Evaluating Definite Integral Of (4x^2 + 2) / √x From 4 To 6

In this article, we will delve into the process of evaluating the definite integral of the function ${ f(x) = \frac{4x^2 + 2}{\sqrt{x}} }$ over the interval [4, 6]. Definite integrals are a fundamental concept in calculus, representing the signed area between a curve and the x-axis within specified limits. This particular integral involves a rational function with a square root in the denominator, requiring us to employ algebraic manipulation and the power rule for integration. Understanding how to solve such integrals is crucial for various applications in physics, engineering, and economics, where calculating areas, volumes, and other cumulative quantities is essential. The steps we will follow include simplifying the integrand, finding the antiderivative, and then applying the Fundamental Theorem of Calculus to evaluate the definite integral. This process not only enhances our understanding of integral calculus but also provides a practical skill for solving real-world problems. By breaking down the problem into manageable steps, we will gain a clear and concise method for tackling similar integration challenges in the future. The ability to accurately evaluate definite integrals is a cornerstone of advanced mathematical analysis, allowing us to model and predict outcomes in a wide range of disciplines. Through this detailed exploration, we aim to provide a comprehensive guide that demystifies the process and equips readers with the confidence to approach integral problems effectively. Ultimately, mastering the evaluation of definite integrals opens doors to deeper insights into the behavior of functions and their applications in diverse fields. Let’s embark on this mathematical journey to unravel the intricacies of this integral and its significance.

Problem Statement

Our task is to evaluate the definite integral:

$$\int_4^6 \frac{4x^2 + 2}{\sqrt{x}} dx$$

This integral represents the area under the curve ${ f(x) = \frac{4x^2 + 2}{\sqrt{x}} }$ from ${ x = 4 }$ to ${ x = 6 }$. To solve this, we will first simplify the integrand, find its antiderivative, and then apply the Fundamental Theorem of Calculus. The integrand, ${ \frac{4x^2 + 2}{\sqrt{x}} }$, is a rational function with a square root in the denominator, which requires us to manipulate the expression algebraically before we can apply the power rule for integration. The limits of integration, 4 and 6, define the interval over which we are calculating the area. Evaluating this definite integral will give us a precise numerical value representing the signed area between the curve and the x-axis within this interval. This type of problem is common in calculus and is a foundational skill for more advanced mathematical concepts and applications. The process involves a combination of algebraic simplification and calculus techniques, highlighting the interconnectedness of these mathematical disciplines. By systematically breaking down the problem, we can gain a deeper understanding of the underlying principles and develop a robust approach to solving similar integrals. The final answer will provide a quantitative measure of the area, which can be interpreted in various contexts depending on the application. Understanding how to evaluate such integrals is essential for anyone working with mathematical models in fields such as physics, engineering, and economics, where areas and cumulative quantities are frequently encountered. This exercise is not just about finding a number; it’s about mastering a fundamental technique in calculus and appreciating its broader significance.

Step 1: Simplify the Integrand

To begin, we simplify the integrand by dividing each term in the numerator by ${ \sqrt{x} }$. Recall that ${ \sqrt{x} = x^{\frac{1}{2}} }$, so we can rewrite the integral as:

$$\int_4^6 \frac{4x^2 + 2}{x^{\frac{1}{2}}} dx = \int_4^6 (4x^{2 - \frac{1}{2}} + 2x^{-\frac{1}{2}}) dx$$

This step is crucial because it transforms the integrand into a form that is easier to integrate using the power rule. By dividing each term, we separate the original fraction into two simpler terms, each involving a power of ${ x }$. This algebraic manipulation is a standard technique in calculus, allowing us to handle more complex expressions by breaking them down into manageable components. The exponents are then simplified by subtracting ${ \frac{1}{2} }$ from the exponent of ${ x }$ in each term. This process highlights the importance of understanding exponent rules in calculus. Simplifying the integrand not only makes the integration process smoother but also reduces the chances of making errors in subsequent steps. The resulting expression, ${ 4x^{\frac{3}{2}} + 2x^{-\frac{1}{2}} }$, is now in a form that we can directly integrate using the power rule, which states that ${ \int x^n dx = \frac{x^{n+1}}{n+1} + C }$, where ${ n \neq -1 }$ and ${ C }$ is the constant of integration. This initial simplification sets the stage for the rest of the solution, making the integral more approachable and demonstrating the power of algebraic manipulation in calculus. The ability to recognize and apply such simplifications is a key skill for anyone studying calculus, enabling them to tackle a wide range of integration problems effectively. This step is not merely a preliminary exercise; it’s a fundamental part of the problem-solving strategy.

Simplifying the exponents, we get:

$$\int_4^6 (4x^{\frac{3}{2}} + 2x^{-\frac{1}{2}}) dx$$

Step 2: Find the Antiderivative

Now, we find the antiderivative of the simplified integrand. Using the power rule for integration, ${ \int x^n dx = \frac{x^{n+1}}{n+1} + C }$, we integrate each term separately:

$$\int (4x^{\frac{3}{2}} + 2x^{-\frac{1}{2}}) dx = 4 \int x^{\frac{3}{2}} dx + 2 \int x^{-\frac{1}{2}} dx$$

Applying the power rule, we have:

$$4 \cdot \frac{x^{\frac{5}{2}}}{\frac{5}{2}} + 2 \cdot \frac{x^{\frac{1}{2}}}{\frac{1}{2}} + C = \frac{8}{5}x^{\frac{5}{2}} + 4x^{\frac{1}{2}} + C$$

Here, we have found the general antiderivative, which includes the constant of integration ${ C }$. However, since we are evaluating a definite integral, the constant ${ C }$ will cancel out when we apply the Fundamental Theorem of Calculus. This step is a direct application of the power rule, which is a cornerstone of integral calculus. The power rule allows us to reverse the process of differentiation, finding a function whose derivative is the given integrand. The antiderivative we have found represents a family of functions that differ only by a constant. The terms ${ \frac{8}{5}x^{\frac{5}{2}} }$ and ${ 4x^{\frac{1}{2}} }$ are the result of applying the power rule to each term in the simplified integrand. The exponents are increased by 1, and the coefficients are adjusted accordingly. This process highlights the inverse relationship between differentiation and integration. Finding the antiderivative is a critical step in evaluating definite integrals, as it allows us to apply the Fundamental Theorem of Calculus, which connects the concept of antiderivatives to the area under a curve. The accuracy of this step is essential for obtaining the correct final answer. A thorough understanding of the power rule and its application is therefore vital for mastering integral calculus. This step is not just a mechanical application of a formula; it’s a demonstration of the fundamental principles that underpin calculus.

Step 3: Apply the Fundamental Theorem of Calculus

To evaluate the definite integral, we apply the Fundamental Theorem of Calculus, which states that:

$$\int_a^b f(x) dx = F(b) - F(a)$$

where ${ F(x) }$ is the antiderivative of ${ f(x) }$. In our case, ${ a = 4 }$, ${ b = 6 }$, and ${ F(x) = \frac{8}{5}x^{\frac{5}{2}} + 4x^{\frac{1}{2}} }$. Thus, we need to evaluate ${ F(6) - F(4) }$:

$$F(6) = \frac{8}{5}(6)^{\frac{5}{2}} + 4(6)^{\frac{1}{2}}$$

$$F(4) = \frac{8}{5}(4)^{\frac{5}{2}} + 4(4)^{\frac{1}{2}}$$

The Fundamental Theorem of Calculus is the bridge that connects the concepts of differentiation and integration, allowing us to compute definite integrals using antiderivatives. This theorem is a cornerstone of calculus, providing a powerful tool for calculating areas, volumes, and other cumulative quantities. Applying the theorem involves evaluating the antiderivative at the upper and lower limits of integration and then finding the difference. In this step, we substitute ${ x = 6 }$ and ${ x = 4 }$ into the antiderivative we found in the previous step. The calculations for ${ F(6) }$ and ${ F(4) }$ involve raising numbers to fractional powers, which may require careful evaluation or the use of a calculator. The expressions ${ (6)^{\frac{5}{2}} }$, ${ (6)^{\frac{1}{2}} }$, ${ (4)^{\frac{5}{2}} }$, and ${ (4)^{\frac{1}{2}} }$ represent the values of ${ x^{\frac{5}{2}} }$ and ${ x^{\frac{1}{2}} }$ at the limits of integration. This step highlights the importance of understanding the properties of exponents and how to work with fractional powers. The accuracy of these calculations is crucial for obtaining the correct final answer. The Fundamental Theorem of Calculus not only provides a method for evaluating definite integrals but also gives a deep insight into the relationship between the rate of change of a function (the derivative) and the accumulation of that function over an interval (the integral). This step is not merely a substitution of values; it’s the culmination of the integration process and the application of a fundamental theorem.

Step 4: Calculate F(6) and F(4)

Let's calculate ${ F(6) }$ and ${ F(4) }$ separately:

$$F(6) = \frac{8}{5}(6)^{\frac{5}{2}} + 4(6)^{\frac{1}{2}} = \frac{8}{5}(6^2 \sqrt{6}) + 4\sqrt{6} = \frac{8}{5}(36\sqrt{6}) + 4\sqrt{6} = \frac{288}{5}\sqrt{6} + 4\sqrt{6}$$

$$F(4) = \frac{8}{5}(4)^{\frac{5}{2}} + 4(4)^{\frac{1}{2}} = \frac{8}{5}(2^2)^{\frac{5}{2}} + 4(2) = \frac{8}{5}(32) + 8 = \frac{256}{5} + 8$$

In this step, we evaluate the antiderivative at the limits of integration, which involves calculating the values of the function ${ F(x) }$ at ${ x = 6 }$ and ${ x = 4 }$. The calculation of ${ F(6) }$ involves simplifying ${ 6^{\frac{5}{2}} }$ as ${ 6^2 \sqrt{6} }$, which is ${ 36\sqrt{6} }$. We then multiply this by ${ \frac{8}{5} }$ and add ${ 4\sqrt{6} }$ to obtain the final value for ${ F(6) }$. The calculation of ${ F(4) }$ involves simplifying ${ 4^{\frac{5}{2}} }$ as ${ (2^2)^{\frac{5}{2}} }$, which simplifies to ${ 2^5 }$ or 32. We then multiply this by ${ \frac{8}{5} }$ and add ${ 4(2) }$ or 8 to obtain the final value for ${ F(4) }$. These calculations require careful attention to the order of operations and the properties of exponents and radicals. The ability to simplify expressions involving fractional exponents is crucial for obtaining accurate results. This step is a direct application of the substitution process in the Fundamental Theorem of Calculus. The numerical values obtained for ${ F(6) }$ and ${ F(4) }$ will be used in the next step to find the definite integral. This step is not just about performing arithmetic; it’s about applying mathematical principles to evaluate a function at specific points, a fundamental skill in calculus and mathematical analysis. The accuracy of these calculations is paramount, as they directly impact the final result of the integral.

Step 5: Compute F(6) - F(4)

Now, we compute ${ F(6) - F(4) }$:

$$F(6) - F(4) = (\frac{288}{5}\sqrt{6} + 4\sqrt{6}) - (\frac{256}{5} + 8)$$

Combine the terms with ${ \sqrt{6} }$:

$$\frac{288}{5}\sqrt{6} + \frac{20}{5}\sqrt{6} = \frac{308}{5}\sqrt{6}$$

Rewrite 8 as a fraction with a denominator of 5:

$$8 = \frac{40}{5}$$

Now subtract:

$$F(6) - F(4) = \frac{308}{5}\sqrt{6} - \frac{256}{5} - \frac{40}{5} = \frac{308}{5}\sqrt{6} - \frac{296}{5}$$

This step involves subtracting the value of the antiderivative at the lower limit of integration from its value at the upper limit, as dictated by the Fundamental Theorem of Calculus. We begin by substituting the expressions for ${ F(6) }$ and ${ F(4) }$ that we calculated in the previous step. The next step is to simplify the expression by combining like terms. We first combine the terms involving ${ \sqrt{6} }$ by adding the coefficients ${ \frac{288}{5} }$ and ${ 4 }$, which we rewrite as ${ \frac{20}{5} }$. This gives us a combined coefficient of ${ \frac{308}{5} }$. We then rewrite the integer 8 as a fraction with a denominator of 5, which is ${ \frac{40}{5} }$, to facilitate the subtraction of the constant terms. Finally, we subtract ${ \frac{256}{5} }$ and ${ \frac{40}{5} }$ from the term involving ${ \sqrt{6} }$, resulting in the expression ${ \frac{308}{5}\sqrt{6} - \frac{296}{5} }$. This step requires careful attention to algebraic manipulation and arithmetic operations. The accuracy of this subtraction is crucial for obtaining the correct final answer. This step is not just a mechanical application of subtraction; it’s the culmination of the entire integration process, bringing together the results of the previous steps to arrive at the value of the definite integral. The final expression represents the signed area under the curve of the integrand between the specified limits of integration.

Final Answer

Therefore, the definite integral is:

$$\int_4^6 \frac{4x^2 + 2}{\sqrt{x}} dx = \frac{308}{5}\sqrt{6} - \frac{296}{5}$$

This final answer represents the signed area between the curve ${ f(x) = \frac{4x^2 + 2}{\sqrt{x}} }$ and the x-axis over the interval [4, 6]. The value ${ \frac{308}{5}\sqrt{6} - \frac{296}{5} }$ is a precise numerical representation of this area, taking into account the regions where the function is above and below the x-axis. The process of obtaining this answer involved several key steps, including simplifying the integrand, finding the antiderivative, applying the Fundamental Theorem of Calculus, and performing algebraic manipulations. Each step required a thorough understanding of calculus principles and algebraic techniques. The final result is a testament to the power of calculus in solving problems involving areas, volumes, and other cumulative quantities. This answer not only provides a numerical value but also encapsulates the entire mathematical journey undertaken to solve the integral. It demonstrates the interconnectedness of various mathematical concepts and the importance of precision in calculations. The ability to arrive at this final answer underscores the mastery of integral calculus and its applications. This is not just a number; it’s a solution that represents a specific area, and its accuracy reflects the rigor and correctness of the entire problem-solving process. The final answer serves as a concrete example of the utility of definite integrals in quantifying real-world phenomena and solving practical problems in various fields such as physics, engineering, and economics.

In conclusion, we have successfully evaluated the definite integral ${ \int_4^6 \frac{4x^2 + 2}{\sqrt{x}} dx }$ and found the result to be ${ \frac{308}{5}\sqrt{6} - \frac{296}{5} }$. This process involved simplifying the integrand, finding the antiderivative using the power rule, and applying the Fundamental Theorem of Calculus. The steps taken highlight the importance of algebraic manipulation and a solid understanding of calculus principles. Evaluating definite integrals is a fundamental skill in calculus with wide-ranging applications in various fields. This exercise demonstrates how complex integrals can be solved systematically by breaking them down into simpler steps. The ability to accurately compute definite integrals is essential for solving problems involving areas, volumes, and other cumulative quantities. The result we obtained provides a precise numerical value for the signed area under the curve of the given function over the specified interval. This process underscores the power of calculus as a tool for quantitative analysis and problem-solving. The steps we followed serve as a model for tackling similar integration problems, emphasizing the importance of clear, methodical approaches. This example also illustrates the interconnectedness of different mathematical concepts, such as algebra and calculus, and the necessity of mastering each to succeed in mathematical problem-solving. Ultimately, the successful evaluation of this definite integral reinforces the significance of calculus as a fundamental discipline in mathematics and its applications in science, engineering, and beyond. The insights gained from this exercise extend beyond the specific problem, providing a framework for approaching a wide range of mathematical challenges. This journey through the evaluation of a definite integral highlights the beauty and utility of calculus in unraveling complex mathematical problems and providing meaningful solutions.