1

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 $f$?

Unknown x
  • 703
  • 2
    Existence of inverse requires injectivity, and existence of the integral requires continuity (at least these are sufficient conditions). – Sarvesh Ravichandran Iyer Jan 06 '21 at 16:07
  • Differentiability is not needed. If $f$ is injective and continuous then it is strictly monotone. Now it is easy to prove the result using partitions and Riemann sums. – RRL Jan 07 '21 at 00:44

0 Answers0