let $f(x)$ be a invertible function staisfying $f(x) + f^{-1}(x)=x$.Find $\int_{-1}^{1}f(x) dx$

This is a puzzling question. Are there any real valued functions that satisfy this? I am assuming that $f: \mathbb{R} \rightarrow \mathbb{R}$ (which is mentioned in the link given by pardesi below)

This means f intersects the line y = x at only one point x = y =0. Since f is invertible, it is monotonic.

Now comes the puzzling part. Suppose for x>0, f(x)>0, then f'(x)>0 as f is monotonic. Also f(x)<0 for x<0.

Since f does not intersect y = x at other than x =0, f(x) < x for x>0 as, if x>0, then $f^{-1}(x) > 0$

This means that $f^{-1}(x) > x$ (as $f^{-1}(x)$ is also an increasing function)

and since f(x) >0 we get $f(x) + f^{-1}(x) > x$

But, if f(x) <0 for x>0 ( and consequently f(x) >0 for x<0), we get $f(x) + f^{-1}(x) < 0 < x$

Is there something wrong with my reasoning?

Maybe if you try f(z) = az+b for complex z, you will get zero for the line y=0 with limits x=-1 and x=1

hm…i think ur proof depends on the fact that if f is invertible then it's monotonic but i don't think that's valid …consider f from

1,2,3 to 4.5.6 f(1)=4 ,f(2)=6.f(3)=5 though f is invertible it's not monotonic

@Pardesi: Do you have a solution?

As for [x], we are able to find an area under the curve, because its fully continuous within a particular subset of its domain. But yours seems to be only by definition of individual points??