How to Evaluate the Integral of an Invertible Function with Given Symmetry?

[juantheron]

If $f(x)$ is a invertible function such that $f(x)+f(-x) = 2a\;,$ Then $\displaystyle \int_{a-x}^{a+x}f^{-1}(t)dt$

$\bf{My\; Try::}$ Using Integration by parts...

$\displaystyle \int_{a-x}^{a+x}f^{-1}(t)dt = \left[f^{-1}(t)\cdot t \right]_{a-x}^{a+x}-\int_{a-x}^{a+x}\frac{d}{dt}\left(f^{-1}(t)\right)\cdot tdt$

Now I did not understand how can we solve it.

Help me

Thanks

 

[I like Serena]

If $f(x)$ is a invertible function such that $f(x)+f(-x) = 2a\;,$ Then $\displaystyle \int_{a-x}^{a+x}f^{-1}(t)dt$

$\bf{My\; Try::}$ Using Integration by parts...

$\displaystyle \int_{a-x}^{a+x}f^{-1}(t)dt = \left[f^{-1}(t)\cdot t \right]_{a-x}^{a+x}-\int_{a-x}^{a+x}\frac{d}{dt}\left(f^{-1}(t)\right)\cdot tdt$

Now I did not understand how can we solve it.

Help me

Thanks

We can write $f(x)$ as the sum of $a$ and a monotone odd function.
The inverse of a monotone odd function is also odd...