Problem 74 - Applied Calculus

Question

Show that for any number \(a>0\) $$ \int_{1}^{a} \frac{1}{x} d x=\ln a $$

Step-by-Step Solution

Answer

For any \( a > 0 \), \( \int_{1}^{a} \frac{1}{x} \, dx = \ln a \).

Step 1: Understand the Problem Statement

We are asked to show that the definite integral \( \int_{1}^{a} \frac{1}{x} \, dx \) equals \( \ln a \) for any positive number \( a \). This involves using integral calculus and understanding the properties of the natural logarithm function.

Step 2: Identify the Antiderivative

The function we are integrating, \( \frac{1}{x} \), has a well-known antiderivative. The antiderivative of \( \frac{1}{x} \) is \( \ln |x| + C \), where \( C \) is a constant of integration. Since we are dealing with positive \( a \), we have \( \ln x \) as the antiderivative.

Step 3: Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if \( F \) is an antiderivative of \( f \) on an interval \([a, b]\), then \[ \int_{a}^{b} f(x) \, dx = F(b) - F(a). \] In our case, \( f(x) = \frac{1}{x} \), so \( F(x) = \ln x \).

Step 4: Calculate the Definite Integral

Using the antiderivative \( \ln x \), evaluate the definite integral from 1 to \( a \): \[ \int_{1}^{a} \frac{1}{x} \, dx = \ln a - \ln 1. \]Since \( \ln 1 = 0 \), this simplifies to:\[ \ln a - 0 = \ln a. \]

Step 5: Conclude the Solution

Therefore, the definite integral \( \int_{1}^{a} \frac{1}{x} \, dx \) yields the result \( \ln a \). This confirms the problem statement: for any \( a > 0 \), \( \int_{1}^{a} \frac{1}{x} \, dx = \ln a \).

Key Concepts

Definite Integral

A definite integral is a fundamental concept in integral calculus. It computes the accumulation of quantities, like areas under a curve, over an interval. Specifically, when we say \( \int_{a}^{b} f(x) \, dx\), we are referring to the definite integral of function \(f(x)\) as \(x\) ranges from \(a\) to \(b\).
This gives us a real number that represents the net area under the curve \(f(x)\) between \(x = a\) and \(x = b\).

• The limits of integration \(a\) and \(b\) show where to start and end on the \(x\)-axis.
• The result may be positive or negative, depending on where the curve is above or below the axis.

In the given exercise, the expression \(\int_{1}^{a} \frac{1}{x} \, dx\) demonstrates this concept, seeking the accumulated value of \(\frac{1}{x}\) from 1 to \(a\).
This is crucial in connecting the geometry of the region with the functional behavior in calculus.

Antiderivative

In calculus, the antiderivative is the reverse process of differentiation. It helps us find a function given its derivative. Formally, the antiderivative of a function \(f(x)\) is a function \(F(x)\) where the derivative \(F'(x) = f(x)\).
This is often used for indefinite integrals, which include the constant 'C'. Here, however, we deal specifically with the definite integral, focusing on exact accumulation.
For the function \(\frac{1}{x}\), its antiderivative is widely known to be \(\ln |x| + C\).

• Since \(x\) is positive in our scenario, this simplifies to \(\ln x\).
• 'C' is eliminated during definite integration due to the limits applying specific values.

Recognizing antiderivatives like \(\ln x\) is crucial, as it allows one to find solutions efficiently and understand the underlying behavior of functions.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a bridge between differential calculus and integral calculus. It provides the connection between antiderivatives and definite integrals, making it one of the key results in calculus.

• It states: If \(F\) is an antiderivative of \(f\) on an interval \([a, b]\), then \(\int_{a}^{b} f(x) \, dx = F(b) - F(a)\).

This theorem justifies why we can use antiderivatives to calculate definite integrals.
For instance, in our given problem, by identifying \(\ln x\) as the antiderivative of \(\frac{1}{x}\), the Fundamental Theorem allows us to conclude \[\int_{1}^{a} \frac{1}{x} \, dx = \ln a - \ln 1 = \ln a,\]showcasing the seamless transition from antiderivatives to precise numerical results.
This principle shows that understanding both parts of calculus is essential for solving a wide range of problems efficiently.