Solving Y(2) Given Integral And Derivative Expression In Calculus

This article provides a comprehensive walkthrough of solving a definite integral and derivative problem. Specifically, we address the equation $\bold{y(x)=\int A dx + \frac{d}{dx}(A)}$, where $\bold{A = \ln\left(\sqrt{\frac{x-1}{2}}\right)}$. The goal is to find the value of $\bold{y(2)}$. This involves simplifying the expression for $\bold{A}$, performing the integration and differentiation, and then evaluating the result at $\bold{x = 2}$. This exploration is crucial for students and enthusiasts alike to deepen their understanding of calculus, logarithmic functions, and problem-solving strategies in mathematical analysis. By dissecting each step, we aim to provide a clear, concise, and educational resource that enhances both the theoretical knowledge and practical skills of our readers.

Breaking Down the Problem

To effectively solve for $\bold{y(2)}$, we will break down the problem into several manageable steps. First, we will simplify the expression for $\bold{A}$ using logarithmic properties. Then, we will compute the indefinite integral of $\bold{A}$ with respect to $\bold{x}$. Following this, we will find the derivative of $\bold{A}$ with respect to $\bold{x}$. Finally, we will substitute these results back into the equation for $\bold{y(x)}$ and evaluate it at $\bold{x = 2}$. This methodical approach ensures that we can tackle each component of the problem with precision and clarity. Understanding these steps is essential not only for this specific problem but also for developing a robust approach to solving calculus problems in general. This detailed breakdown will serve as a valuable guide for anyone looking to master the intricacies of calculus and improve their problem-solving abilities.

Step 1 Simplify A

The function $\bold{A}$ is given by $\bold{A = \ln\left(\sqrt{\frac{x-1}{2}}\right)}$. To simplify this expression, we can use properties of logarithms and exponents. The square root can be written as a power of $\frac{1}{2}$, so we have $\bold{A = \ln\left(\left(\frac{x-1}{2}\right)^{\frac{1}{2}}\right)}$. Using the power rule of logarithms, which states that $\bold{\ln(a^b) = b \ln(a)}$, we can rewrite $\bold{A}$ as $\bold{A = \frac{1}{2} \ln\left(\frac{x-1}{2}\right)}$. Now, using the quotient rule of logarithms, which states that $\bold{\ln(\frac{a}{b}) = \ln(a) - \ln(b)}$, we can further simplify $\bold{A}$ to $\bold{A = \frac{1}{2} [\ln(x-1) - \ln(2)]}$. This simplified form of $\bold{A}$ will make it easier to compute both the integral and the derivative. This initial simplification is a crucial step, as it reduces the complexity of the expression and sets the stage for more straightforward calculations in the subsequent steps. By mastering these simplification techniques, one can significantly enhance their ability to solve complex mathematical problems.

Step 2 Compute the Integral of A

Next, we need to compute the indefinite integral of $\bold{A}$ with respect to $\bold{x}$, which is $\bold{\int A dx = \int \frac{1}{2} [\ln(x-1) - \ln(2)] dx}$. We can split this integral into two parts: $\bold{\frac{1}{2} \int \ln(x-1) dx - \frac{1}{2} \int \ln(2) dx}$. For the first integral, we use integration by parts. Let $\bold{u = \ln(x-1)}$ and $\bold{dv = dx}$. Then, $\bold{du = \frac{1}{x-1} dx}$ and $\bold{v = x}$. Using the integration by parts formula, $\bold{\int u dv = uv - \int v du}$, we get $\bold{\int \ln(x-1) dx = x \ln(x-1) - \int \frac{x}{x-1} dx}$. To solve $\bold{\int \frac{x}{x-1} dx}$, we can rewrite the integrand as $\bold{\frac{x}{x-1} = \frac{x-1+1}{x-1} = 1 + \frac{1}{x-1}}$. Thus, $\bold{\int \frac{x}{x-1} dx = \int (1 + \frac{1}{x-1}) dx = x + \ln|x-1| + C_1}$. Substituting this back, we have $\bold{\int \ln(x-1) dx = x \ln(x-1) - (x + \ln|x-1|) + C_1}$. For the second integral, $\bold{\int \ln(2) dx = \ln(2) \int dx = x \ln(2) + C_2}$. Combining these results, we get $\bold{\int A dx = \frac{1}{2} [x \ln(x-1) - x - \ln|x-1| - x \ln(2)] + C}$, where $\bold{C}$ is the constant of integration. This step-by-step approach highlights the intricacies of integration by parts and demonstrates the importance of breaking down complex integrals into simpler components. The result provides a crucial part of our solution for $\bold{y(x)}$, showcasing the power of integral calculus in problem-solving.

Step 3 Compute the Derivative of A

Now, let's compute the derivative of $\bold{A}$ with respect to $\bold{x}$. Recall that $\bold{A = \frac{1}{2} [\ln(x-1) - \ln(2)]}$. The derivative of $\bold{A}$ with respect to $\bold{x}$, denoted as $\bold{\frac{dA}{dx}}$, can be found by differentiating each term separately. The derivative of $\bold{\ln(x-1)}$ with respect to $\bold{x}$ is $\bold{\frac{1}{x-1}}$, using the chain rule. The derivative of $\bold{\ln(2)}$ with respect to $\bold{x}$ is $\bold{0}$, since $\bold{\ln(2)}$ is a constant. Therefore, we have $\bold{\frac{dA}{dx} = \frac{1}{2} [\frac{1}{x-1} - 0] = \frac{1}{2(x-1)}}$. This derivative is a crucial component of our original equation for $\bold{y(x)}$. Understanding how to compute derivatives of logarithmic functions is essential in calculus, and this step provides a clear example of applying the chain rule and constant rule in differentiation. This straightforward computation simplifies the process of finding $\bold{y(x)}$ and demonstrates the elegance of differential calculus.

Step 4 Substitute and Evaluate y(2)

Having computed both the integral of $\bold{A}$ and the derivative of $\bold{A}$, we can now substitute these results into the original equation for $\bold{y(x)}$, which is $\bold{y(x) = \int A dx + \frac{dA}{dx}}$. We found that $\bold{\int A dx = \frac{1}{2} [x \ln(x-1) - x - \ln|x-1| - x \ln(2)] + C}$ and $\bold{\frac{dA}{dx} = \frac{1}{2(x-1)}}$. Substituting these into the equation, we get $\bold{y(x) = \frac{1}{2} [x \ln(x-1) - x - \ln|x-1| - x \ln(2)] + \frac{1}{2(x-1)} + C}$. Now, we need to evaluate $\bold{y(2)}$, so we substitute $\bold{x = 2}$ into the expression: $\bold{y(2) = \frac{1}{2} [2 \ln(2-1) - 2 - \ln|2-1| - 2 \ln(2)] + \frac{1}{2(2-1)} + C}$. Simplifying, we have $\bold{y(2) = \frac{1}{2} [2 \ln(1) - 2 - \ln(1) - 2 \ln(2)] + \frac{1}{2} + C}$. Since $\bold{\ln(1) = 0}$, this further simplifies to $\bold{y(2) = \frac{1}{2} [-2 - 2 \ln(2)] + \frac{1}{2} + C}$. Finally, $\bold{y(2) = -1 - \ln(2) + \frac{1}{2} + C = -\frac{1}{2} - \ln(2) + C}$. Thus, the value of $\bold{y(2)}$ is $\bold{-\frac{1}{2} - \ln(2) + C}$, where $\bold{C}$ is the constant of integration. This final substitution and evaluation demonstrate the culmination of all the previous steps, providing a complete solution to the problem. The precise calculation of $\bold{y(2)}$ showcases the practical application of calculus principles and the importance of careful algebraic manipulation.

Final Answer

In conclusion, by simplifying the given expression, performing the integral and derivative calculations, and substituting $\bold{x = 2}$, we have found that the value of $\bold{y(2)}$ is $\bold{-\frac{1}{2} - \ln(2) + C}$, where $\bold{C}$ is the constant of integration. This comprehensive solution demonstrates a step-by-step approach to solving calculus problems involving integrals and derivatives of logarithmic functions. This detailed walkthrough provides valuable insights and reinforces key concepts, making it an excellent resource for students and anyone interested in mastering calculus.