Problem 19 - Thomas Calculus

Question

In Exercises \(15-22,\) graph the integrands and use areas to evaluate the integrals. $$ \int_{-2}^{1}|x| d x $$

Step-by-Step Solution

Answer

The integral evaluates to 2.5.

Step 1: Understanding Absolute Value Graph

The function we want to graph is the absolute value function, \(|x|\). The absolute value of any number \(x\) is \(x\) itself if \(x\) is positive or zero, and \(-x\) if \(x\) is negative. The graph of \(|x|\) is a V-shaped figure with the vertex at the origin (0,0). It has two linear segments: \(y=x\) for \(x \geq 0\) and \(y=-x\) for \(x < 0\).

Step 2: Sketching the Graph

On the coordinate plane, sketch the graph of \(y = |x|\) within the region from \(x = -2\) to \(x = 1\). The graph will intersect the x-axis at the origin (0,0) and extend to (-2,2) for negative \(x\) and (1,1) for positive \(x\).

Step 3: Determine Area Under the Curve from -2 to 0

Since the integrand function \(|x|\) from \(-2\) to \(0\) is \(-x\), it forms a right triangle with vertices at (-2,2), (0,0), and (-2,0). The base of this triangle is 2 (from -2 to 0), and the height is 2. The area of this triangle is \(\frac{1}{2} \times \, \text{base} \, \times \, \text{height} = \frac{1}{2} \times 2 \times 2 = 2\).

Step 4: Determine Area Under the Curve from 0 to 1

For \(x\) from 0 to 1, the integrand \(|x|\) is simply \(x\), forming a right triangle with vertices at (0,0), (1,1), and (0,1). The base and height of this triangle are both 1. The area is \(\frac{1}{2} \times 1 \times 1 = 0.5\).

Step 5: Calculate Total Area

The integral \(\int_{-2}^{1} |x| \, dx\) is the sum of the areas found: the triangle from \(-2\) to \(0\) and the triangle from \(0\) to \(1\). Therefore, the total area is \(2 + 0.5 = 2.5\).

Key Concepts

Absolute Value Function

The absolute value function, denoted as \(|x|\), is essential when dealing with expressions that include both positive and negative numbers. It determines how far a number is from zero on the number line. This function is defined as:

• \(x\) itself if the number is positive or zero,
• \(-x\) if the number is negative.

This results in a V-shaped graph that is symmetrical around the y-axis. The vertex of this function is at the origin \(0,0\). As you plot the graph of \(|x|\), it consists of two linear segments:

• The line \(y = x\) for \(x \geq 0\),
• The line \(y = -x\) for \(x < 0\).

Understanding the behavior of the absolute value ensures that we can effectively analyze how the function behaves over specific intervals, crucial for calculating areas under its curve.

Area Under a Curve

When you hear about the area under a curve, it's often in the context of integrals. Integrals allow us to find the total area enclosed by the curve and the x-axis over a specified interval. For the absolute value function \(|x|\), the integration task often involves finding the geometric area of shapes like triangles under each segment of the curve.In the problem given, we evaluate the integral \(\int_{-2}^{1}|x| \, dx\), which involves two intervals:

• From \(x = -2\) to \(x = 0\), where the curve forms part of a right triangle with vertices at (-2,2), (0,0), and (-2,0). The triangle's base and height are both 2, and its area is given by \(\frac{1}{2} \times 2 \times 2 = 2\).
• From \(x = 0\) to \(x = 1\), where the graph forms a smaller right triangle with vertices at (0,0), (1,1), and (0,1). Here, the base and height are both 1, so its area is \(\frac{1}{2} \times 1 \times 1 = 0.5\).

When calculating definite integrals of piecewise functions like \(|x|\), make sure to assess each segment separately and sum up their areas. This ensures you account for all portions of the curve under consideration.

Graphing Functions

Graphing functions provides a visual representation that helps us understand the underlying mathematical relationships. For an absolute value function like \(|x|\), the graph shows a clear V-shape with a key vertex at the origin, where the function changes from negative to positive.Considerations when graphing functions:

• Start by identifying key points, like intercepts at the y-axis and where the function changes direction.
• Use symmetry properties for the absolute value function, noting that the line is mirrored across the y-axis.
• Sketch the linear segments that make up the graph, ensuring they meet at the vertex.

By carefully laying out each piece of the function and observing the vertex, graphing becomes a reliable tool for simplifying complex integrals. In problems like these, visual aids help identify sections of the graph to focus on for calculating areas, particularly when using geometric methods.