Problem 44 - Calculus: Early Transcendentals

Question

Evaluate the integral. \( \displaystyle \int^{2}_{0} \mid 2x - 1 \mid \,dx \)

Step-by-Step Solution

Answer

The integral evaluates to 2.5.

Step 1: Understand the Absolute Value Function

The absolute value function transforms inputs according to whether they are positive or negative. In this case, \( |2x - 1| \) breaks into two parts based on the condition \( 2x - 1 = 0 \). Solving \( 2x - 1 = 0 \) leads to \( x = 0.5 \). This tells us where the behavior of the function changes.

Step 2: Split the Integral at the Critical Point

Since the critical point is \( x = 0.5 \), we divide the integral into two parts from \( 0 \) to \( 0.5 \) and from \( 0.5 \) to \( 2 \). This gives us: \[ \int^{2}_{0} |2x - 1| \,dx = \int^{0.5}_{0} |2x - 1| \,dx + \int^{2}_{0.5} |2x - 1| \,dx \]

Step 3: Evaluate Each Part of the Integral

**For \( 0 \leq x \leq 0.5 \):**Since \( 2x - 1 \leq 0 \), \(|2x - 1| = -(2x - 1)\).Calculate: \[ \int^{0.5}_{0} -(2x - 1) \,dx = \int^{0.5}_{0} (-2x + 1) \,dx \]Using the power rule for integration:\[ = \left[ -x^2 + x \right]^{0.5}_{0} = \left[ -(0.5)^2 + 0.5 \right] - [0] \]\[ = -0.25 + 0.5 = 0.25 \]**For \( 0.5 \leq x \leq 2 \):**Since \( 2x - 1 \geq 0 \), \(|2x - 1| = 2x - 1\).Calculate: \[ \int^{2}_{0.5} (2x - 1) \,dx \]\[ = \left[ x^2 - x \right]^{2}_{0.5} \] \[ = (4 - 2) - (0.25 - 0.5) \] \[ = 2 - (-0.25) = 2 + 0.25 = 2.25 \]

Step 4: Add the Results of the Two Integrals

Add the results of the two parts to find the total integral:\[ 0.25 + 2.25 = 2.5 \]

Key Concepts

Understanding the Absolute Value Function

In mathematics, the absolute value function is a straightforward yet powerful concept. It measures the "distance" a number is from zero, which is always non-negative. Absolute value changes the sign of negative numbers to positive, while positive numbers remain unchanged. For example, the absolute value of both -3 and 3 is 3. This function is denoted by vertical bars like this: \(|x|\).

In the case of functions like \(|2x - 1|\), the expression inside affects how we handle the function. We look for points where the expression switches from negative to positive, which are called critical points. Here, we solve \(2x - 1 = 0\) to find the critical point \(x = 0.5\).

Once we identify these critical points, we understand where the graph changes direction and use these points to split our integral for easier evaluation.

The Process of Integral Evaluation

Integral evaluation is the method of finding the integral of a function across a specified interval. It's like determining the area under a curve, between a range of values on the x-axis. For definite integrals, these intervals are fixed, providing a specific value as a result.

In our example, we split the original integral \(\int_{0}^{2} |2x - 1| \, dx\) into two separate problems, each evaluated over a region determined by critical values. The integral is divided as follows:

• From \(x = 0\) to \(x = 0.5\), where \(|2x - 1| = -(2x - 1)\).
• From \(x = 0.5\) to \(x = 2\), where \(|2x - 1| = 2x - 1\).

This splitting allows us to handle each segment according to the behavior of the absolute value function within that interval. It simplifies the computation by ensuring we're always dealing with a linear expression, either negative or positive.

Applying the Power Rule for Integration

One of the fundamental rules of integration is the power rule. It's especially useful for integrating basic polynomial functions. This rule states that when you integrate a function of the form \(x^n\), the result is \(\frac{x^{n+1}}{n+1} + C\), where \(C\) is the constant of integration.

In definite integral evaluation, this constant of integration cancels out, resulting in a net change calculated over the interval of interest. Applying the power rule to each part of our split integral:

• For \(\int_{0}^{0.5} -(2x - 1) \, dx\), use the rule to integrate \(-2x + 1\): give \(-x^2 + x\).
• For \(\int_{0.5}^{2} (2x - 1) \, dx\), the result is \(x^2 - x\).

After integrating each part, you apply the limits of the interval to evaluate the difference \(\left[ f(b) - f(a) \right]\), where \(b\) and \(a\) are the upper and lower bounds of each section. This results in two separate integrals: 0.25 and 2.25. Adding these together provides the final integral result: 2.5.