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 June 25th, 2014, 05:20 AM #1 Senior Member   Joined: Apr 2013 Posts: 425 Thanks: 24 Two integrals Hello! Either $\displaystyle f:[0.1] \to [1,2]$ a continuous function with the property $\displaystyle \int_0^1 f(x) dx = 2$.To show that $\displaystyle \int _ 0 ^1 \frac { 1 }{ f(x) } dx = \frac{ 1 }{ 2 }$.
 June 25th, 2014, 05:45 AM #2 Member   Joined: Apr 2014 From: norwich Posts: 84 Thanks: 9 not quite understanding the question If you can explain why you wrote "Either...", at the beginning of the sentence and I understand, I'll give this problem a shot... ... shouldn't there be an "or", after an "either"?
 June 25th, 2014, 06:19 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,937 Thanks: 2265 Math Focus: Mainly analysis and algebra It should be "if". $$\newcommand{\d}[1]{\,\mathbb{d}{#1}} \\ f:[0,1] \mapsto [1, 2] \implies f(x) \le 2 \qquad \forall x \in [0,1] \\ \text{Thus} \qquad \int \limits_0^1 f(x)\d{x} \le \int \limits_0^1 2 \d{x} = 2x \bigg|_0^1 = 2 \\ \text{with equality when} \qquad f(x) = 2 \qquad \forall x \in [0,1] \\[12pt] \text{Thus} \qquad \int \limits_0^1 \frac{1}{f(x)}\d{x} = \int \limits_0^1 \frac{1}{2}\d{x} = \frac{1}{2} \\$$ I don't need f(x) to be continuous. It is sufficient that is be discontinuous at finitely many points. Last edited by v8archie; June 25th, 2014 at 06:36 AM.
June 28th, 2014, 04:56 AM   #4
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 Originally Posted by v8archie It should be "if". $$\newcommand{\d}[1]{\,\mathbb{d}{#1}} \\ f:[0,1] \mapsto [1, 2] \implies f(x) \le 2 \qquad \forall x \in [0,1] \\ \text{Thus} \qquad \int \limits_0^1 f(x)\d{x} \le \int \limits_0^1 2 \d{x} = 2x \bigg|_0^1 = 2 \\ \text{with equality when} \qquad f(x) = 2 \qquad \forall x \in [0,1] \\[12pt] \text{Thus} \qquad \int \limits_0^1 \frac{1}{f(x)}\d{x} = \int \limits_0^1 \frac{1}{2}\d{x} = \frac{1}{2} \\$$ I don't need f(x) to be continuous. It is sufficient that is be discontinuous at finitely many points.
It may be that $\displaystyle f(x)=\tan(ax+b)$?

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