 Suggested languages for you:

Europe

Problem 186

# Suppose $$\mathrm{f}$$ is defined on $$[0,2]$$ by $$\mathrm{f}(\mathrm{x})=\mathrm{x}$$ if $$0 \leq \mathrm{x}<1$$, $$\mathrm{f}(\mathrm{x})=\mathrm{x}-1$$ if $$1 \leq \mathrm{x} \leq 2$$. Show that $$\mathrm{f}$$ is integrable.

Expert verified
We are given a piecewise linear function $$f(x) = x$$ if $$0 \leq x < 1$$, and $$f(x) = x - 1$$ if $$1 \leq x \leq 2$$. We set the partition $$P = \{0, 1, 2\}$$ and calculate the lower sum $$L(P,f) = 0$$ and the upper sum $$U(P, f) = 2$$. As these values do not change regardless of partition refinement, we can conclude that f(x) is integrable on the interval [0, 2].
See the step by step solution

## Step 1: Examine the given function

The function is provided as: f(x) = x, if 0 ≤ x < 1, f(x) = x - 1, if 1 ≤ x ≤ 2 This is a piecewise linear function on the interval [0, 2].

## Step 2: Establish the partition

Set the partition P = {0, 1, 2} which consists of two intervals: [0, 1] and [1, 2]. This partition is chosen because the function changes its definition at x = 1.

## Step 3: Determine the lower and upper sums

For each interval in the partition, determine the infimum (lowest value) and supremum (highest value) of the function on that interval. Then calculate the lower sum (L(P, f)) and the upper sum (U(P, f)). Interval [0, 1]: Infimum (m1) = 0 Supremum (M1) = 1 - 0 = 1 Interval [1, 2]: Infimum (m2) = 1 - 1 = 0 Supremum (M2) = 2 - 1 = 1 Lower sum (L(P, f)): L(P, f) = m1 * (1-0) + m2 * (2-1) = 0 * 1 + 0 * 1 = 0 Upper sum (U(P, f)): U(P, f) = M1 * (1-0) + M2 * (2-1) = 1 * 1 + 1 * 1 = 2

## Step 4: Refine the partition and recalculate the lower and upper sums

As the two intervals are linear functions, the infimum and supremum values remain constant for any refinement of the partition. Therefore, lower and upper sums will also remain the same regardless of the refinement.

## Step 5: Show the function is integrable

As the lower and upper sums of the function do not change regardless of the partition refinement, and we can find a partition such that for any predetermined ε > 0, the difference between the lower and upper sums is less than ε (in this case, L(P, f) = 0 and U(P, f) = 2, and U(P, f) - L(P, f) = 2 - 0 = 2 < ε), we can conclude that f(x) is integrable on the interval [0, 2].

We value your feedback to improve our textbook solutions.

## Access millions of textbook solutions in one place

• Access over 3 million high quality textbook solutions
• Access our popular flashcard, quiz, mock-exam and notes features ## Join over 22 million students in learning with our Vaia App

The first learning app that truly has everything you need to ace your exams in one place.

• Flashcards & Quizzes
• AI Study Assistant
• Smart Note-Taking
• Mock-Exams
• Study Planner 