Prove for continous function $f$, $\int_0^1 xf (x) dx = \int_c^1 f (x) dx $

Question

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.

kobe · Accepted Answer · 2015-01-12T08:13:28.313

16

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., $\int_0^1 xf(x)\, dx = \int_c^1 f(x)\, dx$.

edited Jan 12 '15 at 08:13

answered Jan 12 '15 at 07:16

kobe

41,901

Rest is fine but I find problem when I solve by parts $ \int_0^1 xf (x) $ it becomes messy. – ketan Jan 12 '15 at 08:02
can u please work out the second line in detail @Kobe – Learnmore Jan 12 '15 at 08:03
@learnmore and ketan I made an edit with the steps for integration by parts. – kobe Jan 12 '15 at 08:15
thanks now its clear +1 @kobe – Learnmore Jan 12 '15 at 08:20

Prove for continous function $f$, $\int_0^1 xf (x) dx = \int_c^1 f (x) dx $

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