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Given that $f$ is a continuous and increasing function on $[a, b]$, $c = f(a), d = f(b)$ and $a, b, c ,d \geq 0$, explain why

$\int_c^d f^{-1}(t)~dt = bd - ac - \int_a^b f(x)~dx$

I am not sure how to treat the inverse function or the functions as the limits of integration. I tried to apply what I found here, but couldn't seem to create the RHS exactly.

$\int_c^d f^{-1}(t)~dt = F^{-1}(d) - F^{-1}(c)$ $= F^{-1}(f(b)) - F^{-1}(f(a))$ $ = b - a$

I am trying to understand exactly what the equality is trying to demonstrate in terms of properties of integrals rather than copying a formula from wikipedia superficially.

EDIT:

So, from the RHS, I would be taking the area of $ab$ (largest rectangle) and subtracting $ca$ (shaded rectangle) and then subtracting the area under $f(t)$ (lightly shaded)?

enter image description here

So I'm finding the area between $f^{-1}$ and the vertical axis from $f(a)$ to $f(b)$.

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    The problem seems to stem from the convention of always graphing functions with the independent variable along the horizontal axis and the function values along the vertical axis. If you draw the picture using the usual convention for $f$, but the opposite convention for $f^{-1}$, then the graphs of $f$ and $f^{-1}$ will be the same! And the integral of $f^{-1}$ will be an area between that graph and the vertical axis.2012-06-08

1 Answers 1

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For a formal proof, start by substituting $t=f(x)$ in $\int_c^d f^{-1}(t)\,dt$.

For a pictorial proof:

Pictorial proof