Let $f:[a,b]\to \mathbb R$ be an injective function as well as continuously differentiable function. Then $\int_a^bf(x)dx+ \int_{f(a)}^{f(b)} f^{-1}(y)dy=bf(b)-af(a).$
stackexchange solution I could have comment below the answer, but author is not active in the site.
Consider 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.
Question
Where did he uses, injectivity and continuous differentiability of $f$?