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let $f : R \rightarrow R$ be a continious and nonnegative function.

which of the following statement is TRue ?

a) if there exist $ c \in (0,1)$ such that $f(c) = 100$ then $\int_{0}^{1} f(x) dx \ge \frac {1}{2}.$

b)$\int_{0}^{1} f(x) dx > \frac {1}{2}$.then $f(c) > \frac{1}{2}$ for some $c \in (0,1).$

c)$\int_{0}^{1} f(x) dx = \frac {1}{2}$ then there exist $c\in (0,1)$ such that $f(c) = \frac {1}{2}$

d) None of these

My answer : option b) and C) is true.. by intermediate theorem

For option a) if i take $f(x) = 200x$ now put $x = \frac{1}{2}$..then $\int_{0}^{1} f(x) dx = \frac {200 x^2}{2} |_0^1$...we will not get $\int_{0}^{1} f(x) dx = \frac {1}{2}.$..so option a) is false

Is its right or wrong ?? Pliz tell me

Any hints/ solution

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  • $\begingroup$ How does your example prove that a) is false? $\endgroup$
    – Jo Mo
    May 17, 2018 at 10:36
  • $\begingroup$ if i take $f(x) = 200x$ now put $x = \frac{1}{2}$..then $\int_{0}^{1} f(x) dx = \frac {200 x^2}{2} |_0^1$...we will not get $\int_{0}^{1} f(x) dx = \frac {1}{2}.$..so option a) is false $\endgroup$
    – user525416
    May 17, 2018 at 10:41
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    $\begingroup$ $a\geq b$ is true if either $a>b$ or $a=b$ $\endgroup$
    – Jo Mo
    May 17, 2018 at 10:46
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    $\begingroup$ I would agree that a) is false but your example does not show that. You need to find a function $f$ and a value of $c$ for which $f(c) = 100$ but the value of the integral is less than $\frac{1}{2}$. For b) and c) I think that you need to say much more. $\endgroup$
    – badjohn
    May 17, 2018 at 10:47
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    $\begingroup$ Equivalently, $a\geq b$ is true if and only if $a < b$ is false. $\endgroup$
    – Jo Mo
    May 17, 2018 at 10:47

2 Answers 2

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In the same vein as GNU Supporter's answer, but not as elementary (and neat :D)

Consider another "peak function", namely $$f(x) = 100\ e^{-a(x-\frac{1}{2})^2}\ .$$ Clearly $f(\frac{1}{2}) = 100.$ We can compute

\begin{align*} \int_0^1 100\ e^{-a(x-1/2)^2} \ dx &= 100 \int_{-\frac{1}{2}}^{\frac{1}{2}} e^{-a x^2} \ dx \\ &= \frac{100}{\sqrt{a}} \int_{-\frac{\sqrt{a}}{2}}^{\frac{\sqrt{a}}{2}} e^{-x^2} \ dx \\ &= \frac{100 }{\sqrt{a}} \sqrt{\pi}\ \mathsf{erf}\left(\frac{\sqrt{a}}{2}\right) \end{align*} where $\mathsf{erf}$ is the error function. Now, all we have to do is find for what values of $a$

\begin{align*} \frac{100 }{\sqrt{a}} \sqrt{\pi}\ \mathsf{erf}\left(\frac{\sqrt{a}}{2}\right) < \frac{1}{2}\ . \end{align*} This can be solved with WolframAlpha, and the answer is roughly $a\geq 125\ 664$.

So an example would be with $a = 100^3$, and we get \begin{align*} \int_0^1 100\ e^{-100^3 (x-1/2)^2} \ dx \approx 0.177 < \frac{1}{2} \end{align*}

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You're in the right direction by invoking the MVT to prove (b) and (c).

It remains to show that (a) is false by providing a counterexample: consider a "peak" function which is piecewise linear by joining the points $(0,0)$, $(c-\delta,0),(c,100),(c+\delta,0)$ and $(1,0)$. This traces a triangle with height and base width $100$ and $2\delta$ respectively. The area under the graph is $100\delta$. To make it smaller than $0.5$, set $\delta < 0.05$.

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