# 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.

## 2 Answers

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!)

• 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 $\int_a^buv^\prime dx=[uv]_a^b-\int_a^bvu^\prime dx$. – J.G. Jul 11 at 10:23
• That prompts a question then: let $g(x)=x$, explain the identity $$\int_a^b xf'(x) dx = \int_{f(a)}^{f(b)} f^{-1}(y) dy$$ – Ben Jul 11 at 14:23
• @BrianTung I appreciate your professional response. After the correction I think this is a great answer which I can whole-heartedly upvote. – Adayah Jul 11 at 19:19
• 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. – 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 $$fdg,\,gdf$$.

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