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Chapter 4: Problem 19

Find all values of c that satisfy the Mean Value Theorem for Integrals on the given interval. $$ f(x)=|x| ; \quad[0,2] $$

Short Answer

Expert verified
The value of \( c \) that satisfies the Mean Value Theorem is \( c = 1 \).

Step by step solution

01

Understand the Mean Value Theorem for Integrals

The Mean Value Theorem for Integrals states that if a function \( f(x) \) is continuous on the closed interval \([a,b]\), then there exists at least one number \( c \) in the interval \([a,b]\) such that \(\int_a^b f(x) \, dx = f(c) (b - a)\).
02

Define the Continuous Function

The function given is \( f(x) = |x| \). On the interval \([0, 2]\), \( f(x) = x \) because \( x \) is non-negative in this interval. Thus, \( f(x) \) is continuous on \([0, 2]\).
03

Calculate the Definite Integral

Calculate \( \int_0^2 x \, dx \).\[ \int_0^2 x \, dx = \left[ \frac{x^2}{2} \right]_0^2 = \frac{2^2}{2} - \frac{0^2}{2} = 2. \]
04

Apply the Mean Value Theorem for Integrals

Substitute into the theorem formula: \( \int_0^2 f(x) \, dx = f(c) (2 - 0) \).So, \( 2 = f(c) \cdot 2 \).Divide by 2: \( f(c) = 1 \).
05

Solve for \( c \)

Since \( f(c) = c \) for \( c \) in \([0, 2]\), solve for \( c \) when \( c = 1 \). Therefore, \( c = 1 \).

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Key Concepts

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

Integrals
Integrals are a fundamental concept in calculus that help us find things like areas under curves and can even determine accumulated quantities. An integral essentially "sums up" an infinite number of infinitely small areas to provide a total value. There are different types of integrals, but the one we are most interested in here is the definite integral.
  • Definite Integral: This is the kind of integral used when we want to calculate the area under a curve between two specific points on the x-axis, which are known as the limits of integration.
  • Notation: A definite integral of a function f(x) from a to b is denoted as \( \int_a^b f(x) \, dx \), where [a, b] is the interval over which we want to find the area.
  • Calculation: To compute the definite integral, we find the antiderivative of the function, and then evaluate it at the upper limit and lower limit, subtracting the latter from the former.
The Mean Value Theorem for Integrals utilizes the definite integral to guarantee that between any two points where a continuous function is measured, there is at least one point where the function's average value matches the definite integral's overall value.
Continuous Function
In mathematics, a continuous function is a function that has no sudden jumps or breaks. Think of a continuous function as a smooth curve that you can draw without lifting your pen from the paper.
  • Definition: Formally, a function f(x) is continuous on an interval [a, b] if there are no gaps, peaks, or asymptotes within the interval.
  • Importance in Integrals: Continuous functions are crucial when performing integration because the concepts of definite integrals and theorems like the Mean Value Theorem rely on this smoothness over a given interval.
  • Interval Example: In the exercise, since \( f(x) = |x| \) converts to \( f(x) = x \) for \([0, 2]\), it remains continuous due to this straightforward linear form within the interval.
A continuous function means that every point within the interval contributes to the integral value, validating the preconditions for applying the Mean Value Theorem for Integrals.
Definite Integral
The definite integral is a key tool in finding accumulated differences between portions of a function over a specified interval. It connects the usage of integrals with real-world applications, like calculating areas or other cumulative quantities.
  • Role in Mean Value Theorem for Integrals: The definite integral serves as the foundation in calculating an 'average value' of the function over the interval \([a, b]\).
  • Calculation Process: Like in the solution, calculating \( \int_0^2 x \, dx \) involves finding the antiderivative \( \frac{x^2}{2} \) and applying the limits to evaluate as: \[ \left[ \frac{x^2}{2} \right]_0^2 = \frac{2^2}{2} - \frac{0^2}{2} = 2. \]
  • Significance: This calculated value represents a tangible quantity, such as area under the curve between the points, and is equated to \( f(c) \times (b - a) \) as per the Theorem.
This makes the definite integral incredibly valuable, as it's not just abstract theory but a way to connect math to tangible outcomes.

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Most popular questions from this chapter

Find the average value of the function on the given interval. $$ f(x)=x \cos x^{2} ; \quad[0, \sqrt{\pi}] $$

Sketch the graph of the given function over the interval \([a, b] ;\) then divide \([a, b]\) into \(n\) equal subintervals. Finally, calculate the area of the corresponding circumscribed polygon. \(f(x)=3 x-1 ; a=1, b=3, n=4\)

Household electric current can be modeled by the voltage \(V=\hat{V} \sin (120 \pi t+\phi),\) where \(t\) is measured in seconds, \(\hat{V}\) is the maximum value that \(V\) can attain, and \(\phi\) is the phase angle. Such a voltage is usually said to be 60 -cycle, since in 1 second the voltage goes through 60 oscillations. The root-mean-square voltage, usually denoted by \(V_{\mathrm{rms}}\) is defined to be the square root of the average of \(V^{2} .\) Hence $$ V_{\mathrm{rms}}=\sqrt{\int_{\phi}^{1+\phi}(\hat{V} \sin (120 \pi t+\phi))^{2} d t} $$ A good measure of how much heat a given voltage can produce is given by \(V_{\mathrm{rms}}\) (a) Compute the average voltage over 1 second. (b) Compute the average voltage over \(1 / 60\) of a second. (c) Show that \(V_{\mathrm{rms}}=\frac{\hat{V} \sqrt{2}}{2}\) by computing the integral for \(V_{\mathrm{rms}}\) Hint: \(\int \sin ^{2} t d t=-\frac{1}{2} \cos t \sin t+\frac{1}{2} t+C\) (d) If the \(V_{\mathrm{rms}}\) for household current is usually 120 volts, what is the value \(\hat{V}\) in this case?

The velocity function for an object is given. \(A s-\) suming that the object is at the origin at time \(t=0\), find the position at time \(t=4\). $$ v(t)=t / 60 $$

Find all values of c that satisfy the Mean Value Theorem for Integrals on the given interval. $$ S(y)=y^{2} ; \quad[0, b] $$
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