If f is continuous real valued function on $[0,1]$ show that there exist a point $c\in (0,1)$ such that $\int_0^1xf(x)dx = \int_c^1 f(x)dx $

I tried to apply mean value theorem for integrals on both sides separately to see if come equal but that didn't work.


Define $F(t) := \int_t^1 f(x)\, dx$. Then $\int_0^1 xf(x)\, dx = \int_0^1 F(x)\, dx$ by integration by parts:

$$\int_0^1 xf(x)\, dx = \int_0^1 x(-F'(x))\, dx = -F(1) + \int_0^1 F(x)\, dx = \int_0^1 F(x)\, dx.$$

By the mean value theorem for integrals, there is a point $c$ in $(0,1)$ such that $\int_0^1 F(x)\, dt = F(c)$, i.e., $\int_0^1 xf(x)\, dx = \int_c^1 f(x)\, dx$.

  • $\begingroup$ Rest is fine but I find problem when I solve by parts $ \int_0^1 xf (x) $ it becomes messy. $\endgroup$ – ketan Jan 12 '15 at 8:02
  • $\begingroup$ can u please work out the second line in detail @Kobe $\endgroup$ – Learnmore Jan 12 '15 at 8:03
  • $\begingroup$ @learnmore and ketan I made an edit with the steps for integration by parts. $\endgroup$ – kobe Jan 12 '15 at 8:15
  • $\begingroup$ thanks now its clear +1 @kobe $\endgroup$ – Learnmore Jan 12 '15 at 8:20

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