Problem 11 - 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. $$ \int_{0}^{2}[2 f(x)+g(x)] d x $$

Step-by-Step Solution

Answer

The integral \( \int_{0}^{2} [2f(x) + g(x)] \, dx \) is equal to 14.

Step 1: Apply Linearity of Integrals

To solve the integral \( \int_{0}^{2} [2f(x) + g(x)] \, dx \), we start by applying the linearity property of integrals, which allows us to separate the integral of a sum into the sum of integrals. Thus, \[ \int_{0}^{2} [2f(x) + g(x)] \, dx = \int_{0}^{2} 2f(x) \, dx + \int_{0}^{2} g(x) \, dx. \] We can also factor constants out of the integral, so the first integral becomes \( 2 \int_{0}^{2} f(x) \, dx \).

Step 2: Use Interval Additivity for \(f(x)\)

Next, we need to evaluate \( \int_{0}^{2} f(x) \, dx \). We use the property of interval additivity, which states that \( \int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx \), to express \( \int_{0}^{2} f(x) \, dx \) as \[ \int_{0}^{2} f(x) \, dx = \int_{0}^{1} f(x) \, dx + \int_{1}^{2} f(x) \, dx. \] Plugging in the given values, we have \( \int_{0}^{1} f(x) \, dx = 2 \) and \( \int_{1}^{2} f(x) \, dx = 3 \), so \[ \int_{0}^{2} f(x) \, dx = 2 + 3 = 5. \]

Step 3: Evaluate the Integral of \(g(x)\)

We are given \( \int_{0}^{2} g(x) \, dx = 4 \). This value can be used directly in our original expression.

Step 4: Combine Results and Solve

Now that we have \( \int_{0}^{2} f(x) \, dx = 5 \) and \( \int_{0}^{2} g(x) \, dx = 4 \), we substitute them into our modified expression: \[ \int_{0}^{2} [2f(x) + g(x)] \, dx = 2 \cdot 5 + 4. \] Simplifying this gives us \( 10 + 4 = 14 \).

Key Concepts

Linearity of Integrals

In calculus, the linearity of integrals is a fundamental concept that makes handling definite integrals much easier. This property allows you to break down complex integrals into more manageable parts. Specifically, it states that the integral of a sum is the sum of the integrals. Here’s how it works:

Consider a function of the form \( a f(x) + b g(x) \), where \( a \) and \( b \) are constants. According to the linearity principle, you can separate and factor out constants as follows:

• \( \int [a f(x) + b g(x)] \, dx = \int a f(x) \, dx + \int b g(x) \, dx \)
• \( = a \int f(x) \, dx + b \int g(x) \, dx \)

This method is not only helpful mathematically; it simplifies the calculation considerably. Instead of working with a complex integral right from the start, linearly decomposing it makes the integration process much neater and more straightforward.

In our exercise, we apply linearity to \( \int_{0}^{2} [2f(x) + g(x)] \, dx \), splitting it into two separate integrals: \( 2 \int_{0}^{2} f(x) \, dx + \int_{0}^{2} g(x) \, dx \). This step eases the computations as we can calculate these integrals individually.

Interval Additivity

Interval additivity is another integral property, crucial for evaluating definite integrals over separated intervals. It states that if you have a continuous function \( f(x) \) on an interval \([a, c] \), you can split the integral at any intermediate point \( b \) between \( a \) and \( c \):

• \( \int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx \)

This property is incredibly useful when you know the values of integrals over sub-intervals, and need to find the integral over a larger interval.

For instance, in our exercise, we are given \( \int_{0}^{1} f(x) \, dx = 2 \) and \( \int_{1}^{2} f(x) \, dx = 3 \). By applying interval additivity, we find the integral over the whole interval \([0, 2]\):

\( \int_{0}^{2} f(x) \, dx = \int_{0}^{1} f(x) \, dx + \int_{1}^{2} f(x) \, dx = 2 + 3 = 5 \).

This simplifies the problem significantly, allowing us to calculate \( \int_{0}^{2} [2f(x) + g(x)] \, dx \) efficiently.

Integral Properties

Understanding the distinct properties of integrals can greatly aid in solving complex integral problems with ease. Let’s look into some essential properties used in the exercise:

• Addition and Subtraction: One can sum or subtract integrals over the same limits, as seen with \( \int_{0}^{2} [2f(x) + g(x)] \, dx \).
• Constant Multiplication: You can factor constants outside the integral, simplifying complex expressions, such as \( 2 \int_{0}^{2} f(x) \, dx \).

Utilizing these properties, we can simplify and solve integrals systematically. In our exercise, knowing these properties enables us to break down the given integral into manageable parts, compute separate values, and combine these values effectively.

Thus, understanding integral properties, including linearity and interval additivity, enhances our capability to tackle integrals in mathematics confidently and successfully.