F(x) is a continuous function on [0,1], Such that $ \int_0^1 F(t)\,dt$=1. Prove that there exists a number c $\in$ (0,1) such that F(c)=3$c^2$.
Asked
Active
Viewed 102 times
0
-
2How are $f$ and $F$ related? Are they supposed to be the same function? What are your thoughts and efforts on the problem so far? – Cameron Buie Dec 17 '13 at 15:21
-
ya ya.. sry . they are same!! – Topology Dec 17 '13 at 15:28
-
Got something from the answer below? – Did Dec 19 '13 at 15:32
1 Answers
3
Otherwise, by the IVT, either $F(x)\gt3x^2$ for every $x$ in $(0,1]$, or $F(x)\lt3x^2$ for every $x$ in $(0,1]$. Since $\int\limits_0^13x^2\mathrm dx=1$, in both cases, $\int\limits_0^1F(x)\mathrm dx\ne1$, which is absurd.
Did
- 279,727