Prove that $\int_a^bf(x)dx+ \int_{f(a)}^{f(b)} f^{-1}(y)dy=bf(b)-af(a)$ [duplicate]

Let f' be continuous and positive on [a,b]. Prove that: $\int_a^bf(x)dx+ \int_{f(a)}^{f(b)} f^{-1}(y)dy=bf(b)-af(a)$

I've got the later steps of this covered using integration by parts, and I know that I need to use substitution on the integral containing the inverse. My question is: how do I set up the substitution on $\int_{f(a)}^{f(b)} f^{-1}(y)dy$ to get it in terms of f(x)?

We have to assume that $f$ is monotone, or $f^{-1}$ would not be well defined. For simplicity assume in addition that $$0\leq a<b,\quad 0\leq f(a)<f(b)\ .$$ Let $\gamma\subset[a,b]\times[f(a),f(b)]$ be the graph of $f$, and at the same time of $f^{-1}$; see the following figure.

Then $$\int_a^b f(x)\ dx$$ is the area bounded by $x=a$, $x=b$, $y=0$, and $\gamma$. In the same way $$\int_{f(a)}^{f(b)} f^{-1}(y)\ dy$$ is the area bounded by $y=f(a)$, $y=f(b)$, $x=0$, and $\gamma$.

Looking at the figure one immediately verifies that the sum of these two areas is $\ =bf(b)-af(a)$.

We need to compute the integral $$\int_{f(a)}^{f(b)} f^{-1}(y) dy.$$

Let $x=f^{-1}(y)$ (inverse of $f$), so $y=f(x)$ with $dy=f'(x)dx$. Thus $\int_{f(a)}^{f(b)} f^{-1}(y) dy=\int_a^b x f'(x)dx$ and applying integration by parts on the integral RHS we get $\int_{f(a)}^{f(b)} f^{-1}(y) dy=xf(x)\bigg|_a^b-\int_a^b f(x)dx$, which, by a simple algebra, implies the desired equality.