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Suppose that $f$ is continuous and maps $[0,1]$ one-to-one onto $[0,1]$. Therefore, it is obvious that inverse function $f^{-1}$ exists. Then, show that $$ \int_{0}^{1} f(x) d x+\int_{0}^{1} f^{-1}(y) d y=1 $$ and give a geometric interpretation of this equation.

There is a similar problem in the following link; however, the major strategy does not work since $f$ is not an increasing function. Moreover, the substitution method does not work as well, since we need compact interval $[0, 1]$ for integration. So, what other alternatives are there for solving this problem?

Show rigorously that the sum of integrals of $f$ and of its inverse is $bf(b)-af(a)$

BlizzardWalker
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The statement is false if $f$ is decreasing. Consider $f(x) = 1 - x^2$ on $[0,1]$. Then $f^{-1}(x) = \sqrt{1-x}$ but $$ \int_0^1 (1-x^2) \ dx + \int_0^1 \sqrt{1-x} \ dx = \frac{2}{3} + \frac{2}{3} = \frac{4}{3} \ne 1 $$

If $f$ is increasing, we can interpret the result geometrically as follows: The area under the graph of $f$ plus the area under the graph of $f^{-1}$ is equal to the area of the unit square, which is $1$.

fwd
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  • If $f$ is not decreasing, would it be necessarily true as well? Which conditions would be enough to ensure this holds true? Is it the fact $f(0) = 0$ and $f(1) = 1$ playing crucial role to conclude the correctness of this problem? – BlizzardWalker May 26 '21 at 17:19
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    It seems that if $f$ was given an increasing function, then $f(0) = 0$ and $f(1) = 1$ would satisfy, and we could proceed on the provided link problem, right?! – BlizzardWalker May 26 '21 at 17:22
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    @Snowflake yes, if $f(0) = 0$ and $f(1) = 0$, then we can apply the linked result. See also the edit to my answer. – fwd May 26 '21 at 17:26
  • Thank you very much for coming up with such counterexample. – BlizzardWalker May 26 '21 at 17:55