Evaluating The Integral J = ∫₀¹ Eˣ (eˣ + 1) Dx A Step-by-Step Guide

Introduction to Integral Evaluation

In the realm of mathematical analysis, integral evaluation stands as a cornerstone, enabling us to compute areas, volumes, and various other quantities. Integrals, both definite and indefinite, play a crucial role in numerous scientific and engineering disciplines. This article delves into the step-by-step evaluation of a specific definite integral, shedding light on the techniques and underlying principles involved. The integral we aim to solve is: J = ∫₀¹ eˣ (eˣ + 1) dx. This integral, while seemingly straightforward, provides an excellent opportunity to illustrate fundamental integration methods and strategies. Understanding how to evaluate such integrals is not only essential for academic pursuits but also for practical applications in fields like physics, economics, and computer science. This article aims to provide a comprehensive guide, ensuring clarity and depth in understanding each step of the solution process. By the end of this exploration, you will be equipped with the knowledge and skills to tackle similar integral problems with confidence.

Understanding the Integral and Its Components

To effectively evaluate the given integral, J = ∫₀¹ eˣ (eˣ + 1) dx, it's crucial to first understand its components. The integral consists of several key elements: the integrand, the limits of integration, and the differential. In this case, the integrand is the function eˣ (eˣ + 1), which represents the function we are integrating. The limits of integration are 0 and 1, indicating the interval over which we are calculating the definite integral. The differential dx signifies that we are integrating with respect to the variable x. Before diving into the solution, let's expand the integrand to simplify the expression. Expanding eˣ (eˣ + 1) gives us e^(2x) + eˣ. This expansion is a critical first step as it transforms the integrand into a sum of simpler terms, each of which is easier to integrate. Recognizing these components and performing initial simplifications are vital for a successful integration process. By breaking down the integral into manageable parts, we can apply appropriate integration techniques and arrive at the correct solution more efficiently. This initial analysis lays the groundwork for the subsequent steps in evaluating the integral. A thorough understanding of the integral's components not only aids in solving the problem at hand but also enhances overall mathematical proficiency.

Step-by-Step Solution

Now, let's proceed with the step-by-step solution of the integral J = ∫₀¹ eˣ (eˣ + 1) dx. As established earlier, the first step involves expanding the integrand. We have:

J = ∫₀¹ (e^(2x) + eˣ) dx

This expansion transforms the integral into a sum of two simpler integrals, making it easier to handle. Next, we apply the linearity of integration, which allows us to split the integral of a sum into the sum of integrals:

J = ∫₀¹ e^(2x) dx + ∫₀¹ eˣ dx

Now, we can evaluate each integral separately. For the first integral, ∫₀¹ e^(2x) dx, we can use a simple u-substitution. Let u = 2x, then du = 2 dx, and dx = (1/2) du. When x = 0, u = 0, and when x = 1, u = 2. Thus, the integral becomes:

∫₀¹ e^(2x) dx = (1/2) ∫₀² e^u du

The integral of e^u with respect to u is simply e^u. Therefore, we have:

(1/2) ∫₀² e^u du = (1/2) [e^u]₀² = (1/2) (e² - e⁰) = (1/2) (e² - 1)

For the second integral, ∫₀¹ eˣ dx, the integration is straightforward:

∫₀¹ eˣ dx = [eˣ]₀¹ = e¹ - e⁰ = e - 1

Finally, we add the results of the two integrals to find the value of J:

J = (1/2) (e² - 1) + (e - 1)

Simplifying this expression, we get:

J = (e²/2) - (1/2) + e - 1

J = (e²/2) + e - (3/2)

Thus, the value of the integral J is (e²/2) + e - (3/2). This step-by-step solution illustrates the application of fundamental integration techniques, including expansion, linearity, and u-substitution. Each step is crucial in arriving at the final result, and understanding these techniques is essential for solving a wide range of integral problems.

Applying U-Substitution for e^(2x)

The u-substitution technique is a fundamental method in integral calculus, particularly useful when dealing with composite functions. In the context of our integral, J = ∫₀¹ eˣ (eˣ + 1) dx, the u-substitution is crucial for evaluating the integral of e^(2x). As demonstrated in the step-by-step solution, we let u = 2x. This substitution simplifies the exponential term and allows us to apply the basic integration rule for exponential functions. The process involves several key steps: first, we identify the appropriate substitution, which in this case is u = 2x. This choice is guided by the presence of the composite function e^(2x), where 2x is the inner function. Next, we find the differential du by differentiating u with respect to x: du/dx = 2, which gives us du = 2 dx. This relationship is vital for changing the variable of integration from x to u. We then express dx in terms of du: dx = (1/2) du. This expression is used to replace dx in the original integral. Additionally, we need to change the limits of integration to reflect the new variable u. When x = 0, u = 2(0) = 0, and when x = 1, u = 2(1) = 2. These new limits are crucial for evaluating the definite integral in terms of u. After substitution, the integral ∫₀¹ e^(2x) dx transforms into (1/2) ∫₀² e^u du. This new integral is much simpler to evaluate. The integral of e^u with respect to u is simply e^u. Therefore, we have (1/2) [e^u]₀². Evaluating this expression at the limits gives us (1/2) (e² - e⁰) = (1/2) (e² - 1). This result is a direct consequence of applying the u-substitution technique. The u-substitution method not only simplifies the integration process but also provides a clear and systematic approach to handling complex integrals. By mastering this technique, one can effectively tackle a wide range of integration problems in calculus and beyond. Understanding the underlying principles of u-substitution is essential for any student or practitioner of mathematics.

Evaluating ∫₀¹ eˣ dx

The integral ∫₀¹ eˣ dx is a fundamental example in calculus, illustrating the basic integration of the exponential function. The exponential function, eˣ, is unique in that its derivative and integral are the same (up to a constant). This property makes its integration particularly straightforward. To evaluate this definite integral, we first find the antiderivative of eˣ, which is simply eˣ. Then, we apply the fundamental theorem of calculus, which states that the definite integral of a function over an interval is the difference of its antiderivative evaluated at the upper and lower limits of integration. In this case, the limits of integration are 0 and 1. Therefore, we have: ∫₀¹ eˣ dx = [eˣ]₀¹. This notation means we evaluate eˣ at x = 1 and x = 0 and subtract the latter from the former. So, we have: [eˣ]₀¹ = e¹ - e⁰. We know that e¹ is simply e, and e⁰ is equal to 1. Thus, the expression becomes: e - 1. This result is the exact value of the definite integral ∫₀¹ eˣ dx. The simplicity of this integral makes it a valuable example for understanding the concept of definite integration and the application of the fundamental theorem of calculus. The ability to quickly and accurately evaluate such integrals is crucial for more complex problems in calculus and related fields. Furthermore, this example highlights the importance of recognizing basic integrals and their antiderivatives, which are essential building blocks for advanced integration techniques. The evaluation of ∫₀¹ eˣ dx not only provides a numerical result but also reinforces the core principles of integral calculus. Grasping this fundamental concept is key to advancing in the study of mathematics.

Final Result and Conclusion

After meticulously evaluating the integral J = ∫₀¹ eˣ (eˣ + 1) dx, we have arrived at the final result. The step-by-step solution involved expanding the integrand, applying the linearity of integration, and using the u-substitution technique for the integral of e^(2x). The integral was broken down into two simpler integrals: ∫₀¹ e^(2x) dx and ∫₀¹ eˣ dx. The first integral, ∫₀¹ e^(2x) dx, was evaluated using u-substitution, where u = 2x, resulting in (1/2) (e² - 1). The second integral, ∫₀¹ eˣ dx, was evaluated directly, yielding e - 1. By adding these results together, we obtained the value of the original integral: J = (1/2) (e² - 1) + (e - 1). This expression can be simplified to: J = (e²/2) - (1/2) + e - 1. Further simplification gives us the final result: J = (e²/2) + e - (3/2). This value represents the exact solution to the definite integral. In conclusion, the evaluation of J = ∫₀¹ eˣ (eˣ + 1) dx demonstrates the application of several fundamental integration techniques. The process highlights the importance of simplifying the integrand, using appropriate substitution methods, and applying the fundamental theorem of calculus. The final result, (e²/2) + e - (3/2), is a precise quantification of the area under the curve of the function eˣ (eˣ + 1) over the interval [0, 1]. This exercise not only provides a numerical answer but also reinforces the principles and methods of integral calculus. Mastery of these techniques is essential for tackling more complex problems in mathematics and related fields. The successful evaluation of this integral underscores the power and versatility of calculus in solving real-world problems.