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Thread: function integrable

  1. #1
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    function integrable

    Suppose $\displaystyle a<c<b<d $. Let $\displaystyle k $ be any non-zero real number. Define $\displaystyle f: \mathbb{R} \to \mathbb{R} $ as follows: $\displaystyle f(x) = \begin{cases} 0 \ \ \ \ \ \text{if} \ x < c \\ k \ \ \ \ \ \text{if} \ c \leq x \leq d \\ 0 \ \ \ \ \ \text{if} \ d < x \end{cases} $

    (a) Prove that $\displaystyle f $ is Riemann Integrable on $\displaystyle [a,b] $. What is its integral?

    (b) If every $\displaystyle < $ in definition of $\displaystyle f $ were changed to $\displaystyle \leq $ and vice versa, how would the answer change?

    For (a) the integral would be $\displaystyle k(d-c) $. Here is a proof. Proof. Let $\displaystyle \epsilon >0 $. Choose $\displaystyle \delta >0 $ such that $\displaystyle \delta < \frac{\epsilon}{2|d-c|} $. Let $\displaystyle P = \{x_{0}, x_{1}, x_{2}, \ldots, x_{n-1}, x_{n} \} $ be a partition of $\displaystyle [a,b] $ with $\displaystyle ||P|| < \delta $. Choose $\displaystyle x_{1}^{*}, x_{2}^{*}, \ldots , x_{n}^{*} $ arbitrarily such that $\displaystyle x_{i-1} \leq x_{i}^{*} \leq x_{i} $. Then $\displaystyle \mathcal{R}(f,P) = \sum_{i=1}^{n} f(x_{i}^{*})(x_{i}-x_{i-1}) $. We have to break this up into three sums? Because ultimately we want to show that $\displaystyle |\mathcal{R}(f,P)-k(d-c)| < \epsilon $.
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  2. #2
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    For any such partition, you can make a finer partition by adding c and d as "break points". Once you do that, the three sums are simple.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    For any such partition, you can make a finer partition by adding c and d as "break points". Once you do that, the three sums are simple.
    Because then two sums are $\displaystyle 0 $ and we are only worried about the "middle" sum?
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