Not recognizing the vertical asymptote

Oct 13, 2025 by magicalceylontours.com 120 views

7. Calculating the Area of an Unbounded Region

In this problem, we are tasked with finding the area of an unbounded region between the graph of a given function and the x-axis. The function in question is ${$ y=\frac{1}{x(\ln x)^2} $}$, and we are interested in the region where ${$ x \geq 3 $}$. To approach this problem, we will first need to understand the behavior of the function and how it relates to the x-axis.

The given function is ${$ y=\frac{1}{x(\ln x)^2} $}$. This function has a few key features that are worth noting. First, the function is defined for all ${$ x > 0 $}$, since the natural logarithm is only defined for positive numbers. Additionally, the function has a vertical asymptote at ${$ x = 1 $}$, since the denominator of the function becomes zero at this point.

To calculate the area of the unbounded region, we will use the definite integral. The area of the region can be represented as:

${$ A = \int_{3}^{\infty} \frac{1}{x(\ln x)^2} dx $}$

This integral represents the area under the curve of the function from ${$ x = 3 $}$ to infinity.

To evaluate this integral, we can use the method of substitution. Let's set ${$ u = \ln x $}$. Then, ${$ du = \frac{1}{x} dx $}$. Substituting these expressions into the integral, we get:

${$ A = \int_{\ln 3}^{\infty} \frac{1}{u^2} du $}$

This integral is now in a form that we can easily evaluate. The antiderivative of ${$ \frac{1}{u^2} $}$ is ${$ -\frac{1}{u} $}$. Applying the fundamental theorem of calculus, we get:

${$ A = \left[ -\frac{1}{u} \right]_{\ln 3}^{\infty} $}$

To simplify this expression, we can evaluate the limit as ${$ u $}$ approaches infinity. As ${$ u $}$ becomes very large, ${$ -\frac{1}{u} $}$ approaches zero. Therefore, the expression simplifies to:

${$ A = 0 - \left( -\frac{1}{\ln 3} \right) $}$

In conclusion, the area of the unbounded region between the graph of ${$ y=\frac{1}{x(\ln x)^2} $}$ and the x-axis for ${$ x \geq 3 $}$ is ${$ -\frac{1}{\ln 3} $}$. This result is obtained by evaluating the definite integral of the function and applying the fundamental theorem of calculus.

The final answer is ${$ -\frac{1}{\ln 3} $}$.
7. Calculating the Area of an Unbounded Region: Q&A

In our previous article, we explored the problem of calculating the area of an unbounded region between the graph of a given function and the x-axis. The function in question was ${$ y=\frac{1}{x(\ln x)^2} $}$, and we were interested in the region where ${$ x \geq 3 $}$. In this article, we will provide a Q&A section to help clarify any questions or doubts that readers may have.

A: The vertical asymptote at ${$ x = 1 $}$ is significant because it represents a point where the function is undefined. In other words, the function approaches infinity as ${$ x $}$ approaches 1 from the right.

A: We used the method of substitution to evaluate the integral because it allowed us to simplify the expression and make it easier to work with. By setting ${$ u = \ln x $}$, we were able to eliminate the ${$ x $}$ term in the denominator and make the integral more manageable.

A: The antiderivative of ${$ \frac{1}{u^2} $}$ is ${$ -\frac{1}{u} $}$. This is a fundamental result in calculus, and it is used to evaluate many types of integrals.

A: We evaluated the limit as ${$ u $}$ approaches infinity because we wanted to determine the behavior of the function as ${$ x $}$ becomes very large. By doing so, we were able to simplify the expression and obtain the final answer.

A: The final answer to the problem is ${$ -\frac{1}{\ln 3} $}$. This result represents the area of the unbounded region between the graph of the function and the x-axis for ${$ x \geq 3 $}$.

Not recognizing the vertical asymptote: One common mistake is not recognizing the vertical asymptote at ${$ x = 1 $}$. This can lead to incorrect conclusions about the behavior of the function.
Not using the method of substitution: Another common mistake is not using the method of substitution to evaluate the integral. This can make the integral more difficult to work with and may lead to incorrect results.
Not evaluating the limit: Finally, it's essential to evaluate the limit as ${$ u $}$ approaches infinity to determine the behavior of the function as ${$ x $}$ becomes very large.

In conclusion, the Q&A section provides additional insight and clarification on the problem of calculating the area of an unbounded region between the graph of a given function and the x-axis. By addressing common mistakes and providing additional explanations, we hope to have helped readers better understand the material.

The final answer is ${$ -\frac{1}{\ln 3} $}$.