Problem 75 - Calculus

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 of \(|x|\) is \(\frac{1}{2} x|x|\). The integral \(\int_{a}^{b} |x| \, dx\) is \(\frac{1}{2} b|b| - \frac{1}{2} a|a|\).

Step 1: Understanding the Problem

We are given the function \(|x|\) and need to show that \(\frac{1}{2} x|x|\) is its antiderivative. An antiderivative is a function whose derivative gives the original function.

Step 2: Differentiate the Antiderivative

First, differentiate \(F(x) = \frac{1}{2}x|x|\) to see if it gives \(|x|\). Use the definition of absolute value: \(|x| = x\) if \(x \geq 0\) and \(|x| = -x\) if \(x < 0\).

Step 3: Case x >= 0

For \(x \geq 0\), \(|x| = x\). Substitute into \(F(x) = \frac{1}{2}x^2\). Differentiating, \(F'(x) = \frac{d}{dx}\left(\frac{1}{2}x^2\right) = x\), which equals \(|x|\) since \(x \geq 0\).

Step 4: Case x < 0

For \(x < 0\), \(|x| = -x\). Substitute into \(F(x) = -\frac{1}{2}x^2\). Differentiating, \(F'(x) = \frac{d}{dx}\left(-\frac{1}{2}x^2\right) = -x\), which equals \(|x|\) since \(x < 0\).

Step 5: Confirm Antiderivative

Since \(F'(x) = |x|\) for both \(x \geq 0\) and \(x < 0\), \(F(x) = \frac{1}{2}x|x|\) is indeed an antiderivative of \(|x|\).

Step 6: Evaluate the Integral

We can use the antiderivative to compute the definite integral: \(\int_{a}^{b}|x| \, dx = F(b) - F(a)\).

Step 7: Substitute the Limits

Substitute \(F(x) = \frac{1}{2}x|x|\) into \(F(b) - F(a)\), giving \(\int_{a}^{b}|x| \, dx = \frac{1}{2}b|b| - \frac{1}{2}a|a|\). Simplify as needed.

Key Concepts

Definite Integral Evaluation

To evaluate a definite integral, we calculate the "signed area" under a curve, between two specified points on the x-axis. Performing a definite integral requires the antiderivative of the function we wish to integrate. When working with properties of definite integrals, keep these points in mind:

• The limits of integration are critical. They specify the start and end points of the calculation.
• Indefinite integrals give us a family of curves or functions, while definite integrals boil down to a single value. This value might represent an area, depending on the context of the problem.

For this exercise, we use the antiderivative \( \frac{1}{2} x|x| \) to find \( \int_{a}^{b}|x| \, dx \). The solution involves evaluating this function at the upper and lower limits and calculating the difference, following the formula F(b) - F(a). This process helps us find the total 'accumulation' of |x| from point 'a' to point 'b'.
This systematic method of working with definite integrals streamlines the calculation and ensures accuracy as long as the correct antiderivative is used.

Differentiation of Piecewise Functions

Piecewise functions define different expressions in different sections of their domain. Understanding how to handle them is key to performing differentiation correctly.When differentiating piecewise functions, it is crucial to address each segment of the function individually:

• For expressions that hold for \( x \geq 0 \), take the derivative assuming the defined expression in that interval.
• The same applies to the section of the domain where \( x < 0 \), ensuring the result reflects the appropriate expression's derivative.

In the case of \( \frac{1}{2} x |x| \), it involves:

• When \( x \geq 0 \), \( |x| = x \) allows us to differentiate \( \frac{1}{2} x^2 \).
• When \( x < 0 \), \( |x| = -x \) leading us to differentiate \( -\frac{1}{2} x^2 \).

Approaching each piece individually ensures the derivative remains consistent with both the function's behavior and continuity across its entire domain, especially at transition points like zero.

Absolute Value Function

The absolute value function takes any real number and outputs its non-negative magnitude. It is notated as \( |x| \). This function shows distinct behavior based on the sign of \( x \):

• If \( x \geq 0 \), then \( |x| = x \).
• If \( x < 0 \), then \( |x| = -x \).

This dual behavior is why it’s categorized as a piecewise function. The absolute value function's graphical representation is a V-shape with its vertex at the origin. Its derivatives can be tricky because the function's slope changes at \( x = 0 \). This is managed by considering separate derivatives on either side of zero.These properties are utilized in solving integrals or finding derivatives, like in the problem given, to ensure accurate mathematical expressions regardless of the curves encountered.