Evaluate Integral Of (x - Α)^(p-1) (x - Β)^(-p-1) A Step-by-Step Solution

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In the realm of calculus, evaluating integrals stands as a fundamental skill, enabling us to solve a myriad of problems across diverse fields. This article delves into the intricacies of evaluating a specific integral, providing a step-by-step guide and insightful explanations to enhance your understanding. We will focus on the integral:

∫ (x - α)^(p-1) (x - β)^(-p-1) dx, where p ≥ 0 and α ≠ β.

This seemingly complex integral can be solved elegantly using a clever substitution technique. Let's embark on this mathematical journey together!

Understanding the Integral

Before diving into the solution, it's crucial to grasp the essence of the integral. The integral represents the area under the curve defined by the function (x - α)^(p-1) (x - β)^(-p-1). The parameters p, α, and β play significant roles in shaping the curve and, consequently, the integral's value. Specifically, p (p ≥ 0) is a non-negative exponent that influences the function's behavior, while α and β (α ≠ β) are distinct constants that determine the function's points of discontinuity or singularities. Recognizing these parameters is vital for choosing the right approach to solve the integral.

The Substitution Technique

To tackle this integral effectively, we'll employ the powerful technique of substitution. The core idea behind substitution is to replace a complex expression within the integral with a simpler variable, making the integration process more manageable. In this case, a strategic substitution can transform the integral into a recognizable form that we can readily solve. The key to a successful substitution lies in choosing the right expression to replace, and in our case, the fraction (x - α) / (x - β) seems like a promising candidate due to its presence in the potential solution forms provided. Let's proceed with the substitution:

y = (x - α) / (x - β)

This substitution introduces a new variable, y, which is a function of x. To proceed further, we need to express dx in terms of dy. This involves differentiating both sides of the equation with respect to x. Applying the quotient rule of differentiation, we get:

dy/dx = [(x - β)(1) - (x - α)(1)] / (x - β)^2

Simplifying the numerator, we have:

dy/dx = (x - β - x + α) / (x - β)^2

dy/dx = (α - β) / (x - β)^2

Now, we can isolate dx:

dx = (x - β)^2 / (α - β) dy

This expression for dx in terms of dy is crucial for transforming the integral from the variable x to the variable y. The next step involves expressing the original integrand in terms of y as well.

Transforming the Integral

Now that we have dx in terms of dy, we need to rewrite the original integral entirely in terms of y. Recall our substitution:

y = (x - α) / (x - β)

We have (x - α) = y(x - β). The integral we want to evaluate is:

∫ (x - α)^(p-1) (x - β)^(-p-1) dx

We can rewrite the integrand using our substitution:

∫ [y(x - β)]^(p-1) (x - β)^(-p-1) dx

∫ y^(p-1) (x - β)^(p-1) (x - β)^(-p-1) dx

∫ y^(p-1) (x - β)^(p-1-p+1) dx

∫ y^(p-1) (x - β)^0 dx

∫ y^(p-1) dx

Now, substitute the expression for dx we found earlier:

∫ y^(p-1) [(x - β)^2 / (α - β)] dy

We need to express (x - β)^2 in terms of y. From the substitution equation, we can write:

y = (x - α) / (x - β)

y(x - β) = x - α

xy - yβ = x - α

xy - x = yβ - α

x(y - 1) = yβ - α

x = (yβ - α) / (y - 1)

Therefore,

x - β = (yβ - α) / (y - 1) - β

x - β = (yβ - α - β(y - 1)) / (y - 1)

x - β = (yβ - α - yβ + β) / (y - 1)

x - β = (β - α) / (y - 1)

Now we can substitute this back into our integral:

∫ y^(p-1) [(β - α)^2 / (y - 1)^2] / (α - β) dy

∫ y^(p-1) (β - α)^2 / (α - β) (y - 1)^2 dy

Since (β - α) = -(α - β), we have (β - α)^2 = (α - β)^2, so

∫ y^(p-1) (α - β)^2 / (α - β) (y - 1)^2 dy

∫ y^(p-1) (α - β) / (y - 1)^2 dy

This looks complicated, and it seems we made an error in our calculation. Let's go back to

∫ y^(p-1) dx = ∫ y^(p-1)  (x - β)^2 / (α - β) dy

Instead of substituting (x-β), let's try another approach. We have:

∫ (x - α)^(p-1) (x - β)^(-p-1) dx

∫ [(x - α) / (x - β)]^(p-1) (x - β)^(p-1) (x - β)^(-p-1) dx

∫ [(x - α) / (x - β)]^(p-1) (x - β)^(-2) dx

Let y = (x - α) / (x - β). Then dy/dx = (α - β) / (x - β)^2 as before. So dx = (x - β)^2 / (α - β) dy. The integral becomes:

∫ y^(p-1) (x - β)^(-2) [(x - β)^2 / (α - β)] dy

∫ y^(p-1) / (α - β) dy

This is much simpler!

Integrating with Respect to y

Now we have a much simpler integral in terms of y:

∫ [y^(p-1) / (α - β)] dy

Since (α - β) is a constant, we can take it out of the integral:

[1 / (α - β)] ∫ y^(p-1) dy

This is a power rule integral. Applying the power rule, we get:

[1 / (α - β)] [y^p / p] + C

where C is the constant of integration.

Substituting Back for x

Finally, we need to substitute back for x using our original substitution:

y = (x - α) / (x - β)

So, the final result is:

[1 / (p(α - β))] [(x - α) / (x - β)]^p + C

The Correct Answer

Comparing our result with the given options, we find that the correct answer is:

a)

[1 / (p(α - β))] [(x - α) / (x - β)]^p + C

Conclusion

Evaluating integrals can be a challenging yet rewarding endeavor. By employing techniques like substitution and carefully manipulating the integrand, we can often transform complex integrals into manageable forms. In this article, we successfully evaluated the integral ∫ (x - α)^(p-1) (x - β)^(-p-1) dx, demonstrating the power and elegance of calculus. Understanding the underlying concepts and practicing various techniques are key to mastering the art of integration.

This step-by-step guide provides a comprehensive approach to solving this particular integral, but the principles and techniques discussed can be applied to a wide range of integration problems. Remember to always analyze the integral carefully, choose appropriate substitution methods, and meticulously perform the calculations. With practice and perseverance, you can unlock the beauty and power of integral calculus.