$f(x) +f^{-1}(x)$

It is interesting to see that $b = f(b) = f(f(b)) = f(f(f(b))) = \overbrace{f(f( ... f(f(b))))}^{'n' \text{ number of times}}$ (similarly with $a$ ).

Now, our integral is

$\int_a^b \left[ f(x) + f^{-1} (x) \right] \,dx$

We know that $f^{-1} (f(y)) = y$ and so by substituting $x = f(y) \implies dx = f'(y) \,dy$ so that $\displaystyle \int_a^b \mapsto \int_{f(a)}^{f(b)} = \int_a^b$ , we get

\begin{aligned} \displaystyle \int_a^b \left[ f(f(y)) + y \right] f'(y) \,dy &= \int_a^b f(f(y)) f'(y) \,dy + \int_a^b y f'(y) \,dy \\ \displaystyle &= \int_a^b f(z) \,dz + \left[ yf(y) {{\huge|}_a^b} - \int_a^b f(y) \,dy \right] & \small\color{#3D99F6}{\text{Substituting } f(y) = z \text{ in the first integral} \\ \text{Integrating second integral by parts}} \\ \displaystyle &= bf(b) - af(a) = b^2 -a^2 \end{aligned}

Thus

$b^2 - a^2 = b^2 - 2a + 1 \implies a^2 - 2a + 1 = 0 \implies {(a-1)}^2 = 0 \implies \boxed{a=1}$

• $\int_{a}^{b} [f(x) +f^{-1}(x)]dx=b^2-2a+1$
• $\color{#D61F06} {\int_{a}^{b} f(x) dx +\int_{f(a) }^{f(b) } f^{-1}(x)dx =bf(b)-af(a)}$ , $\color{#D61F06} {f(x) and f^{-1}(x) are functions}$
• $\therefore\int_{a}^{b} [f(x) +f^{-1}(x)]dx=b^2-a^2$
• $\therefore b^2-a^2=b^2-2a+1 \rightarrow a^2-2a+1\Rightarrow a=1$