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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 ?

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  • $\begingroup$ 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. $\endgroup$ – Ethan Alwaise Mar 22 '16 at 22:07
  • $\begingroup$ 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 $\endgroup$ – idan di Mar 23 '16 at 12:45
  • $\begingroup$ Ah, you're right. My mistake $\endgroup$ – Ethan Alwaise Mar 23 '16 at 19:36
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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, $h(1) = 2 \int_0^1 f(x) dx > \int_0^1 f(x) dx$.

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  • $\begingroup$ So,h(c) is one-to-one and continues on a closed interval. So by the Intermediate value theorem h(c) getting all values between h(0) to h(1). Is this correct? But how you concluded that h(c) is one-to-one? $\endgroup$ – idan di Mar 23 '16 at 11:31
  • $\begingroup$ Why is $h(c)$ increasing ? $\endgroup$ – idan di Mar 23 '16 at 12:48
  • $\begingroup$ Since the derivative of $h(c)$ is strictly positive, the function is increasing and therefore it is injective. $\endgroup$ – Joel Mar 23 '16 at 12:48
  • $\begingroup$ 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 ? $\endgroup$ – idan di Mar 23 '16 at 12:50
  • $\begingroup$ 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. $\endgroup$ – Joel Mar 23 '16 at 12:57

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