$\int_0^{f(a)} f^{-1}(t)\,dt + \int_0^a f(t)\,dt = af(a)?$ If $f: [0, a] \to \mathbb{R}$ is continuous and $1$-to-$1$ with $f(0) = 0$, how do see that $$\int_0^{f(a)} f^{-1}(t)\,dt + \int_0^a f(t)\,dt = af(a)?$$
let $f(a) = b.$  complete the rectangle with corners at $(0,0), (a,0), (a, b)$ and $(0,b).$ you see  that the rectangle is partitioned into two almost triangular looking regions:

(1) one bounded  by $y = 0, x = a$ and $y = f(x)$ which has area $\int_0^a f(x) \, dx$  

(2) and the other one bounded  by $x = 0, x = f^{-1}(y)$ and $y = b$ which has area $\int_0^b f^{-1}(y) \, dy$

but these two areas make up the area of the rectangle, therefore we have 
$$\int_0^a f(x) \, dx +  \int_0^b f^{-1}(y) \, dy = ab = af(a).$$
Hint:Let $H(x)=\int_0^{f(x)}f^{-1}(t)dt+\int_0^xf(t)dt$, $H'(x)=f^{-1}(f(x))f'(x)+f(x)=(xf(x))'$ $H(0)=(xf(x))(0)$ done.