Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

\( c = \frac{b}{\sqrt{3}} \) satisfies the theorem on \([0, b]\).

Step 1: Recall the Mean Value Theorem for Integrals

The Mean Value Theorem for Integrals states: If a function \( f(x) \) is continuous on \([a, b]\), there exists at least one number \( c \) in \([a, b]\) such that \[ f(c) \times (b-a) = \int_{a}^{b} f(x) \, dx. \]

Step 2: Identify \( f(x) \) and the Interval

In this problem, the function given is \( S(y) = y^2 \) and the interval is \([0, b]\). Thus, \( f(y) = y^2 \), \( a = 0 \), and \( b \) is an unspecified upper limit.

Step 3: Calculate the Definite Integral

Calculate the definite integral of \( f(y) = y^2 \) over \([0, b]\):\[ \int_{0}^{b} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{0}^{b} = \frac{b^3}{3} - 0 = \frac{b^3}{3}. \]

Step 4: Set Up the Equation from the Theorem

Set up the equation from the Mean Value Theorem for Integrals: \[ f(c) \times (b - 0) = \int_{0}^{b} y^2 \, dy, \]which simplifies to \[ c^2 \times b = \frac{b^3}{3}. \]

Step 5: Solve for \( c \)

Divide both sides by \( b \) (assuming \( b eq 0 \)) to isolate \( c^2 \):\[ c^2 = \frac{b^2}{3}. \]Taking the square root of both sides gives:\[ c = \pm \frac{b}{\sqrt{3}}. \]

Step 6: Check Validity within the Interval

Since \( c \) must lie within \([0, b]\), only the positive root \( c = \frac{b}{\sqrt{3}} \) is valid, as long as it does not exceed \( b \).

Key Concepts

Definite Integral

The definite integral is a fundamental concept in calculus used to determine the accumulated value of a function over a specified interval. It's represented mathematically as \[ \int_{a}^{b} f(x) \, dx \], where \( a \) and \( b \) are the limits of integration and \( f(x) \) is the function being integrated.

• The lower limit \( a \) is where the integration starts.
• The upper limit \( b \) is where the integration stops.
• The result is the "net area" under the curve of \( f(x) \) from \( a \) to \( b \).

In the context of the Mean Value Theorem for Integrals, the definite integral gives the exact value to compare with the product of \( f(c) \) and the interval \((b-a)\), ensuring there's at least one value \( c \) where this equation holds true. This theorem plays a crucial role in understanding how averages work over intervals in continuous functions.

Continuity on Interval

Continuity on an interval means that a function behaves predictably from start to end without any breaks, jumps, or gaps. A function \( f(x) \) is continuous on an interval \( [a, b] \) if at every point \( x \) in this interval, the limit of \( f(x) \) as \( x \) approaches any point \( c \) in the interval equals \( f(c) \) itself. There are no abrupt changes in value; the graph of \( f(x) \) is unbroken across the interval. For applying the Mean Value Theorem for Integrals:

• The function must be continuous on the closed interval \([a, b]\).
• Continuity ensures that the integral and the specified equation can be meaningfully calculated.
• Without continuity, calculating an integral accurately isn't possible.

In our exercise, \( S(y) = y^2 \) is continuous for all real numbers, making it perfect for this theorem provided \( b \) is a real number, ensuring the application holds true over \([0, b]\). Continuity thus forms the backbone for integration and applying the Mean Value Theorem for Integrals.

Square Root Function

The square root function is crucial for solving equations resulting from the Mean Value Theorem for Integrals when isolating \( c \). This function basically undoes squaring and is written as \( \sqrt{x} \). For any non-negative number, the square root yields a non-negative value. When doing calculations like finding \( c \) in our equation:\[ c^2 = \frac{b^2}{3} \]We take the square root on both sides:\[ c = \pm \frac{b}{\sqrt{3}} \]Here, the operation outputs possible values for \( c \) that solve the equation. Yet, within a bounded interval like \([0, b]\), the non-negative root is generally the valid one, specifically \( c = \frac{b}{\sqrt{3}} \).

• The root chosen must satisfy the interval condition \([0, b]\).
• In cases where \( c \) must be realistic (potentially non-negative in this instance), the correct root fits naturally.

The square root function thus helps us find valid \( c \) values that align with the integral's mean value requirements.