Evaluating Definite Integrals A Step By Step Guide

Jun 16, 2025 by ADMIN 51 views

Evaluating Definite Integrals: A Step-by-Step Guide

In the realm of calculus, definite integrals play a crucial role in determining the area under a curve between two specified limits. This article delves into the evaluation of a specific definite integral, leveraging the properties of linearity and given integral values. We will walk through the process step-by-step, ensuring a clear understanding of the underlying concepts and techniques. Understanding how to solve integrals, especially definite integrals, is crucial for various applications in mathematics, physics, engineering, and other fields. The ability to manipulate and solve integrals allows for the calculation of areas, volumes, probabilities, and other important quantities. In this comprehensive guide, we will explore how to solve a definite integral problem using the properties of linearity, focusing on the integral of a linear combination of functions. We will also delve into the properties that make this manipulation possible, providing a solid foundation for tackling similar problems in calculus and beyond. This article aims to make the process accessible and understandable, even for those who may be new to the concept of definite integrals. So, let's embark on this journey to master the art of evaluating definite integrals and unlock their potential in solving real-world problems.

Problem Statement

Given the following definite integrals:

∫[-73, -45] f(x) dx = 13
∫[-73, -45] g(x) dx = 29
∫[-73, -45] h(x) dx = 28

We aim to determine the value of the following integral:

∫[-73, -45] [9f(x) + 3g(x) - 2h(x)] dx

This problem exemplifies how we can combine known integrals to find the value of a more complex integral, making it a valuable exercise for grasping the fundamentals of calculus. By understanding the properties of linearity and how to apply them, we can efficiently solve this and similar problems. The strategy involves breaking down the complex integral into simpler parts using the properties of linearity. This allows us to use the given values of the individual integrals of f(x), g(x), and h(x) over the same interval to compute the desired result. This is a common technique in calculus that simplifies the evaluation of integrals and provides a systematic approach to solving integration problems. Understanding these principles is fundamental for anyone studying calculus or related fields.

Key Concepts: Properties of Definite Integrals

Before diving into the solution, it's essential to grasp the fundamental properties of definite integrals that make this evaluation possible. Two key properties are particularly relevant:

Linearity of Integrals: The integral of a sum (or difference) of functions is equal to the sum (or difference) of their individual integrals. Mathematically, this is expressed as:
```
∫[a, b] [f(x) ± g(x)] dx = ∫[a, b] f(x) dx ± ∫[a, b] g(x) dx
```
This property allows us to break down complex integrals into simpler components, making them easier to manage and solve. By separating the integral of a sum or difference of functions into individual integrals, we can focus on each part separately and then combine the results. This is a powerful tool in calculus, especially when dealing with complex functions or expressions. The linearity property is a cornerstone of integral calculus and is used extensively in various applications.
Constant Multiple Rule: The integral of a constant multiplied by a function is equal to the constant multiplied by the integral of the function. This is expressed as:
```
∫[a, b] [c * f(x)] dx = c * ∫[a, b] f(x) dx
```
where 'c' is a constant. This property allows us to pull constants out of the integral, simplifying the calculation. By factoring out constants, we can focus on the essential part of the integral, often making the problem more manageable. This rule is especially useful when dealing with coefficients in front of functions within an integral. The constant multiple rule is another fundamental property of definite integrals that facilitates the simplification and evaluation of integrals.

These properties are the building blocks for solving the problem at hand. By understanding and applying these rules, we can manipulate integrals and find solutions efficiently. The linearity and constant multiple rules are not just theoretical concepts; they are practical tools that can be applied to a wide range of integration problems. Mastering these properties is crucial for anyone working with calculus and its applications.

Step-by-Step Solution

Now, let's apply these properties to solve the given problem step-by-step. This methodical approach will illustrate how these concepts translate into practical solutions.

Step 1: Apply the Linearity Property

Using the linearity property, we can break down the given integral as follows:

∫[-73, -45] [9f(x) + 3g(x) - 2h(x)] dx = ∫[-73, -45] 9f(x) dx + ∫[-73, -45] 3g(x) dx - ∫[-73, -45] 2h(x) dx

This step separates the integral of the sum and difference of functions into individual integrals, each involving a single function. This makes the problem more approachable by breaking it into smaller, more manageable parts. The linearity property is a powerful tool for simplifying integrals, and this step demonstrates its application in a practical context. By separating the integral in this way, we can now focus on each term individually, making the overall evaluation process smoother and more efficient. This initial breakdown is crucial for the subsequent steps in the solution.

Step 2: Apply the Constant Multiple Rule

Next, we apply the constant multiple rule to each integral:

∫[-73, -45] 9f(x) dx = 9 * ∫[-73, -45] f(x) dx
∫[-73, -45] 3g(x) dx = 3 * ∫[-73, -45] g(x) dx
∫[-73, -45] 2h(x) dx = 2 * ∫[-73, -45] h(x) dx

This step simplifies the integrals further by factoring out the constants. By moving the constants outside the integral signs, we reduce the complexity of the expressions inside the integrals. This step is another example of how we use the properties of integrals to make the problem easier to solve. Factoring out constants is a common technique in integration and is particularly useful when dealing with linear combinations of functions. This simplification allows us to substitute the given values directly in the next step.

Step 3: Substitute the Given Values

Now, we substitute the given values of the integrals:

9 * ∫[-73, -45] f(x) dx = 9 * 13
3 * ∫[-73, -45] g(x) dx = 3 * 29
2 * ∫[-73, -45] h(x) dx = 2 * 28

This step directly incorporates the information provided in the problem statement. By substituting the known values of the integrals, we are moving closer to the final numerical answer. This step is a straightforward application of the given data and demonstrates how the previous simplifications have set us up for this direct substitution. This substitution transforms the integral problem into a simple arithmetic calculation.

Step 4: Calculate the Result

Finally, we perform the arithmetic operations:

9 * 13 = 117
3 * 29 = 87
2 * 28 = 56

Thus,

∫[-73, -45] [9f(x) + 3g(x) - 2h(x)] dx = 117 + 87 - 56 = 148

Therefore, the value of the integral is 148. This final calculation provides the answer to the problem. By following the steps outlined, we have successfully evaluated the definite integral using the properties of linearity and the constant multiple rule. This process demonstrates the power of these properties in simplifying and solving complex integrals. The result, 148, represents the definite integral of the given linear combination of functions over the specified interval.

Conclusion

In this article, we successfully evaluated the definite integral ∫[-73, -45] [9f(x) + 3g(x) - 2h(x)] dx by leveraging the properties of linearity and the constant multiple rule. The step-by-step solution demonstrates how these properties allow us to break down complex integrals into simpler components, making the evaluation process more manageable. This exercise not only provides a solution to a specific problem but also reinforces the importance of understanding and applying the fundamental properties of definite integrals. These properties are essential tools in calculus and are used extensively in various applications.

The key takeaway from this discussion is the power of linearity and the constant multiple rule in simplifying integral calculations. By mastering these properties, students and practitioners can approach complex integrals with confidence and solve them efficiently. The ability to manipulate integrals using these properties is a crucial skill in calculus and its applications in physics, engineering, economics, and other fields.

Furthermore, this example highlights the importance of a systematic approach to problem-solving in calculus. By breaking down the problem into smaller, manageable steps, we can avoid confusion and ensure accuracy. This methodical approach is not only valuable in calculus but also in other areas of mathematics and science. The ability to organize and simplify problems is a key skill for success in any quantitative field.

In summary, the evaluation of definite integrals is a fundamental skill in calculus, and the properties of linearity and the constant multiple rule are powerful tools for simplifying these calculations. By understanding and applying these concepts, we can solve a wide range of integration problems and gain a deeper appreciation for the beauty and power of calculus. This article has provided a comprehensive guide to solving a specific integral problem, but the principles and techniques discussed are applicable to a wide range of similar problems, making this a valuable resource for anyone studying or working with calculus.