Chapter 5: Problem 19
Sketch the integrand of the given definite integral over the interval of integration. Evaluate the integral by calculating the area it represents. $$ \int_{-1}^{3}|x| d x $$
Short Answer
Expert verified
The integral \( \int_{-1}^{3} |x| \, dx \) evaluates to 5.
Step by step solution
01
Understanding the Integrand
The integrand is the absolute value function, \(|x|\), which takes the value \(x\) when \(x \geq 0\) and \(-x\) when \(x < 0\). This creates a V-shaped graph that is symmetric about the y-axis.
02
Sketching the Graph
Draw the graph of \(|x|\) over the interval \([-1, 3]\). It will be a V-shaped graph with the point at the origin (0,0). From \(-1\) to 0, the graph is increasing, and from 0 to 3, it continues with a positive slope line.
03
Identifying Areas Under the Sections
The area needed is split into two regions: one from \([-1,0]\) and another from \([0,3]\). Compute these areas to evaluate the integral.
04
Calculating Area from -1 to 0
In the interval \([-1,0]\), the function \(|x| = -x\). The area under this line is the area of a right triangle with a base of 1 and height of 1: \( \text{Area} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}\).
05
Calculating Area from 0 to 3
In the interval \([0,3]\), the function \(|x| = x\). The area under this line is the area of a right triangle with a base of 3 and height of 3: \( \text{Area} = \frac{1}{2} \times 3 \times 3 = \frac{9}{2}\).
06
Adding Areas from Both Sections
Add the areas computed in the previous steps. The total area under the curve, representing the integral, is \( \frac{1}{2} + \frac{9}{2} = 5 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Absolute Value Function
The absolute value function, denoted as \(|x|\), is a fundamental mathematical function that measures the distance of a number from zero on the number line. Essentially, it converts negative numbers into positive ones while leaving positive numbers unaffected.
- Mathematically, it is defined as \(|x| = x\) if \(x \geq 0\) and \(|x| = -x\) if \(x < 0\).
- Graphically, the plot of \(|x|\) results in a V-shaped curve that is symmetric about the y-axis.
- This V-shape is due to the piecewise nature of the function, transitioning smoothly from one piece to another at the origin.
Area Under a Curve
Evaluating a definite integral often involves finding the area under a curve on a graph. For the absolute value function, this concept is visually interpreted as the sum of areas under the line segments that form the V-shape.
- The process begins with identifying the intervals where different expressions of the function apply.
- In this problem, the function \(|x|\) has two segments on the interval \([-1,3]\); one for \([-1,0]\) and another for \([0,3]\).
- For \([-1,0]\), the area of the triangle is computed using \(\frac{1}{2} \times \text{{base}} \times \text{{height}} \).
- From \([0,3]\), the function simplifies to a similar calculation for another triangular area.
Graph Sketching
Sketching a graph is a practical way to visualize functions and their properties. For the function \(|x|\), sketching the graph helps in clearly identifying the areas under the curve.
- The graph for \(|x|\) over an interval \([-1,3]\) reflects a symmetrical V-shape centered at the origin.
- From points \([-1,0]\), the graph ascends linearly with a negative slope as \(|x| = -x\).
- From \(0\) onward to \(3\), it continues with a positive slope where \(|x| = x\).
- Make sure to accurately plot the vertex at the origin and properly scale axis marks.
- Mark intervals correctly to ensure accurate calculations of areas under curves.
Piecewise Function
The absolute value function serves as a quintessential example of a piecewise function. In mathematics, piecewise functions are defined by multiple sub-functions, each applying to a certain interval or portion of the domain.
- In the case of \(|x|\), there are two pieces: \(|x| = -x\) for negative x-values and \(|x| = x\) for non-negative x-values.
- This division creates distinct linear segments with different slopes, explaining the V-shape.
- Analyze each segment separately for its defined interval, focusing on where and how the function switches between pieces.
- For integrals, this segmentation aids in determining which formula to apply over specific intervals.