# Inverse function on definite integral

• October 20th 2010, 08:03 AM
nvwxgn
Inverse function on definite integral
The function $\displaystyle f$ and its inverse $\displaystyle f^{-1}$ are continuous. If $\displaystyle f(0) = 0$, find $\displaystyle\int^5_0 f(x)\ dx + \int^{f(5)}_0 f^{-1}(t)\ dt$

I really have no idea how to proceed.

• October 20th 2010, 08:48 AM
Ackbeet
I would interpret the problem geometrically, in terms of area. Obviously,

$\displaystyle\int_{0}^{5}f(x)\,dx$

is the area under the function $f(x),$ if we assume $f(x)\ge 0\;\forall\,x\in[0,5],$ which we will do for now. Now the graph of the function $f^{-1}$ is just the graph of $f$, but flipped over the line $y=x.$ So the integral

$\displaystyle\int_{0}^{f(5)}f^{-1}(t)\,dt$

is the area under the function $f^{-1}.$ If you plot this up for a few functions, you might notice that if you flip the area represented by

$\displaystyle\int_{0}^{f(5)}f^{-1}(t)\,dt$

over the line $y=x$, that will just equal the area required to fill the rectangle whose area is $5f(5).$

The result follows for positive functions, at least.

I should point out that one assumption here, that is justified by the hypothesis that both $f$ and $f^{-1}$ are continuous, is that $f^{-1}$ exists. That means $f$ would be 1-1.
• October 20th 2010, 08:56 AM
TheEmptySet
Quote:

Originally Posted by nvwxgn
The function $\displaystyle f$ and its inverse $\displaystyle f^{-1}$ are continuous. If $\displaystyle f(0) = 0$, find $\displaystyle\int^5_0 f(x)\ dx + \int^{f(5)}_0 f^{-1}(t)\ dt$

I really have no idea how to proceed.

Hint in the 2nd integral let

$u=f^{-1}(t) \iff f(u)=t \implies f'(u)du=dt$

and the new limits of integration are

$t=0 \implies u=0 \text{ and } t=f(5) \implies u=5$

then use integration by parts.

Also this can be seen on a graph.

The first integral is in blue the inverse is in brown.

Attachment 19399

Edit: Wow I am really slow at posting today