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Problem 187

Given that a bounded function \(\mathrm{f}(\mathrm{x})\) is Riemann integrable in \([a, b]\) if and only if given any \(\varepsilon>0\) there exists a partition with upper and lower sums \(\mathrm{U}\) and \(\mathrm{L}\) such that \(\mathrm{U}-\mathrm{L}<\varepsilon\), prove that a continuous function \(\mathrm{f}(\mathrm{x})\) in \([\mathrm{a}, \mathrm{b}]\) is Riemann integrable in \([\mathrm{a}, \mathrm{b}]\).

Short Answer

Expert verified
To prove that a continuous function \(f(x)\) in \([a, b]\) is Riemann integrable, follow these steps: 1. Understand that a continuous function on \([a, b]\) is uniformly continuous, so for every \(\varepsilon>0\), there is a \(\delta>0\) for which \(|x-y|<\delta\) implies \(|f(x)-f(y)|<\frac{\varepsilon}{b-a}\). 2. Divide the interval \([a, b]\) into equally spaced subintervals using a partition with \(\Delta x = \frac{b-a}{n}<\delta\). 3. Find the minimum and maximum values of \(f(x)\) in each subinterval. 4. Calculate the lower sums \(L\) and upper sums \(U\) using these minimum and maximum values. 5. Show that \(U-L<\varepsilon\) by observing that \(U-L \leq \Sigma \frac{\varepsilon}{b-a} * \Delta x =\varepsilon\). 6. Conclude that the continuous function \(f(x)\) is Riemann integrable, as we have shown that there exists a partition with upper and lower sums \(\mathrm{U}\) and \(\mathrm{L}\) such that \(U-L<\varepsilon\).
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Step 1: Understand the definitions and properties of continuous functions and Riemann integrable functions.

A continuous function is a function whose graph has no breaks or gaps. A function is Riemann integrable if and only if for every ε>0, there exists a partition with upper and lower sums U and L such that U-L<ε. A continuous function on a closed interval [a, b] is uniformly continuous. Uniform continuity states that for every ε>0, there exists a δ>0 such that if the distance between any two points x and y in [a, b] is less than δ, then the distance between f(x) and f(y) is less than ε.

Step 2: Divide the interval [a, b] into equally spaced subintervals.

Choose a partition of the interval [a, b] such that the length of each partition is: Δx = (b - a) / n, where n is a positive integer. Since the function is uniformly continuous, there must exist a δ>0 such that, whenever |x-y|<δ, |f(x)-f(y)|<ε/(b-a). Now, select n sufficiently large such that: Δx = (b - a) / n < δ.

Step 3: Find upper and lower bounds for each subinterval.

For each subinterval [x_i, x_(i+1)], find the minimum value m_i and the maximum value M_i of the function f(x) in that subinterval. Since f(x) is continuous, the minimum and maximum values exist.

Step 4: Calculate the lower sum L and the upper sum U.

Using the m_i and M_i found in step 3, calculate the lower sum L and the upper sum U for the partition as follows: L = Σ (m_i * Δx) for i=1 to n, U = Σ (M_i * Δx) for i=1 to n. Here, the symbol Σ denotes taking the sums of the products for each subinterval.

Step 5: Show that U-L < ε.

Using the results of steps 2 and 4, observe that the difference U-L for the partition can be derived as follows: U - L = Σ (M_i * Δx) - Σ (m_i * Δx) = Σ (M_i - m_i) * Δx Since |f(x)-f(y)|<ε/(b-a) for all x and y in the same subinterval, we have that: U - L ≤ Σ ε/(b-a) * Δx = (ε/(b-a)) * Σ Δx = (ε/(b-a)) * (b - a) = ε This means that for the given partition and any ε>0, U-L < ε.

Step 6: Conclude that the continuous function f(x) is Riemann integrable.

Since for any ε>0, there exists a partition with upper and lower sums U and L such that U-L<ε, it follows that the continuous function f(x) in [a, b] is Riemann integrable in [a, b].

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Most popular questions from this chapter

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