Problem 9 - 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 9-16. $$ \int_{1}^{2} 2 f(x) d x $$

Step-by-Step Solution

Answer

The integral \( \int_{1}^{2} 2 f(x) \, dx = 6 \).

Step 1: Identify the given integral

We need to solve the integral \( \int_{1}^{2} 2 f(x) \, dx \). Notice that this is a definite integral scaled by a constant factor of 2.

Step 2: Use the property of linearity

According to the linearity property of integrals, \( \int_{a}^{b} c \cdot f(x) \, dx = c \cdot \int_{a}^{b} f(x) \, dx \), where \( c \) is a constant. In our case, \( c = 2 \).

Step 3: Apply the linearity property

Use the linearity property: \( \int_{1}^{2} 2 f(x) \, dx = 2 \cdot \int_{1}^{2} f(x) \, dx \).

Step 4: Substitute known integral values

From the problem statement, we know that \( \int_{1}^{2} f(x) \, dx = 3 \). Substitute this value into the equation from Step 3 to get: \( 2 \cdot 3 \).

Step 5: Calculate the result

Perform the multiplication: \( 2 \cdot 3 = 6 \). This is the value of the integral \( \int_{1}^{2} 2 f(x) \, dx \).

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus, used to find the area under a curve within a specific interval. The notation \( \int_{a}^{b} f(x) \, dx \) represents a definite integral where \( a \) and \( b \) are the limits of integration. These limits define the interval over which the area is calculated. The function \( f(x) \) is the integrand, which is integrated over the interval from \( a \) to \( b \).

One essential feature of definite integrals is that they provide a numerical value, representing the net area under the curve between the two limits. This process not only accounts for the area above the x-axis but also considers areas below it, which are treated as negative.

To efficiently use definite integrals, understanding the fundamental theorem of calculus is key, as it links the process of differentiation with integration. This enables us to evaluate definite integrals using antiderivatives of the given function.

Properties of Integrals

Integrals possess several important properties that can simplify calculations immensely. Two of the most useful properties are linearity and additivity, both of which are pivotal in solving integral problems.

• Linearity: The linearity property states that the integral of a sum of functions is the sum of the integrals. This means \( \int_{a}^{b} [f(x) + g(x)] \, dx = \int_{a}^{b} f(x) \, dx + \int_{a}^{b} g(x) \, dx \). Furthermore, if a function is multiplied by a constant, we can factor out that constant: \( \int_{a}^{b} c \, f(x) \, dx = c \int_{a}^{b} f(x) \, dx \).
• Additivity: This property allows us to split the integral over an interval into parts. If we know \( \int_{a}^{b} f(x) \, dx \) and \( \int_{b}^{c} f(x) \, dx \), then \( \int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx \).

Utilizing these properties helps to break down complex integrals into more manageable pieces, simplifying the calculation process.

Linearity of Integrals

Linearity of integrals is a powerful property that can make the process of integration more straightforward. This property is particularly useful when dealing with integrals of functions that are scaled or adjusted by constants.

According to the linearity of integrals, the integral of a constant times a function equals the constant times the integral of the function itself. Mathematically, this is expressed as \( \int_{a}^{b} c \cdot f(x) \, dx = c \cdot \int_{a}^{b} f(x) \, dx \), where \( c \) is a constant factor and does not depend on the variable of integration.

This means that when you multiply a function by a constant, you can calculate the integral of the function first, and then simply scale the result by the constant. This significantly reduces the complexity of calculating definite integrals involving constant multiples.