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 9-16. $$ \int_{0}^{2} 2 f(x) d x $$

Step-by-Step Solution

Answer

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

Step 1: Identify the given integrals

We have the following definite integrals provided: \( \int_{0}^{1} f(x) \, dx = 2 \) and \( \int_{1}^{2} f(x) \, dx = 3 \). We aim to find \( \int_{0}^{2} 2f(x) \, dx \).

Step 2: Use interval additivity

Use the interval additivity property of definite integrals: \( \int_{0}^{2} f(x) \, dx = \int_{0}^{1} f(x) \, dx + \int_{1}^{2} f(x) \, dx \). Substitute the given integrals: \( \int_{0}^{2} f(x) \, dx = 2 + 3 = 5 \).

Step 3: Apply linearity property

Use the linearity property of definite integrals, which says \( \int_{a}^{b} c\,f(x) \, dx = c\int_{a}^{b} f(x) \, dx \) for a constant \( c \). In this case, \( c = 2 \) and the original integral was \( \int_{0}^{2} f(x) \, dx = 5 \). So, \( \int_{0}^{2} 2f(x) \, dx = 2\int_{0}^{2} f(x) \, dx = 2 \times 5 = 10 \).

Step 4: Compute the final integral

The computation from Step 3 gives us the final value: \( \int_{0}^{2} 2f(x) \, dx = 10 \).

Key Concepts

Linearity Property in Definite Integrals

The linearity property is a fundamental concept in calculus, especially when dealing with definite integrals. It states that for a given constant \( c \) and a function \( f(x) \), the scale of the entire integral can be stretched by multiplying it by \( c \). Mathematically, the linearity property can be expressed as follows:

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

This means if you multiply a function by a constant before integrating it over an interval, it is equivalent to multiplying the result of the integral of the original function by that constant.

The linearity property is also applicable when you have a sum of functions. For example, if you have functions \( f(x) \) and \( g(x) \), the linearity property can spread over them as:

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

Understanding this property is crucial for simplifying and solving complex integral problems by breaking them into manageable parts.

Interval Additivity in Definite Integrals

The concept of interval additivity helps us understand how integrals can be combined over adjacent intervals. This principle allows us to integrate across a larger range by breaking it into smaller, more manageable pieces. The additivity property can be summarized as:

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

Here, the interval \([a, c]\) is divided into \([a, b]\) and \([b, c]\). The integral over the entire range is the sum of integrals over these smaller intervals.

For instance, in our exercise, knowing that \( \int_{0}^{1} f(x) \, dx = 2 \) and \( \int_{1}^{2} f(x) \, dx = 3 \), we can easily find the integral over the interval \([0, 2]\) by adding these two results:

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

Using interval additivity simplifies the process of solving definite integrals over complex intervals by reducing them into simpler, adjacent ones.

Role of Calculus in Solving Integrals

Calculus provides us with powerful tools to analyze and solve complex problems, and definite integrals are a shining example of this. A definite integral provides the accumulated area under a curve described by a function \( f(x) \), from one point to another. This area calculation has numerous real-world applications, from determining distances and areas to solving problems in physics and engineering.

The definite integral is computed based on the principle of limits, fundamental to calculus, which allows for precise computation of areas under curves, especially when the function might be complex or varies along the interval. The calculation of definite integrals can be made more efficient by applying properties like linearity and interval additivity, allowing for streamlining the solution process.

• Linearity lets us apply constants to integrals and sum integrals of function combinations directly.
• Interval additivity allows integration over various sections of the domain.

These strategies ensure that even as functions become more intricate, the use of calculus principles helps in maintaining accuracy and simplicity in integral computation.