/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Sketch the integrand of the give... [FREE SOLUTION] | 91Ó°ÊÓ

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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.
Understanding the absolute value function is crucial in this exercise, as it sets the foundation for evaluating the definite integral of such a function over a specified interval.
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 \([0,3]\).
Within each interval, compute the area by considering the general shapes present, typically triangles in this case.
  • 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.
The sum of these triangular areas gives the total value of the integral, making it an essential approach for solving such problems.
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 \(|x| = x\).
When sketching such a graph:
  • 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.
Graph sketching not only aids in a deeper understanding of the function's behavior but is also a step to ensure accurate integration outcomes.
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.
Piecewise functions allow modeling of complex behaviors in simpler segments, making them crucial for solving real-world problems.
  • 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.
Understanding how to work with piecewise functions improves the skills needed to handle a variety of mathematical tasks, from graphing to integration.

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Most popular questions from this chapter

Treat the \(y\) variable as the independent variable and the \(x\) variable as the dependent variable. By integrating with respect to \(y,\) calculate the area of the region that is described. The region between the curves \(x=y^{2}\) and \(x=-y^{2}+4\)

Calculate the area of the region between the pair of curves. $$ y=(x-3) / 2 \quad x=y^{2} $$

The graphs of \(y=f(x)\) and \(y=g(x)\) intersect in more than two points. Find the total area of the regions that are bounded above and below by the graphs of \(f\) and \(g\). $$ f(x)=x^{4}-5 x^{2} \quad g(x)=-4 $$

Calculate the area of the region between the pair of curves. $$ x=y^{2}+6 \quad x=-y^{2}+14 $$

In a particular regional climate, the temperature varies between \(-22^{\circ} \mathrm{C}\) and \(40^{\circ} \mathrm{C}\), averaging \(\mu=11^{\circ} \mathrm{C}\). The number of days \(F(T)\) in the year on which the temperature remains below \(T\) degrees centigrade is given (approximately) by $$F(T)=\int_{-22}^{T} f(x) d x \quad(-22 \leq T \leq 40)$$ where $$f(x)=12.72 \exp \left(-\frac{(x-11)^{2}}{266.4}\right) $$ Notice that \(F(T)\) is the sort of area integral that we studied in Section 5.4 . a. Use Simpson's Rule with \(N=20\) to approximate \(F(40) .\) What should the exact value of \(F(40)\) be? b. Heat alerts are issued when the daily high temperature is \(36^{\circ} \mathrm{C}\) or more. On about how many days a year are heat alerts issued? c. Suppose that global warming raises the average temperature by \(1^{\circ} \mathrm{C}\), shifting the graph of \(f\) by 1 unit to the right. The new model may be obtained by simply replacing \(\mu\) with 12 and using [-21,41] as the domain (see Figure 13). What is the percentage increase in heat alerts that will result from this \(1^{\circ} \mathrm{C}\) shift in temperature?

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