Problem 79 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that \(\frac{1}{2} x|x|\) is an antiderivative of \(|x|\), and use this fact to get a simple formula for \(\int_{a}^{b}|x| d x\).

Step-by-Step Solution

Answer

The antiderivative is verified, and the integral is \(\frac{1}{2}b|b| - \frac{1}{2}a|a|\).

Step 1: Understand the Function

We are given the function \(f(x) = \frac{1}{2} x|x|\) and need to verify it serves as an antiderivative of \(|x|\). This means we need to differentiate \(\frac{1}{2} x|x|\) and check if it equals \(|x|\).

Step 2: Differentiating \(\frac{1}{2} x|x|\)

The function \(\frac{1}{2} x |x|\) can be split into two cases: \(x \geq 0\) where \(|x| = x\), and \(x < 0\) where \(|x| = -x\). For \(x \geq 0\), \(\frac{1}{2}x|x| = \frac{1}{2}x^2\). For \(x < 0\), \(\frac{1}{2}x|x| = -\frac{1}{2}x^2\). Now, differentiate for both cases.

Step 3: Differentiate for \(x \geq 0\)

For \(x \geq 0\), the derivative of \(\frac{1}{2}x^2\) is \(x\). Since \(x = |x|\) when \(x \geq 0\), the derivative is \(|x|\).

Step 4: Differentiate for \(x < 0\)

For \(x < 0\), the derivative of \(-\frac{1}{2}x^2\) is \(-x\). Since \(x < 0\), \(-x = |x|\). Thus, the derivative in this case is also \(|x|\).

Step 5: Conclude the Antiderivative Verification

Since in both \(x \geq 0\) and \(x < 0\) cases, the derivative of \(\frac{1}{2} x|x|\) is \(|x|\), we confirm that \(\frac{1}{2} x|x|\) is indeed an antiderivative of \(|x|\).

Step 6: Setup the Integral for \(\int_{a}^{b}|x| dx\)

Use the fundamental theorem of calculus. Since \(\frac{1}{2}x|x|\) is an antiderivative of \(|x|\), \(\int_{a}^{b}|x| \, dx = \left[ \frac{1}{2} x|x| \right]_a^b\).

Step 7: Evaluate the Definite Integral

Compute \(\left[ \frac{1}{2} x|x| \right]_a^b\) as \(\frac{1}{2}b|b| - \frac{1}{2}a|a|\). This uses the expression for the antiderivative at limits \(b\) and \(a\).

Key Concepts

Absolute Value Function

The absolute value function, denoted as

• \(|x|\),

is a mathematical function that measures the distance of a number from zero on a number line. It returns the non-negative value of any real number. This function is characterized by two key cases:

• For any non-negative number \(x\), the absolute value is simply \(x\) itself.
• For any negative number \(x\), it is \(-x\), effectively making it positive.

This piecewise nature makes the absolute value function special, as it changes form depending on the sign of the input. It's essential in many fields of mathematics and real-world applications where distance, regardless of direction, is a factor. The absolute value also plays a key role in our exercise, where we explore its antiderivative. Understanding how to manipulate the absolute value function algebraically is fundamental in calculus, especially for evaluating integrals.

Definite Integral

A definite integral is a powerful tool in calculus that calculates the net area under a curve

• between two points on a graph.

The notation for a definite integral from \(a\) to \(b\) of a function \(f(x)\) is written as: \[ \int_{a}^{b} f(x) \, dx \] This concept isolates a specific segment of the function to analyze its behavior or cumulative value across that interval. The boundaries, \(a\) and \(b\), tell us where to start and stop on the x-axis.In our exercise, the definite integral is used to find the area under the absolute value function

• \(|x|\) between two points, \(a\) and \(b\).

By calculating this, we determined that the definite integral \[ \int_{a}^{b} |x| \, dx = \left[ \frac{1}{2} x|x| \right]_a^b,\]requires evaluating the antiderivative at both endpoints and subtracting the two results. Definite integrals are crucial for solving real-world problems involving areas, volumes, and other quantities represented by this 'net signed area.'

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone in calculus that bridges the concept of differentiation and integration, two principal operations in calculus. It can be expressed in two parts:

• The first part states that the integral of a function over an interval can be found by using any of its antiderivatives.
• The second part says that the derivative of an integral function is the original function \(f(x)\).

This theorem plays a crucial role in our exercise by enabling us to evaluate the definite integral of the absolute value function. Here's how it works:By finding an antiderivative

• (like \(\frac{1}{2}x|x|\) for \(|x|\)),

we can compute the integral by taking the antiderivative's value at the boundary \(b\), then subtracting its value at the boundary \(a\). This application simplifies complex calculations and connects the concepts of areas under curves with the slopes of their functions. Using this fundamental understanding, we effectively computed the integral of \(|x|\), linking antiderivatives to definite integrals.