$ @1b3b As I said, IBP is just the product rule integrated. $\endgroup$ – J.G. Jul 12 at 12:21

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