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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