Evaluating ∫√(2x + 1) Dx A Step-by-Step Guide

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In the realm of calculus, evaluating integrals is a fundamental skill. Integrals represent the area under a curve and have wide applications in physics, engineering, and economics. This article delves into the step-by-step process of evaluating the integral ∫√(2x + 1) dx, providing a clear and concise guide for students and enthusiasts alike. Understanding integral calculus is crucial for various scientific and engineering disciplines, as it allows us to model and solve problems involving rates of change and accumulation. This specific integral, ∫√(2x + 1) dx, is a classic example that demonstrates the power of u-substitution, a technique widely used in integral calculus.

To begin, we must first understand the basic concepts of integration. Integration is the reverse process of differentiation, and it involves finding a function whose derivative is equal to the given function. The function we are trying to integrate, √(2x + 1), is a composite function, which means it's a function within a function. In such cases, a technique called u-substitution is often the most effective method. U-substitution simplifies the integral by replacing a complex expression with a single variable, making it easier to integrate. This technique is particularly useful when the integrand contains a function and its derivative, or a close variant thereof. For example, in our case, the expression inside the square root, 2x + 1, is a good candidate for substitution. By letting u = 2x + 1, we can transform the integral into a simpler form that we can readily integrate. This process involves differentiating the substitution equation to find the relationship between dx and du, which is a critical step in the transformation. Once we integrate with respect to u, we then substitute back the original expression in terms of x to obtain the final result. The constant of integration, denoted by C, is also a crucial part of the final answer, as it represents the family of functions that have the same derivative as the integrand. Understanding these fundamental concepts and techniques is essential for mastering integral calculus and applying it to solve a wide range of problems.

Understanding the U-Substitution Method

The u-substitution method is a powerful technique for simplifying integrals that involve composite functions. It's based on the chain rule in differentiation, but applied in reverse. The core idea is to identify a part of the integrand, which we call 'u', and substitute it, along with its differential 'du', to transform the integral into a simpler form. In our case, the integral ∫√(2x + 1) dx presents a perfect opportunity for u-substitution. The expression 2x + 1 within the square root makes the integral complex, but by setting u = 2x + 1, we can significantly simplify it. This substitution allows us to treat the square root term as a simpler √u, which is much easier to integrate. The key to successful u-substitution lies in choosing the right 'u'. Typically, we look for a function whose derivative is also present in the integral, or can be easily obtained. This is because the differential 'du' will involve the derivative of 'u', and we need to be able to replace the dx term in the original integral with an expression involving du. In this case, the derivative of 2x + 1 is 2, which is a constant and easily handled. Once we've chosen 'u' and found 'du', we substitute both into the original integral, aiming to rewrite the entire integral in terms of 'u'. This often results in a simpler integral that we can directly integrate using standard integration rules. After integrating with respect to 'u', the final step is to substitute back the original expression for 'u' in terms of x, to obtain the answer in terms of the original variable. This back-substitution is crucial to ensure that our final answer is consistent with the original problem. The u-substitution method is a versatile tool that can be applied to a wide variety of integrals, and mastering this technique is essential for success in integral calculus.

Step-by-Step Solution: Evaluating ∫√(2x + 1) dx

Let's embark on the journey of solving the integral ∫√(2x + 1) dx step-by-step. This process will clearly illustrate the application of u-substitution and the fundamental principles of integration. Each step is crucial, and careful attention to detail will ensure an accurate solution. We start by identifying the appropriate substitution, then differentiate to find the relationship between dx and du, substitute these into the integral, perform the integration, and finally, substitute back to express the result in terms of x.

  1. Identify the substitution: Our first step is to recognize the composite function within the integral. The expression inside the square root, 2x + 1, is the ideal candidate for our substitution. Let u = 2x + 1. This substitution aims to simplify the square root term, making the integral easier to handle. By replacing 2x + 1 with u, we transform the integral from involving a complex expression to a simpler form involving the square root of a single variable. This is the essence of u-substitution: to simplify the integrand by replacing a complex part with a single variable.

  2. Find du: Now, we need to find the differential of u, which is denoted as du. Differentiating u = 2x + 1 with respect to x, we get du/dx = 2. This means that du = 2 dx. However, our original integral contains dx, not 2 dx. Therefore, we need to solve for dx in terms of du. Dividing both sides of the equation du = 2 dx by 2, we get dx = (1/2) du. This relationship between dx and du is crucial for substituting into the original integral. It allows us to replace the dx term with an equivalent expression involving du, thereby transforming the entire integral into a form that is expressed in terms of u.

  3. Substitute into the integral: With our substitution u = 2x + 1 and dx = (1/2) du, we can now rewrite the original integral in terms of u. The integral ∫√(2x + 1) dx becomes ∫√u (1/2) du. The constant (1/2) can be moved outside the integral, simplifying it further: (1/2) ∫√u du. This step is where the power of u-substitution becomes evident. The original integral, which involved a square root of a more complex expression, has been transformed into a much simpler integral involving the square root of a single variable. This simplification makes the integration process significantly easier.

  4. Integrate: We now have the integral (1/2) ∫√u du. To integrate √u, we can rewrite it as u^(1/2). Using the power rule for integration, which states that ∫x^n dx = (x^(n+1))/(n+1) + C, we can integrate u^(1/2) with respect to u. Applying the power rule, we get ∫u^(1/2) du = (u^(3/2))/(3/2) + C, which simplifies to (2/3)u^(3/2) + C. Remember, C is the constant of integration, which is always included in indefinite integrals. Now, we multiply this result by the (1/2) that we factored out earlier: (1/2) * (2/3)u^(3/2) + C, which simplifies to (1/3)u^(3/2) + C. This step represents the core of the integration process, where we find the antiderivative of the simplified integrand.

  5. Substitute back: The final step is to substitute back the original expression for u in terms of x. We recall that u = 2x + 1. Substituting this back into our result, we get (1/3)(2x + 1)^(3/2) + C. This is the final answer to the integral ∫√(2x + 1) dx. It represents the family of functions whose derivative is √(2x + 1). The constant of integration, C, indicates that there are infinitely many such functions, differing only by a constant value. This back-substitution is crucial because it expresses the result in terms of the original variable, x, as required by the original problem. Without this step, our answer would be in terms of u, which is not what we were asked to find.

The Final Result and Its Significance

Therefore, the final result of evaluating the integral ∫√(2x + 1) dx is (1/3)(2x + 1)^(3/2) + C. This result is not just a mathematical solution; it represents a family of functions whose derivative is the original integrand, √(2x + 1). The constant of integration, C, is a crucial part of this result, as it signifies that there are infinitely many functions that satisfy this condition, differing only by a constant vertical shift. This concept is fundamental to understanding indefinite integrals and their applications.

The significance of this result extends beyond the realm of pure mathematics. Integrals like this one appear frequently in various fields of science and engineering. For example, they can be used to calculate areas, volumes, and average values. In physics, integrals are used to determine displacement from velocity, and work done by a force. In engineering, they are used in circuit analysis, fluid dynamics, and structural mechanics. Understanding how to evaluate integrals like ∫√(2x + 1) dx is therefore essential for students and professionals in these fields.

Moreover, the process of solving this integral using u-substitution highlights the importance of choosing the right technique for a given problem. U-substitution is a powerful tool, but it's not always the most appropriate method. Recognizing when to use it, and how to apply it effectively, is a key skill in integral calculus. This example serves as a valuable illustration of the power and versatility of u-substitution, and it reinforces the importance of mastering this technique for tackling more complex integrals.

In conclusion, evaluating ∫√(2x + 1) dx provides a comprehensive example of u-substitution in action. The step-by-step solution demonstrates the logic behind the method and the importance of each step. The final result, (1/3)(2x + 1)^(3/2) + C, is a testament to the power of calculus in solving real-world problems. Understanding the process and the result is a valuable asset for anyone studying or working in mathematics, science, or engineering.

Practice Problems

To solidify your understanding of u-substitution and integral evaluation, try these practice problems:

  1. ∫x√(x^2 + 1) dx
  2. ∫sin(x)cos(x) dx
  3. ∫x/(x^2 + 1) dx

By working through these problems, you'll gain confidence in your ability to apply u-substitution and solve a variety of integrals. Remember to focus on identifying the appropriate substitution and carefully following the steps outlined in this guide.