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Chapter 5: Problem 194

In the following exercises, identify the roots of the integrand to remove absolute values, then evaluate using the Fundamental Theorem of Calculus, Part 2 . $$ \int_{-2}^{3}|x| d x $$

Short Answer

Expert verified
The integral evaluates to \(\frac{13}{2}\).

Step by step solution

01

Identify Points where the Integrand Changes

The integrand changes at the root of the expression within the absolute value. Since the integrand is \(|x|\), we need to identify where \(x = 0\). This root divides the integration into two parts: from \(-2\) to \(0\) and from \(0\) to \(3\).
02

Remove Absolute Values and Set Up Integration

We split the integral into two separate integrals: one for the interval \([-2, 0]\) and another for the interval \([0, 3]\). Since \(|x| = -x\) when \(x < 0\) and \(|x| = x\) when \(x \geq 0\), we have:\[\int_{-2}^{0} -x \, dx + \int_{0}^{3} x \, dx\]
03

Evaluate Each Integral

Evaluate the first integral \(\int_{-2}^{0} -x \, dx\):\[\int -x \, dx = -\frac{x^2}{2} + C\]Apply the limits:\[= \left[ -\frac{x^2}{2} \right]_{-2}^{0} = \left(-\frac{0^2}{2}\right) - \left(-\frac{(-2)^2}{2}\right) = 0 + 2 = 2\]Evaluate the second integral \(\int_{0}^{3} x \, dx\):\[\int x \, dx = \frac{x^2}{2} + C\]Apply the limits:\[= \left[ \frac{x^2}{2} \right]_{0}^{3} = \left(\frac{3^2}{2}\right) - \left(\frac{0^2}{2}\right) = \frac{9}{2}\]
04

Sum Up the Results

Add the results of the two integrals together:\[2 + \frac{9}{2} = \frac{13}{2}\]

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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 concept in mathematics. It describes the distance of a number from zero on the number line, regardless of direction. This means it always outputs a non-negative result.
When dealing with absolute value, the function alters based on the sign of the input:
  • If \(x \geq 0\), then \(|x| = x\).
  • If \(x < 0\), then \(|x| = -x\). This essentially "flips" the negative input into a positive output.
Understanding the behavior of the absolute value function is essential in calculus, especially when setting up integrals. In the given exercise, the root of the function, \(x = 0\), is identified to remove absolute values and divide the integral into separate parts where \(|x|\) behaves linearly as \(x\) and \(-x\) depending on the interval.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a cornerstone of integral calculus and connects two primary concepts: differentiation and integration. Part 2 of the theorem specifically provides a powerful method to evaluate definite integrals by leveraging antiderivatives.
When applying the theorem, consider these key steps:
  • The definite integral of a function over an interval \([a, b]\) is calculated using any antiderivative \(F(x)\) of the function \(f(x)\).
  • The evaluation involves finding the difference \(F(b) - F(a)\), where \(F(x)\) is the antiderivative.
In our example, after splitting the integral due to the absolute value, we calculated each separate part using antiderivatives. By applying this theorem, we elegantly transitioned from the function under the integral \(|x|\) to its antiderivatives \(-x\) and \(x\) based on its behavior on different intervals.
Evaluating Integrals
Evaluating integrals, especially those involving absolute values, is a systematic process. Here鈥檚 a breakdown of how to effectively approach such problems:
  • Identify Change Points: Start by determining where the integrand changes behavior, typically at roots or critical points. For \(|x|\), this happens at \(x = 0\).
  • Split the Integral: Divide the integral into segments based on the identified points. This allows for the removal of absolute values, simplifying the integrand to \(x\) or \(-x\) depending on the interval.
  • Find Antiderivatives: Calculate the antiderivative for each segment. For linear functions like \(x\) and \(-x\), this results in \(-\frac{x^2}{2}\) and \(+\frac{x^2}{2}\) respectively.
  • Apply Limits: Use the Fundamental Theorem of Calculus to evaluate each integral segment by applying the upper and lower limits and subtracting their antiderivatives.
  • Sum Results: Add the evaluated integrals together to obtain the final result, as seen in \(\frac{13}{2}\) for this problem.
This structured approach not only simplifies integration but also verifies accuracy as each step builds towards the final solution.

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Most popular questions from this chapter

Suppose that \(\int_{0}^{4} f(x) d x=5\) and \(\int_{0}^{2} f(x) d x=-3,\) and \(\int_{0}^{4} g(x) d x=-1\) and \(\int_{0}^{2} g(x) d x=2 .\) In the following exercises, compute the integrals. \(\int_{0}^{4}(f(x)+g(x)) d x\)

In the following exercises, use the identity \(\int_{-A}^{A} f(x) d x=\int_{-A}^{0} f(x) d x+\int_{0}^{A} f(x) d x\) to compute the integrals. \(\int_{1}^{3}(2-x) d x(\text { Hint: Look at the graph of } f .)\)

In the following exercises, use the evaluation theorem to express the integral as a function \(F(x)\) . $$ \int_{0}^{x} \cos t d t $$

In the following exercises, evaluate the indefinite integral \(\int f(x) d x\) with constant \(C=0\) using u-substitution. Then, graph the function and the antiderivative over the indicated interval. If possible, estimate a value of \(C\) that would need to be added to the antiderivative to make it equal to the definite integral \(F(x)=\int_{a}^{x} f(t) d t,\) with a the left endpoint of the given interval. $$ \int \frac{\sin x}{\cos ^{3} x} d x \text { over }\left[-\frac{\pi}{3}, \frac{\pi}{3}\right] $$

Find the antiderivative. $$\int \frac{d x}{(x+4)^{3}}$$
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