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Chapter 5: Problem 32

Determine whether the statement is true or false. Explain your answer. The function \(f(x)=\left\\{\begin{array}{ll}0, & x \leq 0 \\ x^{2}, & x>0\end{array}\right.\) is integrable over every closed interval \([a, b]\).

Short Answer

Expert verified
True, the function is integrable over every closed interval \\[a, b\\].

Step by step solution

01

Understanding the Concept of Integrability

A function is said to be integrable over a closed interval \([a, b]\) if the definite integral \(\int_a^b f(x) \, dx\) exists. According to the fundamental theorem of calculus, functions that are bounded and have a finite number of discontinuities over a closed interval are usually integrable.
02

Analyzing the Function's Properties

The function \(f(x)\) is defined as \(f(x) = 0\) when \(x \leq 0\) and \(f(x) = x^2\) when \(x > 0\). This is a piecewise function. Examine each part: \(x \leq 0\) is constant at 0, and \(x^2\) is continuous for \(x > 0\).
03

Checking for Discontinuities

The function is continuous for \(x > 0\) and constant for \(x \leq 0\). At \(x = 0\), the function moves from 0 to \(0^2 = 0\). Since there is no jump or vertical discontinuity at \(x=0\), the function is continuous.
04

Evaluating Integrability over any Interval \([a, b]\)

Given the function is continuous over any interval it goes through (including \([a, b]\)), and its individual sections are either constant or a polynomial (both of which are integrable), the function \(f(x)\) is integrable over every closed interval \([a, b]\).
05

Conclusion

The function \(f(x)\) satisfies the conditions for integrability over any closed interval due to its continuity and boundedness. Therefore, the statement is true.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integral
The concept of a definite integral revolves around finding the area under a curve between two points on the x-axis. This is given as \( \int_a^b f(x) \, dx \), which computes the exact accumulation of values provided by the function \( f(x) \) from \( x = a \) to \( x = b \).

Definite integrals require the function to be bounded on the interval and have only a finite number of discontinuities. Generally, if these conditions are met, the definite integral exists. In this exercise, checking definite integrability involves ensuring that the function behaves well鈥攎eaning it's continuous and doesn鈥檛 spiral into infinity鈥攎aking the computation of the area under the curve possible.
Piecewise Functions
Piecewise functions are functions that have different expressions or rules depending on the input value of \( x \). They "switch" definitions at certain points and can represent complex behaviors in simpler chunks.

For example, in the exercise, \( f(x) = \left\{\begin{array}{ll} 0, & x \leq 0 \ x^2, & x > 0 \end{array}\right.\) exemplifies a piecewise function. Here, for any \( x \leq 0 \), the function outputs a constant 0, and for \( x > 0 \), it follows the quadratic rule \( x^2 \). This piecewise distinction allows a function to remain flexible and adapt to different conditions or domains within the same structure.

Understanding the behavior of each "piece" assists in analyzing the overall function's properties, such as continuity and integrability, across the chosen intervals.
Continuity
Continuity of a function over an interval means there are no abrupt shifts or gaps in its graphical representation over that range. A continuous function smoothly transitions from one point to another without "breaking" or "jumping."

In our exercise, the function \( f(x) \) is continuous since at the transition point \( x = 0 \), the value changes smoothly from \( 0 \) on the left to \( 0 \) from the right, avoiding any discontinuity. This seamless connection of values ensures that \( f(x) \) can be considered continuous across any closed interval \([a, b]\).

The focus on continuity matters because the conditions for a definite integral heavily depend on it. A function that's continuous on a specified interval is more likely to be integrable, helping to confirm the statement's truth regarding integrability in this exercise.

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Most popular questions from this chapter

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Electricity is supplied to homes in the form of alternating current, which means that the voltage has a sinusoidal waveform described by an equation of the form $$V=V_{p} \sin (2 \pi f t)$$ (see the accompanying figure). In this equation, \(V_{p}\) is called the peak voltage or amplitude of the current, \(f\) is called its frequency, and \(1 / f\) is called its period. The voltages \(V\) and \(V_{p}\) are measured in volts (V), the time \(t\) is measured in seconds (s), and the frequency is measured in hertz (Hz). (1 \(\mathrm{Hz}=1\) cycle per second; a cycle is the electrical term for one period of the waveform.) Most alternating-current voltmeters read what is called the rms or root -mean-square value of \(V .\) By definition, this is the square root of the average value of \(V^{2}\) over one period. (a) Show that $$V_{\mathrm{rms}}=\frac{V_{p}}{\sqrt{2}}$$ [Hint: Compute the average over the cycle from \(t=0\) to \(t=1 / f,\) and use the identity \(\sin ^{2} \theta=\frac{1}{2}(1-\cos 2 \theta)\) to help evaluate the integral. \(]\) (b) In the United States, electrical outlets supply alternating current with an rms voltage of \(120 \mathrm{V}\) at a frequency of \(60 \mathrm{Hz}\). What is the peak voltage at such an outlet? (Check your book to see figure)

It can be shown that every interval contains both rational and irrational numbers. Accepting this to be so, do you believe that the function $$ f(x)=\left\\{\begin{array}{lll} 1 & \text { if } & x \text { is rational } \\ 0 & \text { if } & x \text { is irrational } \end{array}\right. $$ is integrable on a closed interval \([a, b] ?\) Explain your reasoning.
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