Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

Suppose that \(\int_{0}^{1} f(x) d x=2, \int_{1}^{2} f(x) d x=3, \int_{0}^{1} g(x) d x=-1\) and \(\int_{0}^{2} g(x) d x=4 .\) Use properties of definite integrals (linearity, interval additivity, and so on) to calculate each of the integrals in Problems. $$ \int_{0}^{2} 2 f(x) d x $$

Step-by-Step Solution

Answer

The integral \( \int_{0}^{2} 2 f(x) \, dx \) is 10.

Step 1: Understand the Problem

The task is to calculate the integral \( \int_{0}^{2} 2 f(x) \, dx \). This requires using properties of definite integrals, specifically the property of linearity.

Step 2: Apply the Linearity Property

The linearity property of integrals states that \( \int_{a}^{b} c \cdot f(x) \, dx = c \cdot \int_{a}^{b} f(x) \, dx \), where \( c \) is a constant. Here, \( c = 2 \), so we can write:\[ \int_{0}^{2} 2 f(x) \, dx = 2 \cdot \int_{0}^{2} f(x) \, dx \]

Step 3: Use Additivity Property for Integrals

We need to find \( \int_{0}^{2} f(x) \, dx \). We have the given integrals \( \int_{0}^{1} f(x) \, dx = 2 \) and \( \int_{1}^{2} f(x) \, dx = 3 \). Using the additivity property of integrals, we can say:\[ \int_{0}^{2} f(x) \, dx = \int_{0}^{1} f(x) \, dx + \int_{1}^{2} f(x) \, dx \]\[ \int_{0}^{2} f(x) \, dx = 2 + 3 = 5 \]

Step 4: Substitute Back into Linearity Equation

Substitute the result from the previous step into the linearity equation calculated in Step 2:\[ \int_{0}^{2} 2 f(x) \, dx = 2 \cdot 5 = 10 \]

Key Concepts

Linearity of Integrals

The linearity of integrals is a fundamental concept in calculus that makes working with integrals more manageable. It states that you can take a constant multiplier outside of the integral. This property is expressed mathematically as \( \int_{a}^{b} c \cdot f(x) \, dx = c \cdot \int_{a}^{b} f(x) \, dx \), where \( c \) is a constant. For instance, if you have \( c = 2 \) and an integral of the form \( \int_{0}^{2} 2 f(x) \, dx \), you can simplify the computation by factoring out the constant. Thus, it becomes \( 2 \cdot \int_{0}^{2} f(x) \, dx \).
This property not only reduces the complexity but also allows you to deal with integrals in a direct manner. Linearity is beneficial because it simplifies multiplication inside the integral. It helps reduce potential errors in calculations with integrals and is often used as a first step in integrating more complicated functions. By understanding this property, you can efficiently solve problems that involve definite integrals.

Additivity Property

The additivity property of integrals is another essential concept. This property allows integrals over adjacent intervals to be combined. If you have two intervals, \([a, b]\) and \([b, c]\), the additivity property of integrals tells us that \( \int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx \).
In practical terms, this means if you know the integral of a function over several sub-intervals, you can find the integral over the whole interval just by summing those individual integrals. For example, if \( \int_{0}^{1} f(x) \, dx = 2 \) and \( \int_{1}^{2} f(x) \, dx = 3 \), by the additivity property, \( \int_{0}^{2} f(x) \, dx = 2 + 3 = 5 \).
This property is particularly helpful when the function or region is complicated but can be broken into simpler, easier-to-evaluate pieces. Comprehending and using the additivity property aids in solving a variety of calculus problems, especially those involving definite integrals.

Calculus Integration

Calculus integration, particularly involving definite integrals, is a core part of analyzing functions and their accumulated values. Integration is often viewed as finding the area under the curve represented by the function between two points on the x-axis. It involves calculating \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration.
Understanding definite integrals is crucial because they provide numerical values representing accumulated quantities. In real-life applications, these may represent total distance traveled, the volume under a surface, or the total charge transferred, among others. Definite integrals can be computed using properties like linearity and additivity, which simplify complex integrations.
Embracing the concept of definite integrals opens the door to solving practical problems across disciplines like physics, engineering, and economics. Mastering integration techniques and properties forms the foundational skill set for students wanting to excel in calculus.