# problem with solving this question involving integration

let $f(x)$ be integrable at $[0,1]$ and suppose that there's a real number

$M>0$ such that $f(x)\ge{M}$ for all $x\in[0,1]$

a)prove that there exists a point $c\in[0,1]$ such that

$$2\int_0^c {f(x)dx} = \int_0^1 {f(x)dx}$$

b)prove that the point $c$ is unique.

I thought to use the intermediate value theorem. Can I get some clues about solving this ?

• Define $$F(x) = 2\int_{0}^{x}f(t)dt.$$ Apply the intermediate value theroem to $F'$. Note that the hypothesis $M > 0$ and $f(x) \geq M$ on $[0,1]$ is critical here. – Ethan Alwaise Mar 22 '16 at 22:07
• why should I differentiate $F(x)$ ? How can I know what is the derivative. I cant use the fundamental theorom of calculus since I havn't been told if $f(x)$ is continues. and the theorem applies only to continues functions – idan di Mar 23 '16 at 12:45
• Ah, you're right. My mistake – Ethan Alwaise Mar 23 '16 at 19:36

Since $f$ is strictly positive, $h(c) = 2 \int_0^c f(x)dx$ is an increasing and continuous function of $c$. We also have $h(0)=0$.
Now since $f(x) \ge M > 0$, we know that $\int_0^1 f(x) dx > 0 = h(0)$. Moreover, $h(1) = 2 \int_0^1 f(x) dx > \int_0^1 f(x) dx$.
• Why is $h(c)$ increasing ? – idan di Mar 23 '16 at 12:48
• Since the derivative of $h(c)$ is strictly positive, the function is increasing and therefore it is injective. – Joel Mar 23 '16 at 12:48
• How can I know what is the derivative of $h(c)$ ? why is it diffrentiable in the first place ? I know it's continues because it's an indefinite integral of $f(x)$ at $[0,1]$ but why is it diffrentiable ? – idan di Mar 23 '16 at 12:50
• I suppose since f is not continuous the fundamental theorem doesn't apply. However, suppose $c<b$. Then $$h(c)-h(b) = \int_c^b 2f(t)dt > 0.$$ this is positive since $f$ is strictly positive. – Joel Mar 23 '16 at 12:57