Question

∣ ∫ ab f ∣∣ ≤∫ ab​ ∣f∣. Give a proof of the statement above by re-ordering the following 9 sentences. Choose from these statments. Suppose f is bounded by B on [a,b], that is ∣f(x)∣≤B for all x∈[a,b]. Since ∥f(x)∥=∣f(x)∣,∣f∣ is bounded by B as well. Next, if f is continuous at c∈[a,b], then by the triangle inequality, we have 0≤∥f(x)−f(c)∥≤∣f(x)−f(c)∣. By continuity of f at c,∣f(x)−f(c)∣→0, and so by the inequality above ||f(x)∣−∣f(c)∥→0, as x→c. This shows that the number of discontinuities of ∣f∣ is at most the number of discontinuities of f and hence must also be finite. so

∣i ∫ ab​ f ∣∣ =max{∫ ab f,−∫ ab f}≤∫ ab f.

Answer 1

The statement ∣ ∫ ab f ∣∣ ≤∫ ab​ ∣f∣ is proved using properties of continuous functions and the boundedness and continuity of f(x) to show that the integral's absolute value is less than or equal to the absolute value's integral.

Step-by-step explanation:

The statement ∣ ∫ ab f ∣∣ ≤∫ ab​ ∣f∣ can be proved by considering properties of absolute values, integrals, and the nature of continuous functions. First, let's assemble the provided sentences in a logical order:

1. Suppose f is bounded by B on [a,b], that is ∣f(x)∣≤B for all x∈[a,b].
2. Since ∣∣f(x)∣∣=∣f(x)∣,∣f∣ is bounded by B as well.
3. Next, if f is continuous at c∈[a,b], then by the triangle inequality, we have 0≤∣∣f(x)−f(c)∣∣≤∣f(x)−f(c)∣.
4. By continuity of f at c,∣f(x)−f(c)∣→0, and so by the inequality above ∣∣f(x)∣∣∣f(c)∣∣∣→0, as x→c.
5. This shows that the number of discontinuities of ∣f∣ is at most the number of discontinuities of f and hence must also be finite.
6. so ∣i ∫ ab​ f ∣∣ =max{∫ ab f,−∫ ab f}≤∫ ab f.

This structured proof outlines how the bounded nature of f(x) and the continuity principles lead to the conclusion that the absolute value of the integral of f from a to b is less than or equal to the integral of the absolute value of f over the same interval.