Question

Find the area under \(y=1 / x\) and above the \(x\) -axis from \(x=1\) to \(x=4\) $$ \int \frac{d x}{x \ln (x) \ln (\ln x)}=\ln (\ln (\ln x))+C $$

Short answer

The area is \(2 \ln 2\).

Step-by-step solution

Identify the Function and the Interval

We need to find the area under the curve defined by the function \(y = \frac{1}{x}\) from \(x = 1\) to \(x = 4\). This can be represented as the integral \( \int_{1}^{4} \frac{1}{x} \, dx \).

Set Up the Integral

Set up the integral that represents the area under the curve from \(x = 1\) to \(x = 4\): \( \int_{1}^{4} \frac{1}{x} \, dx \).

Determine the Antiderivative

The antiderivative of \(\frac{1}{x}\) is \(\ln |x| + C\), where \(C\) is the constant of integration.

Evaluate the Definite Integral

Using the Fundamental Theorem of Calculus, evaluate the definite integral by calculating: \[ \left[ \ln |x| \right]_{1}^{4} = \ln 4 - \ln 1. \] Since \(\ln 1 = 0\), this simplifies to \(\ln 4\).

Final Calculation

The value of \(\ln 4\) can be simplified further to \(2 \ln 2\), since \(4 = 2^2\). Thus, the area under the curve is \(2 \ln 2\).

Key concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus, used to calculate the area under curves over a specified interval on the x-axis. In our original exercise, we are tasked with finding the area under the curve of the function \( y = \frac{1}{x} \) from \( x = 1 \) to \( x = 4 \). This is expressed as the definite integral \( \int_{1}^{4} \frac{1}{x} \, dx \).
Combining integration and limits, definite integrals yield a numerical result that represents the area. For the function \( \frac{1}{x} \), it involves integrating over the interval [1, 4]. Once integrated, the upper and lower bounds are substituted into the antiderivative, and the difference is calculated to get the area. It's important to note that the integral gives the net area, considering any portions of the curve that lie below the x-axis as negative, but in this case, \( \frac{1}{x} \) always stays positive.
This approach simplifies the process of finding areas under complex curves using calculus, and is a powerful tool for solving real-world problems involving physical areas, probability, and accumulations.

Antiderivatives

Antiderivatives, or indefinite integrals, are functions that reverse differentiation. For any given function \( f(x) \), the antiderivative is another function \( F(x) \) such that \( F'(x) = f(x) \). In our example, the function \( \frac{1}{x} \) has the antiderivative \( \ln |x| + C \), where \( C \) represents a constant.
Antiderivatives are crucial in integral calculus because they are used to find definite integrals. In the step-by-step solution provided, the antiderivative \( \ln |x| \) is determined for the function \( \frac{1}{x} \), which then allows us to evaluate the definite integral over the interval [1, 4].

• To find the antiderivative, look for a function whose derivative matches the integrand.
• The constant \( C \) becomes irrelevant when calculating definite integrals, due to the subtraction of \( F(b) - F(a) \) cancelling it out.

Understanding antiderivatives aids in seamlessly transitioning from differentiating functions to integrating them, thereby enabling the calculation of areas and solving other related problems.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone of calculus, intertwining differentiation and integration, two main operations of calculus. It comprises two parts, with the first part establishing that the integral of a function over an interval can be calculated using its antiderivative
In our context, this theorem simplifies the process of evaluating the definite integral \( \int_{1}^{4} \frac{1}{x} \, dx \). By taking the antiderivative \( F(x) = \ln |x| + C \) of the function \( \frac{1}{x} \), the Fundamental Theorem of Calculus allows us to substitute the limits to find:
\[ F(4) - F(1) = (\ln 4 - \ln 1) = \ln 4. \]
Since \( \ln 1 = 0 \), the result simplifies to \( \ln 4 \).

• The theorem bridges the gap between finding antiderivatives and evaluating integrals.
• It enables a quick calculation of areas and accumulated quantities, using limits and functions defined on intervals.

This theorem is fundamental for students to grasp, as it provides the mechanism to relate integrals directly to antiderivatives, transforming complex integration problems into manageable calculations.