I) How To Solve The Integral Of (√x + 1/√x)² Dx? Ii) How To Solve The Integral Of (6x + 8)√(3x² + 8x + 2) Dx? B) If The Marginal Revenue Function Of A Product Is MR = 8 + 5x - X², How To Find The Total Revenue And The Average Revenue, Given TR = 82 When X = 5?

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This article delves into solving two integral problems and determining total and average revenue functions using marginal revenue. We will explore the step-by-step solutions, highlighting key calculus concepts and techniques. Whether you're a student, educator, or simply someone interested in applied mathematics, this guide provides clear explanations and practical applications.

Part 1: Solving Integral Problems

In this section, we will tackle two integral problems. The first involves integrating the square of a sum involving square roots, while the second deals with integrating a product involving a square root function. Let’s dive in!

i) ∫(√x + 1/√x)² dx

Understanding the Integral

The integral ∫(√x + 1/√x)² dx requires us to find the antiderivative of the function (√x + 1/√x)². This involves several steps, including expanding the square, simplifying the terms, and applying the power rule for integration. Mastering such problems is crucial for understanding more complex calculus concepts.

Step-by-Step Solution

  1. Expand the Square: First, we need to expand the square of the binomial (√x + 1/√x)². Using the formula (a + b)² = a² + 2ab + b², we get:

    (√x + 1/√x)² = (√x)² + 2(√x)(1/√x) + (1/√x)²

    = x + 2 + 1/x

  2. Rewrite the Integral: Now, we rewrite the integral using the expanded form:

    ∫(√x + 1/√x)² dx = ∫(x + 2 + 1/x) dx

  3. Apply the Power Rule and Basic Integration Rules: We can now integrate each term separately. Recall the power rule for integration, ∫xⁿ dx = (x^(n+1))/(n+1) + C, and the integral of 1/x is ln|x|. Thus,

    ∫(x + 2 + 1/x) dx = ∫x dx + ∫2 dx + ∫(1/x) dx

    = (x²)/2 + 2x + ln|x| + C

Final Answer

Therefore, the solution to the integral ∫(√x + 1/√x)² dx is:

(x²)/2 + 2x + ln|x| + C

This result showcases the application of basic algebraic expansion and fundamental integration rules. Practicing such problems enhances proficiency in calculus.

ii) ∫(6x + 8)√(3x² + 8x + 2) dx

Understanding the Integral

This integral, ∫(6x + 8)√(3x² + 8x + 2) dx, requires a different approach due to the presence of the square root function and the term (6x + 8). The key to solving this integral is using substitution. Substitution is a powerful technique in calculus, allowing us to simplify complex integrals by replacing a part of the integrand with a new variable.

Step-by-Step Solution

  1. Identify a Suitable Substitution: Let’s try the substitution u = 3x² + 8x + 2. This choice is strategic because the derivative of u with respect to x is:

    du/dx = 6x + 8

    This term appears in the integral, which suggests that this substitution will simplify the integral.

  2. Find du: Multiply both sides by dx to get:

    du = (6x + 8) dx

  3. Rewrite the Integral in Terms of u: Now, we substitute u and du into the integral:

    ∫(6x + 8)√(3x² + 8x + 2) dx = ∫√u du

    = ∫u^(1/2) du

  4. Apply the Power Rule for Integration: We can now integrate u^(1/2) with respect to u. Using the power rule, ∫uⁿ du = (u^(n+1))/(n+1) + C, we get:

    ∫u^(1/2) du = (u^(3/2))/(3/2) + C

    = (2/3)u^(3/2) + C

  5. Substitute Back: Finally, substitute back u = 3x² + 8x + 2 to get the result in terms of x:

    (2/3)(3x² + 8x + 2)^(3/2) + C

Final Answer

Therefore, the solution to the integral ∫(6x + 8)√(3x² + 8x + 2) dx is:

(2/3)(3x² + 8x + 2)^(3/2) + C

This example demonstrates the power of substitution in simplifying integrals. By choosing an appropriate substitution, we transformed a seemingly complex integral into a straightforward application of the power rule.

Part 2: Finding Total and Average Revenue Functions

In this section, we will explore the application of calculus in economics, specifically in determining the total and average revenue functions given the marginal revenue function. Revenue functions are crucial in business and economics for understanding and predicting financial performance.

b) Finding Total and Average Revenue

Understanding the Concepts

In economics, the marginal revenue (MR) function represents the additional revenue generated by selling one more unit of a product. The total revenue (TR) function represents the total income from selling a certain quantity of the product, and the average revenue (AR) function is the revenue per unit sold. Understanding the relationships between these functions is essential for business decision-making.

Problem Statement

The marginal revenue function of a product is given by MR = 8 + 5x - x². We need to find the total revenue (TR) and the average revenue (AR), given that TR = 82 when x = 5. This problem combines calculus and economic principles to derive important financial metrics.

Step-by-Step Solution

  1. Find the Total Revenue Function (TR): The total revenue function is the integral of the marginal revenue function. So, we need to integrate MR = 8 + 5x - x² with respect to x:

    TR = ∫MR dx = ∫(8 + 5x - x²) dx

    Applying the power rule for integration, we get:

    TR = 8x + (5x²)/2 - (x³)/3 + C

    Here, C is the constant of integration, which we need to determine using the given condition.

  2. Determine the Constant of Integration (C): We are given that TR = 82 when x = 5. Substitute these values into the total revenue function:

    82 = 8(5) + (5(5)²)/2 - (5³)/3 + C

    82 = 40 + (125)/2 - (125)/3 + C

    82 = 40 + 62.5 - 41.67 + C

    82 = 60.83 + C

    C = 82 - 60.83

    C ≈ 21.17

    So, the total revenue function is:

    TR = 8x + (5x²)/2 - (x³)/3 + 21.17

  3. Find the Average Revenue Function (AR): The average revenue function is the total revenue divided by the quantity x:

    AR = TR/x

    Substitute the total revenue function we found:

    AR = (8x + (5x²)/2 - (x³)/3 + 21.17) / x

    AR = 8 + (5x)/2 - (x²)/3 + 21.17/x

Final Answers

The total revenue function is:

TR = 8x + (5x²)/2 - (x³)/3 + 21.17

The average revenue function is:

AR = 8 + (5x)/2 - (x²)/3 + 21.17/x

This example illustrates how calculus can be used to derive important economic functions from marginal revenue. Understanding these functions helps businesses make informed decisions about pricing and production levels.

Conclusion

In this article, we explored solving integral problems and deriving revenue functions using calculus. We tackled two integral problems, demonstrating the power rule and substitution techniques. Additionally, we found the total and average revenue functions from a given marginal revenue function, showcasing the practical application of calculus in economics. Mastering these concepts is crucial for anyone studying calculus or working in fields that require mathematical modeling and analysis. The step-by-step solutions provided here offer a clear and comprehensive guide to these topics.