Problem 20 - 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. $$ f(x)=|x| ; \quad[-2,2] $$

Step-by-Step Solution

Answer

The values of \(c\) are 1 and -1.

Step 1: Understanding the Theorem's Requirement

The Mean Value Theorem for Integrals states that for a continuous function on a closed interval \([a, b]\), there exists at least one \(c\) in \([a, b]\) such that: \[f(c) \cdot (b - a) = \int_{a}^{b} f(x) \, dx.\] Here, \(f(x) = |x|\) and the interval is \([-2, 2]\). Since \(f(x)\) is continuous on \([-2, 2]\), we can apply the theorem.

Step 2: Calculate the Definite Integral

Compute the integral \(\int_{-2}^{2} |x| \, dx\). Since the absolute value function \(f(x) = |x|\) splits into two parts, write it as \(f(x) = -x\) for \(x < 0\) and \(f(x) = x\) for \(x \geq 0\). Therefore, \[\int_{-2}^{2} |x| \, dx = \int_{-2}^{0} -x \, dx + \int_{0}^{2} x \, dx.\] Evaluate each integral separately: \[\int_{-2}^{0} -x \, dx = \left[-\frac{x^2}{2}\right]_{-2}^{0} = 0 - \frac{(-2)^2}{2} = -2,\] \[\int_{0}^{2} x \, dx = \left[\frac{x^2}{2}\right]_{0}^{2} = \frac{4}{2} - 0 = 2.\] Thus, \(-2 + 2 = 4\) is the value of the definite integral.

Step 3: Set up the Equation with MVT

According to the Mean Value Theorem for Integrals, we need \[f(c) \cdot (b - a) = 4,\] where \(b = 2\) and \(a = -2\). This simplifies to \[f(c) \cdot 4 = 4.\] So we have \(f(c) = 1\).

Step 4: Solve for the Value of \(c\)

Given that \(f(c) = |c| = 1\), the equation is satisfied when \(|c| = 1\). Thus, the possible values for \(c\) are \(c = 1\) and \(c = -1\). Both values satisfy \(c \in [-2, 2]\).

Key Concepts

Absolute Value Function

The absolute value function, denoted as \(f(x) = |x|\), is a common mathematical function that measures the distance of a number from zero on the real number line. This distance is always a non-negative value.
Key properties of the absolute value function include:

• It outputs non-negative numbers. For any real number \x, |x| \geq 0\.
• The function is symmetric around the y-axis, meaning that \|x| = |-x|\.
• It changes behavior at the point where \x = 0\. For values less than zero, the absolute value negates them (\