Problem 74 - Calculus

Question

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

Step-by-Step Solution

Answer

The value of the integral is \(-4\).

Step 1: Understand the Function Components

The function to integrate is \( 2[x] - 3|x| \). The notation \([x]\) represents the floor function, which returns the greatest integer less than or equal to \(x\). The other component \( |x| \) is the absolute value function that returns \(x\) if \(x \geq 0\) and \(-x\) if \(x < 0\).

Step 2: Divide the Integral Interval

The integral from \(-2\) to \(4\) includes both negative and positive numbers, and changes in the floor function occur at integer values. Therefore, we split the integral at integers within the interval: \([-2,0)\), \([0,1)\), \([1,2)\), \([2,3)\), and \([3,4]\).

Step 3: Evaluate the Integral on \([-2,0)\)

From \(-2\) to \(0\), the floor of \(x\) is \(-2\) and the absolute value of \(x\) is \(-x\). Therefore, the expression becomes: \(2(-2) - 3(-x) = -4 + 3x\). Integrate \(-4 + 3x\) from \(-2\) to \(0\) to get \([-4x + \frac{3}{2}x^2]_{-2}^{0} = 4\).

Step 4: Evaluate the Integral on \([0,1)\)

From \(0\) to \(1\), \([x] = 0\) and \(|x| = x\). Thus, the expression becomes: \(2(0) - 3x = -3x\). Integrate \(-3x\) from \(0\) to \(1\) to get \([-\frac{3}{2}x^2]_{0}^{1} = -\frac{3}{2}\).

Step 5: Evaluate the Integral on \([1,2)\)

From \(1\) to \(2\), \([x] = 1\) and \(|x| = x\). Therefore, the expression simplifies to: \(2(1) - 3x = 2 - 3x\). Integrate \(2 - 3x\) from \(1\) to \(2\) to find \([2x - \frac{3}{2}x^2]_{1}^{2} = -\frac{5}{2}\).

Step 6: Evaluate the Integral on \([2,3)\)

For the range \(2\) to \(3\), \([x] = 2\) and \(|x| = x\), turning our expression into: \(2(2) - 3x = 4 - 3x\). Integrate \(4 - 3x\) from \(2\) to \(3\) to obtain \([4x - \frac{3}{2}x^2]_{2}^{3} = -\frac{1}{2}\).

Step 7: Evaluate the Integral on \([3,4]\)

In the interval \(3\) to \(4\), the floor function is \([3]\) and the absolute is \(|x|=x\). The expression becomes \(2(3) - 3x = 6 - 3x\). The integral of \(6 - 3x\) from \(3\) to \(4\) is \([6x - \frac{3}{2}x^2]_{3}^{4} = -\frac{3}{2}\).

Step 8: Sum All Integrals

Add all results obtained from integrating over each subinterval: \(4 + (-\frac{3}{2}) + (-\frac{5}{2}) + (-\frac{1}{2}) + (-\frac{3}{2}) = -4\).

Key Concepts

Floor Function

The floor function, denoted as \([x]\), is a mathematical function that returns the largest integer less than or equal to a given number \(x\). It's a step function that creates a discontinuous graph, resembling a series of flat steps. Each step represents an interval where the function value remains constant. For example, if \( x \) is 3.7, \([x]\) equals 3. If \(x\) is -2.5, \([x]\) equals -3. This function is pivotal in problems dealing with intervals since each integer marks a boundary where the value of the floor function changes.

• **Discontinuity**: It's critical to account for these jumps at each integer value when integrating a piecewise function.
• **Application**: The floor function is regularly used in computational and numerical methods where discrete outputs are needed.

Understanding the behavior of \([x]\) helps us accurately partition an integral into manageable subintervals, making it simpler to solve.

Absolute Value Function

The absolute value function, represented as \(|x|\), describes the distance of a number from zero on the number line, effectively removing any negative sign. It's defined as:

• \(|x| = x\) if \(x \geq 0\)
• \(|x| = -x\) if \(x < 0\)

This function transforms negative inputs into positive ones and keeps positive inputs the same, resulting in a V-shaped graph. While integrating, the absolute value function requires piecewise consideration over intervals where \(x\) changes sign.
For example, in the range from -2 to 4, the function behaves as \(-x\) from -2 to 0, then changes to \(x\) from 0 to 4. This dual behavior necessitates reevaluating the problem over each segment of the integration to ensure accuracy in calculations.

Piecewise Integration

Piecewise integration refers to the process of integrating functions that have different expressions over different intervals. This technique becomes essential when dealing with functions like the floor and absolute value functions which exhibit abrupt changes at specific points.
To handle piecewise functions, you must:

• **Divide the interval** into smaller segments where the function is continuous and well-defined.
• **Integrate each piece** separately, applying the function's specific expression within each subinterval.
• **Sum the results** of these integrations to get the total integral over the main interval.

In the given problem, dividing the integral from -2 to 4 into specific subintervals like [-2,0), [0,1), etc., allows for a simpler evaluation process. This way, each section accommodates the distinct behavior of the functions involved, giving accurate results that account for every change within the interval.

Calculus Problem Solving

Calculus problem solving often involves breaking down complex expressions into simpler parts, especially when dealing with integrals involving non-linear functions. It's about methodically understanding function behaviors, splitting intervals, and calculating fragment by fragment.
Steps include:

• **Analyzing the components** of the function, identifying behaviors like those from the floor and absolute functions.
• **Determining key points** where these functions change or behave differently, such as at integer values in the context of the floor function.
• **Applying integration rules** or techniques designed for linear and non-linear functions over proper intervals.
• **Summing partial integrations** to achieve the full integral value across the entire domain.

This systematic approach helps solve integrals that might initially seem daunting. Understanding each component's behavior aids in crafting a structured pathway to accurate solutions.