Evaluate The Definite Integral J = ∫₀^(ln 2) (eˣ + E²ˣ)(e⁻ˣ - E⁻²ˣ) Dx. How Do You Solve This Integral?

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Introduction

In this article, we will delve into the detailed evaluation of the definite integral J, which is defined as J = ∫₀^(ln 2) (eˣ + e²ˣ)(e⁻ˣ - e⁻²ˣ) dx. This problem falls under the domain of mathematical analysis, specifically focusing on the techniques of integration. Integrals of this nature often appear in various fields, including physics, engineering, and computer science, making it a crucial skill for students and professionals alike. The integrand, (eˣ + e²ˣ)(e⁻ˣ - e⁻²ˣ), might seem daunting at first glance, but by employing algebraic simplification and the fundamental theorem of calculus, we can arrive at a solution. Our exploration will not only provide the answer but also illuminate the underlying principles and steps involved, enabling a deeper understanding of integral calculus. We will break down the problem into manageable parts, starting with simplifying the integrand, then finding the antiderivative, and finally evaluating the definite integral using the limits of integration. This systematic approach will help you tackle similar problems with confidence. Furthermore, we will discuss the importance of understanding exponential functions and their properties in solving such integrals. The exponential function, eˣ, plays a vital role in many mathematical models and its mastery is essential for anyone pursuing advanced studies in STEM fields. Throughout this article, we aim to provide a clear and concise explanation, making the process of solving this integral accessible to a wide audience. Whether you are a student preparing for an exam or a professional looking to refresh your calculus skills, this guide will serve as a valuable resource. So, let's embark on this journey and unravel the intricacies of this intriguing integral.

Simplifying the Integrand

Before we embark on the actual integration, simplifying the integrand (eˣ + e²ˣ)(e⁻ˣ - e⁻²ˣ) is a crucial step. This simplification not only makes the integration process easier but also reduces the chances of making errors. The integrand is a product of two binomial expressions, and we can use the distributive property (also known as the FOIL method) to expand this product. By applying the distributive property, we multiply each term in the first binomial by each term in the second binomial. This gives us: (eˣ + e²ˣ)(e⁻ˣ - e⁻²ˣ) = eˣ * e⁻ˣ - eˣ * e⁻²ˣ + e²ˣ * e⁻ˣ - e²ˣ * e⁻²ˣ. Now, we can use the property of exponents that states aᵐ * aⁿ = a^(m+n) to simplify each term further. Applying this property, we get: e^(x-x) - e^(x-2x) + e^(2x-x) - e^(2x-2x). This simplifies to: e⁰ - e⁻ˣ + eˣ - e⁰. We know that any non-zero number raised to the power of 0 is 1, so e⁰ = 1. Substituting this into our expression, we have: 1 - e⁻ˣ + eˣ - 1. The 1 and -1 cancel each other out, leaving us with the simplified integrand: eˣ - e⁻ˣ. This simplified form is much easier to integrate than the original expression. It consists of two simple exponential terms, each of which has a well-known integral. The process of simplifying the integrand highlights the importance of algebraic manipulation in calculus. By carefully expanding and simplifying expressions, we can often transform complex integrals into simpler, more manageable forms. This skill is not only useful for integration but also for other areas of mathematics and science. In the next section, we will proceed to find the antiderivative of this simplified integrand. Remember, this step-by-step approach is key to mastering integral calculus. So, keep practicing and refining your skills in algebraic manipulation and simplification.

Finding the Antiderivative

Now that we have successfully simplified the integrand to eˣ - e⁻ˣ, the next step is to find its antiderivative. The antiderivative of a function is a function whose derivative is the original function. In other words, it is the reverse process of differentiation. To find the antiderivative of eˣ - e⁻ˣ, we will use the linearity property of integrals, which states that the integral of a sum (or difference) is the sum (or difference) of the integrals. This means we can find the antiderivative of each term separately and then combine them. The antiderivative of eˣ is simply eˣ, as the derivative of eˣ is eˣ. For the term -e⁻ˣ, we need to recall the chain rule of differentiation. The derivative of e⁻ˣ is -e⁻ˣ, so the antiderivative of -e⁻ˣ is e⁻ˣ. Therefore, the antiderivative of eˣ - e⁻ˣ is eˣ + e⁻ˣ + C, where C is the constant of integration. The constant of integration arises because the derivative of a constant is always zero. This means that there are infinitely many functions that have the same derivative. For example, the derivatives of eˣ + e⁻ˣ, eˣ + e⁻ˣ + 1, and eˣ + e⁻ˣ - 5 are all eˣ - e⁻ˣ. The constant of integration, C, represents this ambiguity. However, when we evaluate a definite integral, the constant of integration cancels out, so we don't need to worry about it at this stage. The process of finding the antiderivative is a fundamental skill in integral calculus. It requires a good understanding of differentiation rules and the ability to recognize patterns. Practice is key to mastering this skill. By working through various examples, you will develop an intuition for finding antiderivatives. In the next section, we will use the antiderivative we have found to evaluate the definite integral using the limits of integration. This will give us the final value of the integral J. Remember, the antiderivative is the bridge between the integrand and the definite integral, so it is essential to understand this concept thoroughly.

Evaluating the Definite Integral

With the antiderivative of the integrand determined to be eˣ + e⁻ˣ, we can now proceed to evaluate the definite integral J = ∫₀^(ln 2) (eˣ - e⁻ˣ) dx. The fundamental theorem of calculus provides the framework for this evaluation. This theorem states that if F(x) is an antiderivative of f(x), then the definite integral of f(x) from a to b is given by F(b) - F(a). In our case, f(x) = eˣ - e⁻ˣ, F(x) = eˣ + e⁻ˣ, a = 0, and b = ln 2. Applying the fundamental theorem of calculus, we have: J = F(ln 2) - F(0) = (e^(ln 2) + e^(-ln 2)) - (e⁰ + e⁻⁰). We know that e^(ln 2) = 2. Also, e^(-ln 2) can be rewritten as e^(ln(2⁻¹)) = e^(ln(1/2)) = 1/2. And, as before, e⁰ = e⁻⁰ = 1. Substituting these values into our expression, we get: J = (2 + 1/2) - (1 + 1) = 2.5 - 2 = 0.5. Therefore, the value of the definite integral J is 0.5. This completes the evaluation of the integral. The final answer, 0.5, represents the signed area between the curve of the integrand and the x-axis over the interval [0, ln 2]. The process of evaluating a definite integral using the fundamental theorem of calculus is a cornerstone of integral calculus. It allows us to connect the concepts of integration and differentiation, and it provides a powerful tool for solving a wide range of problems. This step-by-step approach, from simplifying the integrand to finding the antiderivative and finally evaluating the definite integral, is crucial for mastering integral calculus. By understanding each step and practicing regularly, you can develop the skills and confidence to tackle more complex integrals. In the conclusion, we will summarize the steps we have taken and highlight the key concepts involved in solving this integral. Remember, calculus is a building-block subject, and each concept builds upon the previous one. So, a solid understanding of the fundamentals is essential for success.

Conclusion

In this comprehensive guide, we have successfully evaluated the definite integral J = ∫₀^(ln 2) (eˣ + e²ˣ)(e⁻ˣ - e⁻²ˣ) dx. We began by simplifying the integrand using algebraic manipulation, specifically the distributive property and the properties of exponents. This simplification transformed the integral into a more manageable form: ∫₀^(ln 2) (eˣ - e⁻ˣ) dx. Next, we found the antiderivative of the simplified integrand, which is eˣ + e⁻ˣ + C, where C is the constant of integration. The constant of integration is important to consider when finding general antiderivatives, but it cancels out when evaluating definite integrals. Finally, we applied the fundamental theorem of calculus to evaluate the definite integral. This involved substituting the limits of integration, 0 and ln 2, into the antiderivative and finding the difference. The result of this evaluation gave us the value of the integral, J = 0.5. Throughout this process, we have emphasized the importance of understanding the underlying concepts and techniques of integral calculus. Algebraic simplification, finding antiderivatives, and the fundamental theorem of calculus are all essential tools for solving integrals. This example highlights how these tools can be used in conjunction to solve a seemingly complex problem. Furthermore, we have demonstrated the significance of understanding exponential functions and their properties in calculus. Exponential functions are ubiquitous in mathematics and science, and a strong grasp of their behavior is crucial for success in many fields. By breaking down the problem into manageable steps and providing clear explanations, we have aimed to make this process accessible to a wide audience. Whether you are a student learning calculus for the first time or a professional looking to refresh your skills, we hope this guide has been helpful. Remember, practice is key to mastering calculus. By working through various examples and applying the techniques we have discussed, you can develop a deeper understanding and appreciation for this powerful branch of mathematics. The journey through calculus is a rewarding one, and the skills you acquire will serve you well in many areas of life.