Evaluate The Integral. 1 0 R3 16 + R2 Dr

Evaluating the Definite Integral: ∫₀¹ r³/(16 + r²) dr

Evaluating definite integrals is a cornerstone of calculus, with applications spanning various fields like physics, engineering, and economics. This article delves into the detailed evaluation of the definite integral ∫₀¹ r³/(16 + r²) dr, illustrating various techniques and providing a comprehensive understanding of the process. We’ll explore different approaches, focusing on clarity and step-by-step explanations suitable for both beginners and those seeking a refresher.

Understanding the Problem

Our task is to evaluate the definite integral:

∫₀¹ r³/(16 + r²) dr

This integral presents a challenge because the numerator's degree is higher than the denominator's. Direct integration isn't straightforward. To proceed effectively, we need to employ a technique that simplifies the integrand. The most suitable method in this case is polynomial long division or a substitution technique.

Method 1: Polynomial Long Division

Polynomial long division allows us to rewrite the integrand as a sum of simpler terms. Let's perform the division:

r³ ÷ (16 + r²) = r - (16r)/(16 + r²)

Therefore, our integral becomes:

∫₀¹ [r - (16r)/(16 + r²)] dr

This integral is now separable into two simpler integrals:

∫₀¹ r dr - ∫₀¹ (16r)/(16 + r²) dr

Let's solve each integral separately.

Solving ∫₀¹ r dr

This is a straightforward integral:

∫₀¹ r dr = [r²/2]₀¹ = (1²/2) - (0²/2) = 1/2

Solving ∫₀¹ (16r)/(16 + r²) dr

This integral requires a substitution. Let's use u-substitution:

Let u = 16 + r² Then du = 2r dr or, (1/2)du = r dr

When r = 0, u = 16 + 0² = 16 When r = 1, u = 16 + 1² = 17

Substituting, we get:

∫₁₆¹⁷ (16/2)(1/u) du = 8 ∫₁₆¹⁷ (1/u) du

Now, we can integrate:

8 ∫₁₆¹⁷ (1/u) du = 8 [ln|u| ]₁₆¹⁷ = 8 (ln|17| - ln|16|) = 8 ln(17/16)

Combining the Results

Now, we combine the results from both integrals:

∫₀¹ r³/(16 + r²) dr = 1/2 - 8 ln(17/16)

This is the exact solution to the definite integral. We can approximate this value using a calculator:

1/2 - 8 ln(17/16) ≈ 0.5 - 8 * 0.06061 = 0.5 - 0.48488 ≈ 0.01512

Therefore:

∫₀¹ r³/(16 + r²) dr ≈ 0.01512

Method 2: Partial Fraction Decomposition (Not Applicable Here)

Partial fraction decomposition is a powerful technique for integrating rational functions where the degree of the numerator is less than the degree of the denominator. However, in our case, the degree of the numerator (3) is greater than the degree of the denominator (2). Therefore, partial fraction decomposition isn't directly applicable without first performing polynomial long division.

Understanding the Significance of the Result

The result, approximately 0.01512, represents the area under the curve of the function r³/(16 + r²) from r = 0 to r = 1. This value has practical implications depending on the context where this integral arises. For example, if this integral represents work done, volume, or any other physical quantity, the result gives a numerical quantification of that quantity within the specified limits.

Error Analysis and Numerical Methods

While we've obtained an analytical solution, it's crucial to understand that numerical methods can be used to approximate definite integrals, particularly when an analytical solution is difficult or impossible to find. Methods like the trapezoidal rule, Simpson's rule, or more advanced techniques provide numerical approximations of the integral. The accuracy of these methods depends on the number of intervals used in the approximation. A comparison of the analytical solution with the results from various numerical methods would allow us to assess the accuracy of those numerical methods.

Applications of Definite Integrals

Definite integrals find widespread applications in diverse fields:

• Physics: Calculating work done by a force, displacement of an object under variable acceleration, center of mass of a body.
• Engineering: Determining the area of irregular shapes, volume of solids of revolution, moment of inertia of mechanical components.
• Economics: Computing total cost, total revenue, consumer surplus, producer surplus.
• Probability and Statistics: Finding probabilities, expected values, and other statistical measures.

Conclusion

Evaluating the definite integral ∫₀¹ r³/(16 + r²) dr involves employing appropriate integration techniques, primarily polynomial long division followed by u-substitution. The solution provides a precise analytical value, approximately 0.01512. This approach highlights the importance of understanding various integration techniques and their applications in solving real-world problems across multiple disciplines. Furthermore, the comparison with numerical methods underscores the power and limitations of both analytical and numerical approaches to solving definite integrals. Remember that selecting the appropriate method hinges on the characteristics of the integrand and the desired level of accuracy.