Solving ∫ T³ (3t² - 4)^(5/2) Dt A Step-by-Step Guide

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Introduction to Integration

In calculus, integration is a fundamental operation that reverses the process of differentiation. It is used extensively in various fields such as physics, engineering, economics, and statistics. Integrals can represent areas under curves, volumes, and other accumulated quantities. Solving integrals often involves a variety of techniques, including substitution, integration by parts, trigonometric substitution, and partial fraction decomposition. In this article, we will explore the solution to the integral ∫ t³ (3t² - 4)^(5/2) dt, showcasing a method that effectively handles this type of problem.

Understanding the Integral ∫ t³ (3t² - 4)^(5/2) dt

The given integral ∫ t³ (3t² - 4)^(5/2) dt presents a unique challenge because it involves a power of t multiplied by a composite function raised to a fractional power. Direct integration isn't straightforward, so we need to employ a suitable method to simplify it. The most appropriate technique for this integral is u-substitution, which allows us to transform the integral into a more manageable form. The key to successful u-substitution is identifying a suitable expression within the integral whose derivative (or a multiple thereof) also appears in the integral. In this case, the expression inside the parentheses, 3t² - 4, looks promising because its derivative involves t, which is present in the integrand. By making this substitution, we aim to reduce the complexity of the integral, making it easier to solve. Successfully solving integrals like this requires not only a strong understanding of calculus principles but also the ability to recognize patterns and choose the correct techniques. Practicing different types of integrals helps in developing this intuition and skill, making complex problems more approachable. Let’s delve into the step-by-step solution to unravel this integral using u-substitution.

Step-by-Step Solution Using u-Substitution

To solve the integral ∫ t³ (3t² - 4)^(5/2) dt, we will use the u-substitution method, a powerful technique for simplifying integrals by changing the variable of integration. Here’s a detailed walkthrough:

  1. Identify a suitable substitution:

    The expression inside the parentheses, 3t² - 4, appears to be a good candidate for substitution. Let's set:

    u = 3t² - 4

  2. Compute the derivative of u with respect to t:

    du/dt = d/dt (3t² - 4) = 6t

    Thus,

    du = 6t dt

  3. Express dt in terms of du:

    dt = du / (6t)

  4. Rewrite the integral in terms of u:

    We need to express t³ dt in terms of u and du. We already have t dt from du, but we also need to handle the remaining t² term. From our substitution u = 3t² - 4, we can solve for t²:

    3t² = u + 4

    t² = (u + 4) / 3

    Now, substitute u and dt into the integral:

    ∫ t³ (3t² - 4)^(5/2) dt = ∫ t² (3t² - 4)^(5/2) t dt

    = ∫ [(u + 4) / 3] * u^(5/2) * (du / 6)

    = (1/18) ∫ (u + 4) * u^(5/2) du

  5. Expand and simplify the integral:

    Now, distribute u^(5/2) across the terms inside the parentheses:

    (1/18) ∫ (u^(7/2) + 4u^(5/2)) du

  6. Integrate with respect to u:

    Apply the power rule for integration, ∫ x^n dx = (x^(n+1)) / (n + 1) + C:

    (1/18) [∫ u^(7/2) du + 4 ∫ u^(5/2) du]

    = (1/18) [(u^(9/2) / (9/2)) + 4(u^(7/2) / (7/2))] + C

    = (1/18) [(2/9) u^(9/2) + (8/7) u^(7/2)] + C

  7. Simplify the expression:

    Distribute the 1/18:

    (1/81) u^(9/2) + (4/63) u^(7/2) + C

  8. Substitute back for t:

    Replace u with 3t² - 4:

    (1/81) (3t² - 4)^(9/2) + (4/63) (3t² - 4)^(7/2) + C

Thus, the solution to the integral ∫ t³ (3t² - 4)^(5/2) dt is:

(1/81) (3t² - 4)^(9/2) + (4/63) (3t² - 4)^(7/2) + C

This step-by-step approach clearly demonstrates how u-substitution can transform a seemingly complex integral into a manageable one. By correctly identifying the substitution and following the algebraic manipulations carefully, we were able to arrive at the final solution. Understanding and practicing such techniques is crucial for mastering integral calculus.

Common Mistakes to Avoid

When solving integrals using u-substitution, several common mistakes can lead to incorrect answers. Recognizing and avoiding these pitfalls is crucial for achieving accuracy. Here are some key errors to watch out for:

  1. Incorrect Substitution Choice: The initial choice of u is critical. Selecting an inappropriate expression can lead to a more complex integral rather than simplifying it. The best u is often the expression inside a composite function, especially if its derivative (or a multiple thereof) is present in the integral. For instance, in the integral ∫ t³ (3t² - 4)^(5/2) dt, choosing u = 3t² - 4 works because its derivative, 6t, is related to the t³ term. A poor choice might be u = t³, which doesn’t simplify the integral effectively.

  2. Forgetting to Substitute Back: After integrating with respect to u, it’s essential to substitute the original expression in terms of t back into the result. This is a common oversight, especially when dealing with complex integrals. The final answer should always be in terms of the original variable unless it's a definite integral where the limits of integration also need to be changed. For example, after finding the antiderivative in terms of u, remember to replace u with 3t² - 4 to get the solution in terms of t.

  3. Improperly Handling the Differential: The differential du is just as important as the substitution u. It's crucial to correctly find du in terms of dt (or vice versa) and substitute it properly into the integral. For example, if u = 3t² - 4, then du = 6t dt, and dt = du / (6t). Failing to correctly express dt in terms of du will lead to an incorrect integral in terms of u.

  4. Algebraic Errors: Algebraic mistakes during the simplification process can easily derail the solution. This includes errors in expanding expressions, factoring, or dividing. For example, when rewriting the integral ∫ t³ (3t² - 4)^(5/2) dt in terms of u, you need to correctly express t² using the substitution u = 3t² - 4. An error in this step can propagate through the rest of the solution.

  5. Forgetting the Constant of Integration: Always remember to add the constant of integration, C, when finding indefinite integrals. This constant represents the family of antiderivatives that differ by a constant. Omitting C means providing only one specific antiderivative instead of the general solution.

  6. Incorrectly Applying Integration Rules: Ensure that you apply the integration rules correctly. For instance, the power rule for integration, ∫ x^n dx = (x^(n+1)) / (n + 1) + C, should be applied accurately. Mistakes in applying such rules can lead to an incorrect antiderivative. In the given integral, when integrating u^(7/2) and u^(5/2), ensure the power rule is applied correctly.

By being mindful of these common mistakes and practicing careful and methodical problem-solving, you can significantly improve your accuracy in solving integrals using u-substitution. A thorough understanding of the underlying calculus principles and consistent practice are key to mastering integration techniques.

Alternative Methods for Solving Integrals

While u-substitution is an effective method for solving integrals like ∫ t³ (3t² - 4)^(5/2) dt, it's beneficial to be aware of other techniques that can be applied to various types of integrals. Here are a few alternative methods that are commonly used in integral calculus:

  1. Integration by Parts:

    Integration by parts is a technique derived from the product rule for differentiation. It is particularly useful for integrals involving products of functions, such as polynomials, exponentials, logarithms, and trigonometric functions. The formula for integration by parts is:

    ∫ u dv = uv - ∫ v du

    The key to using this method effectively is choosing appropriate parts for u and dv. Generally, you select u as a function that becomes simpler when differentiated, and dv as a function that is easy to integrate. For example, if you had an integral like ∫ x sin(x) dx, you might choose u = x and dv = sin(x) dx.

  2. Trigonometric Substitution:

    Trigonometric substitution is used for integrals that contain expressions of the form √(a² - x²), √(a² + x²), or √(x² - a²), where a is a constant. The idea is to eliminate the square root by making a substitution involving trigonometric functions. The specific substitution depends on the form of the expression:

    • For √(a² - x²), use x = a sin(θ).
    • For √(a² + x²), use x = a tan(θ).
    • For √(x² - a²), use x = a sec(θ).

    For instance, if you encounter an integral with √(4 - x²), you would substitute x = 2 sin(θ) to simplify the expression under the square root.

  3. Partial Fraction Decomposition:

    Partial fraction decomposition is a technique used to integrate rational functions (ratios of polynomials). It involves breaking down a complex rational function into simpler fractions that are easier to integrate. This method is particularly useful when the denominator of the rational function can be factored. For example, consider the integral ∫ (1 / (x² - 1)) dx. The denominator can be factored as (x - 1)(x + 1), and the rational function can be decomposed into partial fractions of the form A / (x - 1) + B / (x + 1). After finding the values of A and B, you can integrate each term separately.

  4. Trigonometric Integrals:

    Integrals involving trigonometric functions often require specific strategies. These strategies usually involve using trigonometric identities to simplify the integrand. Common trigonometric integrals include powers of sine and cosine, products of sine and cosine, and integrals involving tangent, cotangent, secant, and cosecant. For example, to solve ∫ sin³(x) cos²(x) dx, you might rewrite sin³(x) as sin(x) sin²(x), then use the identity sin²(x) = 1 - cos²(x) to simplify the integral.

  5. Definite Integrals:

    Definite integrals have limits of integration, and their evaluation often involves the Fundamental Theorem of Calculus. This theorem states that if F(x) is an antiderivative of f(x), then:

    ∫[a to b] f(x) dx = F(b) - F(a)

    When using substitution for definite integrals, it's important to change the limits of integration to correspond to the new variable. For example, if you substitute u = g(x), the new limits of integration would be g(a) and g(b).

Understanding these alternative methods expands your toolkit for tackling a wide range of integrals. Each technique has its strengths and is suited to particular types of problems. The ability to recognize which method to apply is a crucial skill in integral calculus. Practice and familiarity with these techniques will improve your problem-solving abilities and make you a more proficient mathematician.

Conclusion

In conclusion, solving the integral ∫ t³ (3t² - 4)^(5/2) dt exemplifies the application of u-substitution, a powerful technique in integral calculus. By carefully selecting u = 3t² - 4 and following the steps of substitution, integration, and back-substitution, we successfully found the antiderivative:

(1/81) (3t² - 4)^(9/2) + (4/63) (3t² - 4)^(7/2) + C

This process highlights the importance of recognizing patterns within integrals and choosing the appropriate method to simplify them. We also discussed common mistakes to avoid, such as incorrect substitution choices, algebraic errors, and forgetting the constant of integration, which can help improve accuracy in problem-solving.

Furthermore, we explored alternative methods for solving integrals, including integration by parts, trigonometric substitution, partial fraction decomposition, and strategies for trigonometric integrals. Each technique offers a unique approach and is suitable for different types of integrals, broadening the scope of problems that can be tackled. Mastering these techniques is essential for anyone studying calculus and its applications in various fields.

Integral calculus is a fundamental tool in mathematics, with applications spanning physics, engineering, economics, and computer science. Understanding and practicing various integration techniques not only enhances mathematical skills but also provides a solid foundation for advanced studies and real-world problem-solving. The journey through calculus is one of continuous learning and refinement, where each solved integral adds to a deeper understanding of the subject. By embracing the challenges and practicing consistently, one can develop a strong command over integral calculus and its many applications.