Is there a more complete way of proving this theorem (that I devised)? OR does this theorem already exist and have a name?

Question

Last night, I was working on re-solidifying my understanding of parametric equations, and came up with a neat trick for evaluating certain definite integrals. I'm not sure if it exists as a formal theorem, though I refuse to believe I'm the first to notice such a relation. The 'theorem' is a useful trick to evaluate the definite integral of a continuous function over an interval where the function is either strictly increasing or strictly decreasing, given that the inverse of the function is 'nicer' to work with than the function itself. My idea goes as follows:

If $y = f(x)$ is a continuous function over the interval $I$, and either:

$f'(x) > 0$

OR

$f'(x) < 0$

for all values of $x \epsilon I$, then:

$\int_{a}^{b}{f(x)dx} = bf(b) - af(a) - \int_{f(a)}^{f(b)}{f^{-1}(x)dx}$

where $a,b \epsilon I$

Quasi-proof:

Clearly, if a function is strictly increasing or decreasing, it is injective. This is clear because no two values in the domain can correspond to a single value in the range. I feel intuitively that it must also be surjective (though I'm not entirely sure how to prove this part). Assuming it is both injective and surjective over $I$, the function must have an inverse.

If we were to draw two rectangles out of our bounds, they will have areas:

$af(a)$ and $bf(b)$, so that

$bf(b) - af(a)$ will be the area of the region we are observing.

This area is equivalent to the sum of the areas connecting the curve $f(x)$ to the $x$-axis and to the $y$-axis. Thus:

$bf(b) - af(a) = \int_{a}^{b}{f(x)dx} + \int_{f(a)}^{f(b)}{f^{-1}(x)dx}$

Solving for $\int_{a}^{b}{f(x)dx}$, we get:

$\int_{a}^{b}{f(x)dx} = bf(b) - af(a) - \int_{f(a)}^{f(b)}{f^{-1}(x)dx}$

A particularly illuminating example is evaluating:

$\int_{a}^{b}{ln(x)dx}$

$ln(x)$ is strictly increasing over any interval, so the 'theorem' applies.

$\int_{a}^{b}{ln(x)dx} = bln(b) - aln(a) - \int_{ln(a)}^{ln(b)}{e^xdx}$

$\int_{a}^{b}{ln(x)dx} = ln(b^b/a^a) - (e^{ln(b)} - e^{ln(a)})$

$\int_{a}^{b}{ln(x)dx} = ln(b^b/a^a) + a - b$

and so we've evaluated the integral without the need of integrating the function itself (by method of integration by parts), but rather by integrating its inverse, which is trivial to evaluate.!

Response to Jason DeVito's answer: [This image was taken from What is integration by parts, really?

(https://i.stack.imgur.com/wvkt2.png)

Answer 1

Here's an alternate proof which uses a substitution together with integration by parts.

We are trying to evaluate $\int_a^b f(x)\; dx$. Because we are using $u$ later, we start with the substitution $x = f^{-1}(w)$. Then $dx = (f^{-1})'(w)\; dw$. Further, when $x = a$, $w = f(a)$, and similarly when $x = b$.

Thus, $$\int_a^b f(x)\; dx = \int_{f(a)}^{f(b)} f(f^{-1}(w)) (f^{-1})'(w)\; dw = \int_{f(a)}^{f(b)} w (f^{-1})'(w)\; dw.$$

Now, we integrate this last integral using integration by parts with $u = w$ and $dv = (f^{-1})'(w)$. Thus, $du = dw$ and $v = f^{-1}(w)$

This gives $$\int_a^b f(x)\; dx = wf^{-1}(w)\Big|_{w=f(a)}^{w=f(b)} - \int_{f(a)}^{f(b)} f^{-1}(w)\; dw.$$

Finally, to get your result, note that $wf^{-1}(w)$ with $w = f(a)$ substituted becomes $f(a)a$, and similarly for $b$. Thus we obtain $$\int_a^b f(x)\; dx = bf(b) - af(a) - \int_{f(a)}^{f(b)} f^{-1}(w)\; dw,$$ as you already discovered.