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Let $f$ be a monotonic differentiable function; $f: (a,b)\to \mathbb{R}$, so that $F'=f$.

I need to prove that:

$\int f^{-1}(x)dx=xf^{-1}(x)-F(f^{-1}(x))+C$.

I tried to use the following formula: $\int u'v=uv-\int v'u+C$, but I can't see how I move on from this.

Any Hints?

Thanks A Lot!

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One idea is to just differentiate both sides and verify. =) –  Srivatsan Nov 22 '11 at 19:01
    
$f^{-1}(x) \neq \frac{1}{f(x)}$! –  N. S. Nov 22 '11 at 19:09
    
@N.S.: Thanks for the correction! –  Jozef Nov 22 '11 at 19:24

2 Answers 2

up vote 2 down vote accepted

Let $f^{-1}(x) = y$. Then $x = f(y)$ which gives us $dx = f'(y) dy$.

Hence, $$\begin{align} \int f^{-1}(x) dx & = \int y d(f(y)) & (\because y = f^{-1}(x))\\ & = y f(y) - \int f(y) dy + C & (\text{Integration by parts})\\ & = f^{-1}(x)x - F(y) + C & (\because F' = f \implies \int fdy = F)\\ & = xf^{-1}(x) - F(f^{-1}(x)) + C & (\because y = f^{-1}(x)) \end{align}$$

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Thanks you for the answer. Can you please explain me why did you write that $ \int f^{-1}(x) dx = \int y df(y)$. Isn't that suppose to be $\int yf'(y)dy$? –  Jozef Nov 22 '11 at 19:21
    
Plug in $x = f(y)$ in the integral to get $\int f^{-1}(x) dx = \int y d(f(y))$ –  user17762 Nov 22 '11 at 19:23
    
So it's basically the same :) Why did I have to know that f is monotonic? –  Jozef Nov 22 '11 at 19:25
1  
@Jozef: You need to know $f$ is monotonic so that $f^{-1}$ is well-defined. –  user17762 Nov 22 '11 at 19:41

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