Evaluating The Definite Integral Of 2624/(x^2+x+1) From 1 To 10

Introduction: The Allure of Definite Integrals

In the realm of calculus, definite integrals stand as powerful tools, allowing us to calculate areas under curves, determine accumulated quantities, and solve a myriad of problems across various scientific disciplines. The integral presented, ${\int_1^{10} \frac{2624}{x^2+x+1} dx}$, is a fascinating example that invites us to explore techniques of integration and delve into the nuances of calculus. This article aims to provide a comprehensive guide to evaluating this integral, breaking down each step and highlighting the underlying mathematical principles.

The integral ${\int_1^{10} \frac{2624}{x^2+x+1} dx}$ presents an interesting challenge due to the quadratic expression in the denominator. Direct integration isn't immediately apparent, necessitating a strategic approach. We will utilize the technique of completing the square to transform the quadratic into a form that allows us to apply a standard integral formula. This process not only simplifies the integration but also showcases the elegance of algebraic manipulation in solving calculus problems. The limits of integration, from 1 to 10, define the interval over which we are calculating the area under the curve of the function ${f(x) = \frac{2624}{x^2+x+1}}$. Understanding these limits is crucial for obtaining the correct numerical value of the definite integral. Definite integrals are foundational in various fields, including physics (calculating displacement from velocity), engineering (determining the volume of solids), and economics (modeling economic growth). Mastering the techniques to solve definite integrals is therefore essential for anyone pursuing studies or careers in these areas.

Transforming the Integrand: Completing the Square

To effectively tackle the integral ${\int_1^{10} \frac{2624}{x^2+x+1} dx}$, the first crucial step involves manipulating the integrand, specifically the denominator ${x^2 + x + 1}$. The technique of completing the square is the key here. This algebraic method allows us to rewrite the quadratic expression in a form that is more amenable to integration.

The initial expression is ${x^2 + x + 1}$. To complete the square, we focus on the quadratic and linear terms, ${x^2 + x}$. We take half of the coefficient of the linear term (which is 1), square it ((1/2)^2 = 1/4), and add and subtract it within the expression. This gives us: ${x^2 + x + \frac{1}{4} - \frac{1}{4} + 1}$. The first three terms now form a perfect square: ${(x + \frac{1}{2})^2}$. Combining the constants, we have: ${(x + \frac{1}{2})^2 + \frac{3}{4}}$. Therefore, the denominator ${x^2 + x + 1}$ can be rewritten as ${(x + \frac{1}{2})^2 + \frac{3}{4}}$. This transformation is significant because it allows us to relate the integral to a standard form involving the inverse tangent function. Completing the square is a versatile technique applicable to various quadratic expressions and is fundamental in simplifying integrals and solving quadratic equations. It’s a testament to the power of algebraic manipulation in making complex problems more tractable. Recognizing when to apply this technique is a crucial skill in calculus. By rewriting the denominator in the completed square form, we've paved the way for a simpler integration process. The transformed integral now has a structure that aligns with the form of the derivative of the arctangent function, which is a crucial observation for the next stage of solving the integral.

Applying the Arctangent Integral: A Crucial Step

Having successfully completed the square in the denominator, the integral ${\int_1^{10} \frac{2624}{x^2+x+1} dx}$ now takes a more manageable form. Our expression is ${\int_1^{10} \frac{2624}{(x + \frac{1}{2})^2 + \frac{3}{4}} dx}$. The key to solving this integral lies in recognizing its resemblance to the standard integral form that yields the arctangent function (also known as the inverse tangent function). The general form we're aiming for is ${\int \frac{1}{u^2 + a^2} du = \frac{1}{a} \arctan(\frac{u}{a}) + C}$, where C is the constant of integration.

To map our integral to this form, we perform a u-substitution. Let ${u = x + \frac{1}{2}}$, then ${du = dx}$. Also, let ${a^2 = \frac{3}{4}}$, which implies ${a = \frac{\sqrt{3}}{2}}$. Substituting these into our integral, we get: ${2624 \int_1^{10} \frac{1}{u^2 + (\frac{\sqrt{3}}{2})^2} du}$. The constant 2624 is factored out for clarity. Now, we can directly apply the arctangent integral formula. The integral becomes: ${2624 \cdot \frac{1}{\frac{\sqrt{3}}{2}} \arctan(\frac{u}{\frac{\sqrt{3}}{2}})}$ evaluated from the transformed limits of integration. Remember that when we perform a u-substitution with definite integrals, we also need to change the limits of integration. When ${x = 1}$, ${u = 1 + \frac{1}{2} = \frac{3}{2}}$. When ${x = 10}$, ${u = 10 + \frac{1}{2} = \frac{21}{2}}$. Therefore, the integral is now evaluated from ${\frac{3}{2}}$ to ${\frac{21}{2}}$ in terms of u. This transformation is a powerful technique in calculus, allowing us to simplify complex integrals by mapping them to known forms. The arctangent function arises frequently in various contexts, including geometry, trigonometry, and complex analysis. Its integral is a crucial tool in solving a wide range of problems. Recognizing the arctangent form and applying the appropriate substitution is a significant step in mastering integration techniques.

Evaluating the Definite Integral: From Limits to Solution

With the integral transformed into the arctangent form, the next step involves evaluating the definite integral using the transformed limits of integration. We have the expression: ${2624 \cdot \frac{2}{\sqrt{3}} \arctan(\frac{2u}{\sqrt{3}})}$ evaluated from ${u = \frac{3}{2}}$ to ${u = \frac{21}{2}}$.

Substituting the limits of integration, we get: ${2624 \cdot \frac{2}{\sqrt{3}} [\arctan(\frac{2(\frac{21}{2})}{\sqrt{3}}) - \arctan(\frac{2(\frac{3}{2})}{\sqrt{3}})]}$. Simplifying the expression inside the arctangent functions: ${2624 \cdot \frac{2}{\sqrt{3}} [\arctan(\frac{21}{\sqrt{3}}) - \arctan(\frac{3}{\sqrt{3}})]}$. Further simplifying: ${2624 \cdot \frac{2}{\sqrt{3}} [\arctan(7\sqrt{3}) - \arctan(\sqrt{3})]}$. Now, we need to evaluate the arctangent functions. ${\arctan(\sqrt{3})}$ is a standard value, which equals ${\frac{\pi}{3}}$ (or 60 degrees). ${\arctan(7\sqrt{3})}$ doesn't have a simple closed-form expression and requires a calculator or computational tool to evaluate. Approximating ${\arctan(7\sqrt{3})}$ gives us approximately 1.4601 radians. Substituting these values, we get: ${2624 \cdot \frac{2}{\sqrt{3}} [1.4601 - \frac{\pi}{3}]}$. Approximating ${\frac{\pi}{3}}$ as 1.0472, the expression becomes: ${2624 \cdot \frac{2}{\sqrt{3}} [1.4601 - 1.0472]}$. This simplifies to: ${2624 \cdot \frac{2}{\sqrt{3}} [0.4129]}$. Calculating the final numerical value: ${2624 \cdot \frac{2}{\sqrt{3}} \cdot 0.4129 \approx 1252.17}$. Therefore, the approximate value of the definite integral ${\int_1^{10} \frac{2624}{x^2+x+1} dx}$ is 1252.17. This numerical evaluation underscores the importance of using accurate approximations or computational tools when dealing with trigonometric functions that don't have simple closed-form values. The process of substituting the limits and simplifying the expression highlights the fundamental theorem of calculus, which connects differentiation and integration, allowing us to evaluate definite integrals by finding the antiderivative and evaluating it at the limits of integration.

Conclusion: The Beauty of Integral Calculus

In conclusion, the evaluation of the definite integral ${\int_1^{10} \frac{2624}{x^2+x+1} dx}$ demonstrates the power and elegance of integral calculus. By employing techniques such as completing the square and u-substitution, we successfully transformed the integral into a recognizable form involving the arctangent function. The final numerical result, approximately 1252.17, represents the area under the curve of the function ${f(x) = \frac{2624}{x^2+x+1}}$ between the limits x = 1 and x = 10.

This journey through the integral highlights several key concepts in calculus. Completing the square is a fundamental algebraic technique that simplifies quadratic expressions, making them more amenable to integration. The u-substitution method allows us to change the variable of integration, mapping complex integrals to simpler forms. Recognizing standard integral forms, such as the arctangent integral, is crucial for efficient problem-solving. The fundamental theorem of calculus provides the theoretical foundation for evaluating definite integrals, connecting antiderivatives to the area under a curve. Beyond the specific example, the process of evaluating definite integrals is a cornerstone of many scientific and engineering applications. From calculating areas and volumes to determining probabilities and modeling physical phenomena, integral calculus provides the tools necessary to solve a vast array of problems. Mastering these techniques not only enhances mathematical proficiency but also unlocks a deeper understanding of the world around us. The integral we've explored serves as a microcosm of the broader field of calculus, showcasing the interplay between algebraic manipulation, trigonometric functions, and the fundamental principles of integration. Its solution underscores the beauty and utility of calculus in addressing both theoretical and practical challenges.