Let's solve both parts step-by-step: --- ## (A) Proof: **Statement:** If on an interval \([a, b]\), \(f(x)\) is continuous and non-negative and \(f(c) > 0\) for some \(c \in [a, b]\), then \(\int_a^b f(x)\,dx > 0\). ### **Step 1: Using Continuity** Because \(f\) is continuous at \(c\) and \(f(c) > 0\), by the definition of continuity, for \(\epsilon = \frac{f(c)}{2} > 0\), there exists \(\delta > 0\) such that for all \(x\) with \(|x - c| < \delta\) and \(x \in [a, b]\): \[ |f(x) - f(c)| < \frac{f(c)}{2} \] So: \[ f(x) > f(c) - \frac{f(c)}{2} = \frac{f(c)}{2} > 0 \] for all \(x\) in \((c-\delta, c+\delta) \cap [a, b]\). ### **Step 2: Consider a Subinterval** Let \(I = [c_1, c_2] \subset [a, b]\) where \(c_1 = \max(a, c-\delta)\) and \(c_2 = \min(b, c+\delta)\). On this interval, \(f(x) > \frac{f(c)}{2}\). ### **Step 3: Lower Bound for the Integral** Therefore, \[ \int_a^b f(x)\,dx \geq \int_{c_1}^{c_2} f(x)\,dx > \int_{c_1}^{c_2} \frac{f(c)}{2} dx = \frac{f(c)}{2}(c_2 - c_1) \] Since \(c_2 > c_1\), \(c_2 - c_1 > 0\), so the above expression is positive. ### **Step 4: Conclusion** \[ \boxed{\int_a^b f(x)\,dx > 0} \] --- ## (B) Deduction: **Statement:** If \(f, g\) are continuous and \(f \leq g\) on \([a, b]\), and there exists \(c \in [a, b]\) such that \(f(c) < g(c)\), then \(\int_a^b f(x)\,dx < \int_a^b g(x)\,dx\). ### **Step 1: Consider \(h(x) = g(x) - f(x)\)** \(h(x)\) is continuous (difference of continuous functions), and \(h(x) \geq 0\) for all \(x\) (since \(f \leq g\)), and \(h(c) = g(c) - f(c) > 0\). ### **Step 2: Apply (A) to \(h(x)\)** By the result from (A), since \(h(x)\) is continuous, non-negative, and positive at some point, we have: \[ \int_a^b h(x) dx > 0 \] But, \[ \int_a^b h(x) dx = \int_a^b (g(x) - f(x)) dx = \int_a^b g(x) dx - \int_a^b f(x) dx \] ### **Step 3: Rearrange** \[ \int_a^b g(x) dx - \int_a^b f(x) dx > 0 \implies \int_a^b f(x) dx < \int_a^b g(x) dx \] ### **Final Answer:** \[ \boxed{\int_a^b f(x)\,dx < \int_a^b g(x)\,dx} \] --- ### **Summary:** **(A)** If \(f(x)\) is continuous, non-negative, and positive at some point, then \(\int_a^b f(x)\,dx > 0\). **(B)** If \(f \leq g\) and at some \(c\), \(f(c) < g(c)\), then \(\int_a^b f(x)\,dx < \int_a^b g(x)\,dx\).