Integral Convergence Evaluation Of ∫[-2 To 14] 2/⁴√(x+2) Dx

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Introduction

In this article, we will delve into the process of determining whether a given integral converges or diverges. Specifically, we will focus on the definite integral of the function 2/⁴√(x+2) from -2 to 14. If the integral converges, we will proceed to evaluate it. This type of problem is a staple in calculus courses and understanding the nuances of improper integrals is crucial for various applications in physics, engineering, and other scientific fields.

Understanding Improper Integrals

Before we tackle the problem at hand, it's essential to understand the concept of improper integrals. Improper integrals arise when either the interval of integration is unbounded (e.g., ∫[1 to ∞] 1/x² dx) or the integrand has a vertical asymptote within the interval of integration. In our case, we have the integral ∫[-2 to 14] 2/⁴√(x+2) dx. Notice that the integrand, 2/⁴√(x+2), has a vertical asymptote at x = -2, which is one of the limits of integration. This makes it an improper integral of the second kind.

To handle improper integrals, we need to use limits. We approach the point of discontinuity and evaluate the limit of the integral as we get arbitrarily close to that point. This process allows us to determine whether the integral converges to a finite value (convergent) or tends to infinity (divergent).

Analyzing the Integral ∫[-2 to 14] 2/⁴√(x+2) dx

Let's analyze our integral, ∫[-2 to 14] 2/⁴√(x+2) dx. As we identified earlier, the integrand 2/⁴√(x+2) has a vertical asymptote at x = -2. To determine if the integral converges, we need to rewrite it using a limit:

∫[-2 to 14] 2/⁴√(x+2) dx = lim[t→-2⁺] ∫[t to 14] 2/⁴√(x+2) dx

Here, we're approaching -2 from the right (denoted by -2⁺) because the interval of integration is [-2, 14]. We replace the lower limit of integration, -2, with a variable t and take the limit as t approaches -2 from the right. This allows us to avoid direct evaluation at the point of discontinuity.

Evaluating the Integral

Now, let's evaluate the integral ∫[t to 14] 2/⁴√(x+2) dx. To do this, we'll use a substitution. Let u = x + 2, then du = dx. We also need to change the limits of integration: when x = t, u = t + 2, and when x = 14, u = 16. So, our integral becomes:

∫[t+2 to 16] 2/⁴√u du = 2 ∫[t+2 to 16] u^(-1/4) du

Now, we can find the antiderivative of u^(-1/4):

∫ u^(-1/4) du = (4/3)u^(3/4) + C

Therefore, our integral evaluates to:

2 [(4/3)u^(3/4)] from t+2 to 16 = (8/3) [u^(3/4)] from t+2 to 16

Plugging in the limits of integration, we get:

(8/3) [16^(3/4) - (t+2)^(3/4)]

Since 16^(3/4) = (2⁴)^(3/4) = 2³ = 8, the expression simplifies to:

(8/3) [8 - (t+2)^(3/4)]

Evaluating the Limit

Now, we need to evaluate the limit as t approaches -2 from the right:

lim[t→-2⁺] (8/3) [8 - (t+2)^(3/4)]

As t approaches -2 from the right, (t+2) approaches 0⁺. Therefore, (t+2)^(3/4) also approaches 0. So, the limit becomes:

(8/3) [8 - 0] = (8/3) * 8 = 64/3

Since the limit exists and is equal to 64/3, the integral converges.

Conclusion

We have successfully determined that the integral ∫[-2 to 14] 2/⁴√(x+2) dx converges, and its value is 64/3. This process involved recognizing the improper nature of the integral, using limits to handle the discontinuity, performing a substitution to simplify the integral, evaluating the integral, and finally, evaluating the limit. Understanding these steps is crucial for handling various improper integrals and their applications.

Further Practice

To solidify your understanding, try evaluating other improper integrals with different types of discontinuities or unbounded intervals. Some examples include:

  • ∫[0 to 1] 1/√x dx
  • ∫[1 to ∞] 1/x³ dx
  • ∫[-∞ to 0] xe^x dx

By working through these examples, you'll gain confidence in your ability to determine convergence and evaluate improper integrals.

Key Concepts Revisited

Let's revisit some of the key concepts we've covered in this article:

  • Improper Integrals: Integrals where the interval of integration is unbounded or the integrand has a vertical asymptote within the interval.
  • Vertical Asymptotes: Points where the function approaches infinity or negative infinity.
  • Limits: A fundamental tool for handling discontinuities in improper integrals.
  • Substitution: A technique for simplifying integrals by changing the variable of integration.
  • Convergence and Divergence: An integral converges if it has a finite value; otherwise, it diverges.

The Importance of Convergence

The convergence or divergence of an integral has significant implications in various fields. For example, in probability theory, the integral of a probability density function over its entire domain must converge to 1. In physics, the energy of a system can be represented by an integral, and its convergence ensures that the energy is finite. In engineering, understanding the convergence of integrals is crucial for analyzing the stability of systems.

Common Mistakes to Avoid

When working with improper integrals, it's essential to avoid some common mistakes:

  • Ignoring Discontinuities: Failing to recognize a vertical asymptote within the interval of integration.
  • Direct Evaluation: Attempting to evaluate the integral directly at the point of discontinuity without using limits.
  • Incorrect Substitution: Making errors in the substitution process, such as forgetting to change the limits of integration.
  • Limit Evaluation Errors: Incorrectly evaluating the limit after integrating.

By being mindful of these potential pitfalls, you can improve your accuracy and understanding of improper integrals.

Conclusion

Mastering the techniques for determining convergence and evaluating improper integrals is a valuable skill in calculus and beyond. By understanding the concepts, practicing regularly, and avoiding common mistakes, you can confidently tackle a wide range of problems involving improper integrals. Remember, the key is to break down the problem into manageable steps, use limits to handle discontinuities, and carefully evaluate the resulting expressions. This detailed exploration provides a comprehensive understanding of how to approach and solve such problems, ensuring a strong foundation in calculus and related fields.

In summary, the integral ∫[-2 to 14] 2/⁴√(x+2) dx converges and its value is 64/3, showcasing the power and elegance of calculus in handling complex problems.