Evaluating The Definite Integral Of 2x√(2x - 3) From 3/2 To 2
How to evaluate the definite integral of 2x times the square root of (2x - 3) from 3/2 to 2?
Introduction
In this article, we will delve into the evaluation of the definite integral of the function 2x√(2x - 3) with respect to x, over the interval from 3/2 to 2. Definite integrals play a crucial role in calculus, representing the net signed area between a function's curve and the x-axis within specified limits. This particular integral presents an interesting challenge that requires a strategic approach to solve. We will explore the necessary steps and techniques to accurately compute the value of this definite integral, providing a comprehensive understanding of the underlying concepts and methodologies involved. This exploration is not only a mathematical exercise but also a demonstration of how calculus can be applied to solve complex problems in various fields, reinforcing the importance of mastering these fundamental mathematical tools.
Setting Up the Integral
The problem at hand is to evaluate the definite integral:
∫3/22 2x√(2x - 3) dx
This integral involves a product of a linear function (2x) and a square root function (√(2x - 3)), making it a suitable candidate for a substitution method. The substitution method is a powerful technique in calculus that simplifies integrals by replacing a complex expression with a simpler variable. This allows us to transform the integral into a form that is easier to handle and solve. By carefully selecting the substitution, we can often break down intricate integrals into manageable parts, making the integration process more straightforward. In this case, the expression under the square root, 2x - 3, appears to be a good choice for substitution, as it simplifies the square root term and potentially simplifies the entire integral. This initial step is crucial in setting the stage for the rest of the solution, as a well-chosen substitution can significantly reduce the complexity of the problem and lead to a more efficient solution.
Applying u-Substitution
To solve this integral, we'll employ the u-substitution method. Let's set:
u = 2x - 3
This substitution choice is strategic because it simplifies the square root term in the integral. By replacing 2x - 3 with u, we transform the √(2x - 3) part of the integral into √u, which is much easier to work with. However, this substitution also necessitates adjusting the rest of the integral to be in terms of u. To do this, we need to find the differential du and express x in terms of u. This is a critical step in the u-substitution process, as it ensures that the entire integral is consistent with the new variable. Failing to properly adjust the integral can lead to incorrect results, highlighting the importance of careful attention to detail in this method. Once we have successfully transformed the integral into a function of u, we can proceed with the integration using standard techniques, making the overall process more manageable and efficient.
Finding du and Expressing x in Terms of u
Differentiating both sides of u = 2x - 3 with respect to x, we get:
du/dx = 2
Thus,
du = 2 dx, which implies dx = du/2
Now, we need to express x in terms of u. From the substitution equation u = 2x - 3, we can solve for x:
2x = u + 3
x = (u + 3) / 2
This step is crucial because it allows us to replace all instances of x in the original integral with expressions involving u. This ensures that the entire integral is now in terms of the new variable u, which is a necessary condition for the u-substitution method to work correctly. The ability to manipulate algebraic expressions and solve for variables is a fundamental skill in calculus, and this step highlights its importance in the context of integration. By expressing x in terms of u, we pave the way for a simpler and more manageable integral that can be solved using standard integration techniques. This transformation is a key step in simplifying the original problem and making it more accessible.
Changing the Limits of Integration
Since we are dealing with a definite integral, we must also change the limits of integration to correspond with the new variable u. The original limits were given in terms of x, so we need to convert them to u values using our substitution equation u = 2x - 3.
When x = 3/2:
u = 2(3/2) - 3 = 3 - 3 = 0
When x = 2:
u = 2(2) - 3 = 4 - 3 = 1
Therefore, the new limits of integration are from u = 0 to u = 1. Changing the limits of integration is a critical step in evaluating definite integrals using substitution. It ensures that the integral is evaluated over the correct interval in the new variable, leading to an accurate result. Failing to change the limits would mean evaluating the integral in terms of u but over the x interval, which would yield an incorrect answer. This step highlights the importance of understanding the relationship between the original and substituted variables and how it affects the limits of integration. By correctly converting the limits, we maintain the integrity of the integral and ensure that our final result accurately represents the area under the curve within the specified bounds.
Rewriting the Integral in Terms of u
Now we can rewrite the integral entirely in terms of u:
∫3/22 2x√(2x - 3) dx = ∫01 2((u + 3) / 2) √u (du/2)
Simplifying, we get:
∫01 (u + 3) √u (du/2) = (1/2) ∫01 (u + 3) √u du
This transformation is a crucial step in the u-substitution method. By replacing all instances of x and dx with their equivalent expressions in terms of u, we have successfully converted the integral into a form that is much easier to handle. The new integral is expressed entirely in terms of the variable u, allowing us to apply standard integration techniques without the complexity of the original expression. This step demonstrates the power of substitution in simplifying integrals and making them more tractable. The algebraic manipulation involved in rewriting the integral is a key skill in calculus, and its proper application is essential for solving a wide range of integration problems. By carefully transforming the integral, we pave the way for a straightforward integration process and an accurate final result.
Integrating with Respect to u
Distribute √u and integrate:
(1/2) ∫01 (u√u + 3√u) du = (1/2) ∫01 (u3/2 + 3u1/2) du
Now, we integrate term by term:
(1/2) [ (2/5)u5/2 + 3(2/3)u3/2 ]01
This step involves applying the power rule for integration, which is a fundamental technique in calculus. The power rule states that the integral of x^n is (x^(n+1))/(n+1), where n is any real number except -1. In this case, we apply the power rule to both u^(3/2) and u^(1/2), increasing their exponents by 1 and dividing by the new exponent. This process transforms the integral into an expression involving powers of u, which can then be evaluated at the limits of integration. The ability to apply the power rule correctly is essential for solving a wide range of integrals, and this step highlights its importance in the context of u-substitution. By successfully integrating with respect to u, we move closer to finding the final value of the definite integral.
Evaluating the Definite Integral
Evaluate the expression at the limits of integration:
(1/2) [ (2/5)(1)5/2 + 2(1)3/2 - (0) ]
= (1/2) [ 2/5 + 2 ]
= (1/2) [ 2/5 + 10/5 ]
= (1/2) [ 12/5 ]
= 6/5
This final step involves substituting the limits of integration into the antiderivative we found in the previous step. The fundamental theorem of calculus states that the definite integral of a function from a to b is equal to the difference between the antiderivative evaluated at b and the antiderivative evaluated at a. In this case, we substitute the upper limit (u = 1) and the lower limit (u = 0) into the expression we obtained after integration. This process yields two values, and we subtract the value at the lower limit from the value at the upper limit to find the definite integral. The arithmetic involved in this step is crucial for obtaining the correct final answer. By carefully evaluating the expression at the limits of integration, we arrive at the numerical value of the definite integral, which represents the net signed area between the curve and the x-axis over the specified interval. This result provides a complete solution to the problem and demonstrates the power of calculus in solving real-world problems.
Conclusion
Therefore, the definite integral ∫3/22 2x√(2x - 3) dx evaluates to 6/5. This problem demonstrates the effectiveness of the u-substitution method in simplifying and solving integrals. By carefully choosing a substitution and changing the limits of integration accordingly, we were able to transform a complex integral into a manageable form. This process highlights the importance of strategic problem-solving techniques in calculus and underscores the power of substitution in simplifying mathematical expressions. The result, 6/5, represents the exact value of the definite integral, providing a clear and concise answer to the problem. This exercise not only reinforces the fundamental concepts of integration but also demonstrates the practical application of calculus in finding the area under a curve. The ability to solve such problems is essential for students and professionals in various fields, emphasizing the continued relevance of calculus in modern science and engineering.
Keywords
Evaluate the definite integral from 3/2 to 2 of 2x times the square root of (2x - 3) with respect to x.