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Math Help - function integrable

  1. #1
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    function integrable

    Suppose  a<c<b<d . Let  k be any non-zero real number. Define  f: \mathbb{R} \to \mathbb{R} as follows:  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  f is Riemann Integrable on  [a,b] . What is its integral?

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

    For (a) the integral would be  k(d-c) . Here is a proof. Proof. Let  \epsilon >0 . Choose  \delta >0 such that  \delta < \frac{\epsilon}{2|d-c|} . Let  P = \{x_{0}, x_{1}, x_{2}, \ldots, x_{n-1}, x_{n} \} be a partition of  [a,b] with  ||P|| < \delta . Choose  x_{1}^{*}, x_{2}^{*}, \ldots , x_{n}^{*} arbitrarily such that  x_{i-1} \leq x_{i}^{*} \leq x_{i} . Then  \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  |\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  0 and we are only worried about the "middle" sum?
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