# Inverse function on definite integral

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

I really have no idea how to proceed.

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

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

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

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

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

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

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

The result follows for positive functions, at least.

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

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

I really have no idea how to proceed.

Hint in the 2nd integral let

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

and the new limits of integration are

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

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