Problem 78 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate \(\int_{-2}^{4}(2[x]-3|x|) d x\).

Step-by-Step Solution

Answer

The evaluated integral is \(-5\).

Step 1: Understand the Function

The given integral is \( \int_{-2}^{4}(2[x]-3|x|) \, dx \). Here, \([x]\) denotes the greatest integer function, also known as the floor function, which returns the largest integer less than or equal to \(x\). \(|x|\) denotes the absolute value of \(x\).

Step 2: Determine Intervals for \([x]\)

Since the greatest integer function \([x]\) changes at integer values, we need to evaluate the integral from \(-2\) to \(4\) by splitting it into intervals based on integer boundaries. These intervals are \([-2, -1)\), \([-1, 0)\), \([0, 1)\), \([1, 2)\), \([2, 3)\), and \([3, 4]\).

Step 3: Evaluate Using Piecewise Function

On each interval, we express \( [x] \) and \( |x| \) as constant functions or well-defined linear functions:- For \([-2, -1)\): \([x] = -2\), \(|x| = -x\)- For \([-1, 0)\): \([x] = -1\), \(|x| = -x\)- For \([0, 1)\): \([x] = 0\), \(|x| = x\)- For \([1, 2)\): \([x] = 1\), \(|x| = x\)- For \([2, 3)\): \([x] = 2\), \(|x| = x\)- For \([3, 4]\): \([x] = 3\), \(|x| = x\)

Step 4: Calculate Each Integral

Evaluate the integral on each interval:- \( \int_{-2}^{-1}(2(-2) - 3(-x)) \, dx = \int_{-2}^{-1}(-4 + 3x) \, dx \)- \( \int_{-1}^{0}(2(-1) - 3(-x)) \, dx = \int_{-1}^{0}(-2 + 3x) \, dx \)- \( \int_{0}^{1}(2(0) - 3x) \, dx = \int_{0}^{1}(-3x) \, dx \)- \( \int_{1}^{2}(2(1) - 3x) \, dx = \int_{1}^{2}(2 - 3x) \, dx \)- \( \int_{2}^{3}(2(2) - 3x) \, dx = \int_{2}^{3}(4 - 3x) \, dx \)- \( \int_{3}^{4}(2(3) - 3x) \, dx = \int_{3}^{4}(6 - 3x) \, dx \)

Step 5: Solve Each Integral

Calculate each individual integral:- For \([-2, -1)\): \( \int_{-2}^{-1}(-4 + 3x) \, dx \) results in \(-(4x - \frac{3}{2}x^2)\bigg|_{-2}^{-1} = -\frac{5}{2} \)- For \([-1, 0)\): \( \int_{-1}^{0}(-2 + 3x) \, dx \) results in \(-(2x - \frac{3}{2}x^2)\bigg|_{-1}^{0} = -\frac{1}{2} \)- For \([0, 1)\): \( \int_{0}^{1}(-3x) \, dx \) results in \(-\frac{3}{2} \)- For \([1, 2)\): \( \int_{1}^{2}(2 - 3x) \, dx \) results in \(-\frac{3}{2} \)- For \([2, 3)\): \( \int_{2}^{3}(4 - 3x) \, dx \) results in \(-\frac{1}{2} \)- For \([3, 4]\): \( \int_{3}^{4}(6 - 3x) \, dx \) results in \(\frac{3}{2} \)

Step 6: Sum the Results

Add up the results of each integral calculation:\(-\frac{5}{2} + (-\frac{1}{2}) + (-\frac{3}{2}) + (-\frac{3}{2}) + (-\frac{1}{2}) + \frac{3}{2} = -5 \).

Key Concepts

Floor Function

The floor function, denoted as \([x]\), is a mathematical function that returns the greatest integer less than or equal to a given number \(x\). This concept is particularly useful in scenarios where you need to map a real number to the nearest lower whole number.
For instance:

• If \(x = 1.7\), then \([x] = 1\).
• If \(x = -1.7\), then \([x] = -2\).
• If \(x = 3\), then \([x] = 3\) because \(3\) itself is an integer.

In integral calculus, the floor function plays an important role in defining step functions on specific intervals. These intervals generally change when \(x\) crosses a whole number. This method of dividing the domain into intervals allows for piecewise integration, simplifying the calculation process.

Absolute Value

The absolute value function, represented by \(|x|\), provides the distance of a number \(x\) from zero on the number line, regardless of its direction. In simpler terms, it turns negative values into positives.

• For positive values, \(|x| = x\).
• For zero, \(|0| = 0\).
• And importantly, for negative values, \(|x| = -x\).

In integral calculus, absolute value functions add complexity because the behavior of the function changes at the zero point. When performing integrations involving absolute values, it's important to consider the number line splits, where \(x\) transitions from negative to positive, or vice versa. By doing so, you're able to address these changes and accurately compute intervals accordingly.

Piecewise Integration

Piecewise integration is a technique utilized when dealing with functions that have different expressions over different intervals. These functions are effectively broken into 'pieces,' where each piece is defined mathematically by a separate formula on certain sub-intervals.
To compute an integral involving a piecewise function, it's essential to:

• Identify the intervals where the function changes its expression.
• Integrate each section separately, using the corresponding formula over its specific interval.
• Finally, sum up the results from each individually computed integral for the entire range.

This approach allows for accurate results even when confronting complex functions like floor or absolute value. Hence, simplifying the process of evaluating definite integrals over such irregular functions.

Definite Integrals

Definite integrals represent the net area under a curve within a specific interval \([a, b]\). It’s computed by taking the integral of the function \(f(x)\) over that interval, denoting it as \(\int_{a}^{b} f(x) dx\).
Unlike indefinite integrals, which produce a general formula, definite integrals yield a numerical value.It's crucial to know that:

• The limits of the integral (\(a\) and \(b\)) define the range over which you’re computing the area.
• Negative values indicate that the function is below the x-axis over this portion of the interval.
• Positive values signify that the function lies above the x-axis.

When dealing with problems involving complex step functions or piecewise functions, it’s essential to evaluate each section within the defined interval before summing the results to get the total area. This understanding aids in handling problems that integrate functions like those involving floor functions and absolute values.