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IBP works both for increasing and decreasing functions.
trying to understand why it gives the correct area in both cases...

As shown, when the point B moves from top to down, IBP exactly removes the unwanted areas and gives the correct area. It seems every formula/expression just works out no matter how the curve is. It doesn't seem that intuitive. Why is it so?

horizontal axis is $u=u(t)$
vertical axis is $v=v(t)$

Consider 2 scenarios.

1) Increasing function
By IBP, area of green region is given by:
$$\int_{t=A}^{t=B} vdu = uv|_{t=A}^{t=B} - \int_{t=A}^{t=B} udv$$ enter image description here

2) Decerasing function
By IBP, area of green region is given by:
$$\int_{t=A}^{t=B} vdu = uv|_{t=A}^{t=B} + \int_{t=B}^{t=A} udv$$ enter image description here

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1 Answer 1

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enter image description here

How about this interpretation? The picture is consistent with the formula.

  1. Increasing function:

$\int_{t=A}^{t=B} vdu=S_{ABCD}=I, uv|_{t=A}^{t=B}=S_{BCDAFG}=I+III,\int_{t=A}^{t=B} udv=S_{BAFG}=III \\ \to I=（I+III）-III$

$\int_{t=A}^{t=B} vdu = uv|_{t=A}^{t=B} - \int_{t=A}^{t=B} udv$

  1. Decreasing function:

$\int_{t=A}^{t=B} vdu=S_{ABCD}=I+IV, uv|_{t=A}^{t=B}=S_{BCEF}-S_{ADEG}=S_{BCDH}-S_{AHFG}=I-III,\\ \int_{t=B}^{t=A} udv=S_{ABFG}=III+IV \\ \to I+IV=(I-III)+(III+IV)$

$\int_{t=A}^{t=B} vdu = uv|_{t=A}^{t=B} + \int_{t=B}^{t=A} udv$

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