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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.

Short Answer

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].
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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].

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Most popular questions from this chapter

Chapter 7

Suppose \(g\) is a continuously differentiable monotonically increasing function on \([a, b]\) and \(f\) is bounded on \([a, b]\). Prove that $$ b \int_{a} f d g=b \int_{a} f(x) g^{\prime}(x) d x $$ Use this result to find the total mass of a linear distribution on \([a, b]\) with a continuous density function \(\rho(x)\).

Chapter 7

Suppose f is defined on \([0,2]\) as follows: $$ \begin{aligned} &\mathrm{f}(\mathrm{x})=1 \text { for } 0 \leq \mathrm{x}<1 \\ &\text { and }=2 \text { for } 1 \leq \mathrm{x} \leq 2 \text { . } \end{aligned} $$ Show that \(\mathrm{f}\) is Riemann integrable.

Chapter 7

Find the derivatives of a) $\mathrm{x} \int_{1} \mathrm{t}^{2} \mathrm{dt} \quad\( with respect to \)\mathrm{x}$. b) ${ }^{\text {(t) } 2} \int_{1} \sin \left(\mathrm{x}^{2}\right) \mathrm{d} \mathrm{x}$ with respect to t.

Chapter 7

Given that \(\mathrm{f}_{1}(\mathrm{x})\) and \(\mathrm{f}_{2}(\mathrm{x})\) are Riemann integrable on \([\mathrm{a}, \mathrm{b}]\) and that $\mathrm{I} \geq \mathrm{J}$ (where I and J represent the upper and lower integrals, respectively, of any Riemann integrable function), prove that \(\mathrm{f}_{1}(\mathrm{x})+\mathrm{f}_{2}(\mathrm{x})\) is Riemann integrable on \([\mathrm{a}, \mathrm{b}]\) and that $\mathrm{b} \int_{\mathrm{a}}\left[\mathrm{f}_{1}(\mathrm{x})+\mathrm{f}_{2}(\mathrm{x})\right] \mathrm{d} \mathrm{x}=\mathrm{b} \int_{\mathrm{a}} \mathrm{f}_{1}(\mathrm{x}) \mathrm{dx}+\mathrm{b} \int_{\mathrm{a}} \mathrm{f}_{2}(\mathrm{x}) \mathrm{dx}$

Chapter 7

Evaluate the Stieltjes integral $^{1} \int_{-1} \mathrm{x} \mathrm{d}|\mathrm{x}|$.

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