Evaluating ∫₀¹ X²(1 + 2x³)⁵ Dx A Step-by-Step Guide

#Introduction

In this article, we will delve into the process of evaluating the definite integral ∫₀¹ x²(1 + 2x³)⁵ dx. This integral falls under the category of calculus problems that can be efficiently solved using the method of u-substitution. U-substitution, also known as substitution, is a powerful technique used to simplify integrals by replacing a complex expression with a simpler variable, thereby making the integration process more manageable. Our focus will be on providing a step-by-step explanation, ensuring clarity and understanding for readers of all levels. We will not only solve the integral but also discuss the underlying concepts and why this particular method is effective in this scenario. Understanding the nuances of u-substitution is crucial for mastering integral calculus, and this example provides an excellent opportunity to strengthen your grasp of the technique. By the end of this article, you will have a thorough understanding of how to approach similar integration problems with confidence and precision.

The integral ∫₀¹ x²(1 + 2x³)⁵ dx presents a classic case for employing the u-substitution method. The complexity of the integrand, particularly the presence of the term (1 + 2x³)⁵, makes direct integration challenging. However, by recognizing that the derivative of the inner function (1 + 2x³) is closely related to the x² term outside the parentheses, we can strategically simplify the integral. This recognition is key to successfully applying u-substitution. The technique involves substituting a new variable, u, for the inner function, which transforms the integral into a more manageable form. This transformation often reveals a basic integral that can be easily evaluated using standard integration rules. Furthermore, when dealing with definite integrals, it's essential to update the limits of integration to correspond to the new variable u. This ensures that the final result accurately reflects the value of the integral over the original interval. Our exploration will cover each of these steps in detail, providing a comprehensive understanding of the entire process.

The importance of mastering techniques like u-substitution in calculus cannot be overstated. Integral calculus forms the foundation for numerous applications in various fields, including physics, engineering, economics, and statistics. Whether you are calculating areas under curves, determining volumes of solids, or modeling physical phenomena, the ability to solve integrals is indispensable. U-substitution, in particular, is a versatile tool that allows us to tackle a wide range of integration problems that would otherwise be intractable. By understanding the underlying principles and practicing with diverse examples, you can develop the skills necessary to confidently approach complex mathematical challenges. This article aims to equip you with the knowledge and skills to not only solve the given integral but also to apply u-substitution effectively in various other contexts. So, let's embark on this journey of unraveling the intricacies of integral calculus and mastering the art of u-substitution.

Step-by-Step Solution

1. Identify the Appropriate Substitution

The first crucial step in solving the integral ∫₀¹ x²(1 + 2x³)⁵ dx using u-substitution is to identify a suitable expression for the substitution. The key to this lies in recognizing a function and its derivative (or a multiple thereof) within the integrand. In this case, we observe the composite function (1 + 2x³)⁵. The inner function is (1 + 2x³), and its derivative is 6x². Notice that we have x² in the integrand, which is a constant multiple of the derivative of (1 + 2x³). This observation makes u = 1 + 2x³ an excellent choice for our substitution. The goal here is to simplify the integral by making it easier to integrate, and this substitution will indeed do that by transforming the complex expression (1 + 2x³)⁵ into a simpler u⁵ term.

By choosing u = 1 + 2x³, we are essentially replacing a more complex part of the integrand with a single variable. This simplifies the algebraic structure of the integral, making it more amenable to standard integration techniques. The presence of x² in the integrand, which is closely related to the derivative of u, is a strong indicator that this substitution will be fruitful. Identifying such relationships is a fundamental skill in applying u-substitution effectively. It requires a keen eye for recognizing patterns and understanding the chain rule in reverse. The correct substitution not only simplifies the integral but also sets the stage for the subsequent steps in the solution process. Therefore, careful consideration of the integrand and its components is essential for a successful application of u-substitution.

2. Compute the Differential du

After identifying the substitution u = 1 + 2x³, the next step is to compute the differential du. The differential du represents the derivative of u with respect to x, multiplied by dx. This step is crucial because it allows us to replace the dx term in the original integral with an expression involving du. To find du, we differentiate u with respect to x:

du/dx = d/dx (1 + 2x³)

Applying the power rule of differentiation, we get:

du/dx = 6x²

Now, we solve for du by multiplying both sides by dx:

du = 6x² dx

This expression for du is vital because it allows us to relate the original variable of integration, x, to the new variable, u. Notice that we have x² dx in the original integral, and our du expression contains 6x² dx. This means we can express x² dx in terms of du by dividing both sides of the equation by 6:

(1/6) du = x² dx

This relationship is key to transforming the original integral into an integral with respect to u. The process of computing the differential du is a fundamental step in u-substitution. It ensures that we can properly change the variable of integration, which is essential for simplifying the integral and making it solvable. The ability to accurately compute du and manipulate the resulting expression is a critical skill in integral calculus.

3. Change the Limits of Integration

Since we are dealing with a definite integral, ∫₀¹ x²(1 + 2x³)⁵ dx, it is essential to change the limits of integration to correspond to the new variable, u. This step is often overlooked but is crucial for obtaining the correct final answer. The original limits of integration, 0 and 1, are the values of x. We need to find the corresponding values of u using our substitution u = 1 + 2x³.

When x = 0, we substitute this value into the equation for u:

u = 1 + 2(0)³ = 1 + 0 = 1

So, the lower limit of integration in terms of u is 1.

Next, we find the upper limit of integration by substituting x = 1 into the equation for u:

u = 1 + 2(1)³ = 1 + 2 = 3

Therefore, the upper limit of integration in terms of u is 3. Now, our new limits of integration are 1 and 3. Changing the limits of integration is a critical step in evaluating definite integrals using u-substitution. It ensures that the final result represents the value of the integral over the correct interval in terms of the new variable. By transforming the limits of integration, we avoid the need to substitute back for x after integrating with respect to u, which simplifies the overall process. This step highlights the importance of careful attention to detail when applying u-substitution to definite integrals.

4. Substitute and Rewrite the Integral

Now that we have our substitution, u = 1 + 2x³, the differential, (1/6) du = x² dx, and the new limits of integration, 1 and 3, we can substitute these into the original integral ∫₀¹ x²(1 + 2x³)⁵ dx. The goal here is to rewrite the integral entirely in terms of u, which will make it easier to evaluate.

We start by replacing (1 + 2x³) with u and x² dx with (1/6) du. We also replace the limits of integration, 0 and 1, with their corresponding u values, 1 and 3. The integral now becomes:

∫₁³ u⁵ (1/6) du

We can simplify this further by factoring out the constant (1/6) from the integral:

(1/6) ∫₁³ u⁵ du

This transformed integral is much simpler than the original. It involves integrating a simple power function, u⁵, with respect to u. The process of substitution has successfully transformed a complex integral into a basic one, which we can easily evaluate using the power rule for integration. The ability to rewrite integrals in a simpler form through substitution is the essence of the u-substitution technique. This step demonstrates the power of this method in simplifying complex mathematical expressions and making them tractable.

5. Evaluate the Integral

With the integral rewritten in terms of u as (1/6) ∫₁³ u⁵ du, we can now proceed to evaluate it. The integral of u⁵ with respect to u is found using the power rule for integration, which states that ∫xⁿ dx = (xⁿ⁺¹)/(n+1) + C, where C is the constant of integration. Applying this rule, we get:

∫ u⁵ du = (u⁶)/6 + C

Since we are dealing with a definite integral, we do not need to include the constant of integration, C. Now, we substitute this result back into our expression:

(1/6) ∫₁³ u⁵ du = (1/6) [(u⁶)/6]₁³

Next, we evaluate the antiderivative at the upper and lower limits of integration and subtract the values:

(1/6) [(3⁶)/6 - (1⁶)/6] = (1/6) [(729)/6 - (1)/6] = (1/6) [(728)/6] = (728)/36

We can simplify the fraction by dividing both the numerator and denominator by their greatest common divisor, which is 4:

(728)/36 = (182)/9

Therefore, the value of the definite integral is 182/9. This step demonstrates the direct application of the power rule for integration after the u-substitution has simplified the integral. The process of evaluating the integral involves finding the antiderivative and then applying the fundamental theorem of calculus to compute the definite integral. The final result, 182/9, represents the exact value of the area under the curve defined by the original integrand over the interval [0, 1].

Final Answer

After completing all the steps of the u-substitution method, we have successfully evaluated the definite integral ∫₀¹ x²(1 + 2x³)⁵ dx. The final answer is:

182/9

This result represents the exact value of the integral. To recap, we began by identifying a suitable substitution, u = 1 + 2x³, which simplified the integrand. We then computed the differential, du = 6x² dx, and changed the limits of integration to correspond to the new variable u. This transformation allowed us to rewrite the integral in a simpler form, (1/6) ∫₁³ u⁵ du. We then applied the power rule for integration to find the antiderivative and evaluated it at the new limits of integration. This process yielded the final result, 182/9. The successful evaluation of this integral demonstrates the power and effectiveness of the u-substitution method in simplifying complex integrals. It also highlights the importance of careful attention to detail and a thorough understanding of the fundamental concepts of integral calculus. This example serves as a valuable illustration of how to approach similar integration problems with confidence and precision. Understanding and mastering such techniques is crucial for anyone delving into advanced mathematics, physics, or engineering.

Conclusion

In conclusion, the evaluation of the definite integral ∫₀¹ x²(1 + 2x³)⁵ dx provides a clear and comprehensive illustration of the u-substitution method. This technique is a cornerstone of integral calculus, enabling us to solve integrals that would otherwise be intractable. By strategically choosing a substitution, we can simplify the integrand and transform the integral into a more manageable form. In this specific example, the substitution u = 1 + 2x³ proved to be highly effective, as it allowed us to leverage the relationship between the inner function and its derivative. The process involved several key steps, including identifying the appropriate substitution, computing the differential, changing the limits of integration, rewriting the integral in terms of the new variable, and finally, evaluating the simplified integral.

The u-substitution method is not merely a mechanical procedure; it requires a deep understanding of the underlying concepts of calculus, particularly the chain rule and the fundamental theorem of calculus. The ability to recognize suitable substitutions comes with practice and a keen eye for patterns within the integrand. Moreover, the careful handling of limits of integration in definite integrals is crucial for obtaining the correct result. This example underscores the importance of attention to detail and a systematic approach when applying u-substitution. The final answer, 182/9, represents the precise value of the definite integral, demonstrating the power of calculus in solving real-world problems.

Furthermore, the u-substitution method serves as a foundation for more advanced integration techniques. As you delve deeper into calculus, you will encounter various other methods, such as integration by parts, trigonometric substitution, and partial fraction decomposition. However, a solid grasp of u-substitution is essential for mastering these more complex techniques. The principles learned in this example can be applied to a wide range of integration problems, making u-substitution a versatile and indispensable tool in your mathematical toolkit. By mastering this technique, you will be well-equipped to tackle a variety of calculus challenges and appreciate the elegance and power of integral calculus.