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Problem 190

# Prove the following: a) If $$\mathrm{f}(\mathrm{x}) \leq \mathrm{g}(\mathrm{x})$$ and $$\mathrm{f}(\mathrm{x})$$ and $$\mathrm{g}(\mathrm{x})$$ are Riemann integrable on $$[\mathrm{a}, \mathrm{b}]$$ then $$\mathrm{b} \int_{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx} \leq \mathrm{b}_{\mathrm{a}} \mathrm{g}(\mathrm{x}) \mathrm{d} \mathrm{x}$$ b) If $$\mathrm{f}(\mathrm{x})$$ is bounded and Riemann integrable and if $$\mathrm{c}$$ is any point such that $$\mathrm{C} \in[\mathrm{a}, \mathrm{b}]$$ then $$\mathrm{b} \int_{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\mathrm{c} \int_{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}+\mathrm{b} \int_{\mathrm{c}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}$$

Expert verified
In summary, we proved the following: a) If $$f(x) \leq g(x)$$, and both functions are Riemann integrable on the interval [a, b], then $$\int_a^b f(x) dx \leq \int_a^b g(x) dx$$. We used the definitions of upper and lower sums and the properties of Riemann integrable functions to show this inequality. b) If $$f(x)$$ is bounded and Riemann integrable and c is any point in [a, b], then $$\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx$$. We demonstrated this by partitioning the interval [a, b] and showing that the sums of the respective partitions give us the integral over the entire interval.
See the step by step solution

## Step 1: Part a: Proving $$\int_a^b f(x) dx \leq \int_a^b g(x) dx$$ given $$f(x) \leq g(x)$$

1. The function f(x) and g(x) are Riemann integrable on the interval [a, b]. Thus, for any partition P, we can define the upper and lower sums of both functions. Let U(f, P) and L(f, P) denote the upper and lower sums of f on P, and let U(g, P) and L(g, P) denote the upper and lower sums of g on P. 2. Since f(x) ≤ g(x) for every x in [a, b], it follows that: $$L(f, P) \leq L(g, P) \leq U(g, P) \leq U(f, P)$$. 3. If we can take any arbitrary partition P, the upper and lower sums of the integrals of f(x) and g(x) can be found as follows: $\int_a^b L(f, P) dx \leq \int_a^b L(g, P) dx \leq \int_a^b U(g, P) dx \leq \int_a^b U(f, P) dx$ 4. Now, we use the property that if a function is Riemann integrable, then the lower integral and upper integral are equal. Since f(x) and g(x) are Riemann integrable: $\int_a^b f(x) dx = \int_a^b L(f, P) dx \quad \text{and} \quad \int_a^b g(x) dx = \int_a^b U(g, P) dx$ 5. Therefore, $$\int_a^b f(x) dx \leq \int_a^b g(x) dx$$, which concludes the proof for part a.

## Step 2: Part b: Proving $$\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx$$ given $$f(x)$$ is Riemann integrable and bounded.

1. Let's consider a partition P on [a, b] that includes the point c, so P contains a, c, and b in the proper sequence. 2. Find the upper sums for f(x) on the subintervals [a, c] and [c, b]: U(f, P, [a,c]) and U(f, P, [c,b]). Do the same for the lower sums: L(f, P, [a,c]) and L(f, P, [c,b]). 3. Then, we have the sums for the whole interval [a, b]: U(f, P) = U(f, P, [a,c]) + U(f, P, [c,b]) and L(f, P) = L(f, P, [a,c]) + L(f, P, [c,b]). 4. This means that upper sums and lower sums on [a, b] can be represented as the sum of their respective values over [a, c] and [c, b]. 5. Again, use the property of Riemann integrability that lower integral equals upper integral. Since f(x) is Riemann integrable on [a, b]: $\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx$ 6. Hence, the result is proven for part b.

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