Problem 16
Question
Evaluate the following. (If you haven't done Problem 14 , do that first.) (a) \(\int_{-3}^{3}\left|x^{2}-4\right| d x\) (b) \(\int_{0}^{5}|(x+3)(x-1)| d x\)
Step-by-Step Solution
Verified Answer
The value of the first integral \(\int_{-3}^{3}|x^2-4| dx\) is 10/3 and the value of the second integral \(\int_{0}^{5}|(x+3)(x-1)| dx\) is 52/3.
1Step 1: Evaluating the first integral
Here, we are integrating \(|x^2 - 4|\) from -3 to 3. To get rid of the absolute value, we need to break this into two integrals based on when \(x^2 - 4\) is negative and when it is positive. This happens when the value of \(x\) is less than -2 and more than 2. Therefore, our integral becomes: \(\int_{-3}^{-2}(-x^2+4) dx + \int_{-2}^{2}(x^2-4) dx + \int_{2}^{3} (-x^2+4) dx\)
2Step 2: Compute the value of the first integral
Compute the value of the integrals using the Power Rule for integrating Polynomials. The result is \(2[|(1/3)x^3-4x|]_{-3}^{-2} + 2[(1/3)x^3-4x]_{-2}^{2} + 2[|(1/3)x^3-4x|]_{2}^{3}\)
3Step 3: Evaluating the second integral
The second integral involves the absolute value of the product of two linear expressions. We'll proceed in similar manner as above, separating the integral into sectors where the expression inside the absolute value is positive and negative. The zero points of \((x+3)(x-1)\) are -3 and 1, so the integral becomes:\(\int_{0}^{-3}-(x+3)(x-1) dx + \int_{-3}^{1} (x+3)(x-1) dx + \int_{1}^{5} -(x+3)(x-1) dx\)
4Step 4: Compute the value of the second integral
Perform integration using the extended power rule for polynomials. This leads to \(2[|1/2x^2+2x|]_{0}^{-3} - 2[1/2x^2+2x]_{-3}^{1} + 2[|1/2x^2+2x|]_{1}^{5}\)
Key Concepts
Absolute Value FunctionsPolynomial IntegrationPiecewise Integration
Absolute Value Functions
Absolute value functions can sometimes make calculus slightly tricky. The absolute value of a function is denoted as \(|f(x)|\), where the value inside depends on whether it is positive or negative. When evaluating integrals involving absolute values, you should split the integral into sections, or intervals, where the expression inside the absolute value maintains a consistent sign. For example, if the expression inside the absolute value is negative over part of the interval and positive in another, you need to compute two separate integrals over those intervals.
Here’s how we handle them:
Here’s how we handle them:
- First, find the points where the inner expression equals zero.
- These points will act as bounds that determine where the expression changes from positive to negative or vice versa.
- Redefine the function without the absolute value for each interval: change the sign inside the absolute value if it’s negative over that interval.
Polynomial Integration
Polynomial integration is an essential part of calculus, and knowing how to integrate polynomials is crucial for evaluating integrals. A polynomial is a mathematical expression consisting of variables and coefficients that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents.
To integrate polynomials, you apply the power rule:
Moreover, polynomials can be combined with constants, where you integrate them separately, letting you express the result as the sum of those antiderivatives. Always remember to add a constant of integration if you are dealing with indefinite integrals. For definite integrals, evaluate each antiderivative expression with the upper and lower bounds and subtract to find the net area under the curve, ensuring that polynomial functions can be handled with clarity and confidence.
To integrate polynomials, you apply the power rule:
- If you have a term of the form \(x^n\), its antiderivative is \(\frac{x^{n+1}}{n+1}\), where \(n\) is not equal to \(-1\).
Moreover, polynomials can be combined with constants, where you integrate them separately, letting you express the result as the sum of those antiderivatives. Always remember to add a constant of integration if you are dealing with indefinite integrals. For definite integrals, evaluate each antiderivative expression with the upper and lower bounds and subtract to find the net area under the curve, ensuring that polynomial functions can be handled with clarity and confidence.
Piecewise Integration
Piecewise integration is a technique used when dealing with functions composed of multiple sub-functions, each valid over a certain interval of the domain. It's especially useful when the function you are integrating consists of different expressions over different ranges. For instance, step functions and functions involving absolute values often require piecewise integration.
The steps include:
The steps include:
- Identify the intervals where each sub-function applies, marking them with their boundaries.
- Integrate each part separately over its designated interval.
- Sum the integrals to cover the entire original function over its entire range of interest.
Other exercises in this chapter
Problem 14
(a) Find \(\int_{-1}^{1}|x| d x\) using the area interpretation of the definite integral. (b) Show that \(\frac{\left|x^{2}\right|}{2}\) is not an antiderivativ
View solution Problem 15
Find the following two definite integrals without using the Fundamental Theorem of Calculus. Instead, use the area interpretation of the definite integral. (a)
View solution Problem 17
Calculate the following definite integrals by calculating the limit of Riemann sums. You'll need to use the formulas provided. Check your answers using the Fund
View solution Problem 20
(a) Write a Riemann sum with 10 equal subdivisions that gives an overestimate for the area under \(\ln x\) on [1,6]. Write your answer in two ways, once with su
View solution