Problem 9 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Let \(f(x)=\int_{1}^{x} \frac{1}{t} d t, x>0\). (a) Find \(f(1), f(5), f(10)\), and \(f(1 / 2)\). (b) What is an alternative formula for \(f(x)\) ? (c) Often mathematicians de ne the natural logarithm by \(\ln x=\int_{1}^{x} \frac{1}{t} d t\) for \(x>0\). Suppose this was the definition you had been given. Use the Fundamental Theorem of Calculus to show that \(\ln x\) is increasing and concave down for \(x>0\).

Step-by-Step Solution

Answer

(a) \(f(1) = 0\), \(f(5) = \ln 5\), \(f(10) = \ln 10\), \(f(1/2) = -\ln 2\). (b) Alternative formula is \(\ln x\). (c) \(\ln x\) is increasing and concave down for \(x>0\).

Step 1: Evaluate \(f(x)\) at given points

(a)Substitute for 'x' as 1, 5, 10 and 1/2 into the integral and calculate the integrals: \(f(1) = \int_{1}^{1} \frac{1}{t} dt = 0\).\(f(5) = \int_{1}^{5} \frac{1}{t} dt = \ln |5| = \ln 5\).\(f(10) = \int_{1}^{10} \frac{1}{t} dt = \ln |10| = \ln 10\).\(f(1/2) = \int_{1}^{1/2} \frac{1}{t} dt = -\int_{1/2}^{1} \frac{1}{t} dt = -\ln |2| = -\ln 2\).

Step 2: Find an alternative formula for \(f(x)\)

(b)It can be observed that \(f(x) = \int_{1}^{x} \frac{1}{t} dt = \ln|x|\) for \(x>0\). You can make this observation base on the evaluation of the definite integrals we computed in Step 1. So, the alternative formula for \(f(x)\) is \(\ln x\).

Step 3: Use Fundamental Theorem to analyze properties of \(\ln x\)

(c)The Fundamental theorem of Calculus states that if a function \(f(x)\) is defined by an integral, the derivative of \(f(x)\) is its integrand. In this scenario \(f'(x) = \frac{1}{x}\). This shows that \(\ln x\) is increasing because \(f'(x) > 0 \) for all \(x > 0\). Moreover, the second derivative \(f''(x) = -\frac{1}{x^2}\) is less than zero for all \(x > 0\), indicating that function is concave down.

Key Concepts

Natural Logarithm

The natural logarithm, often denoted as \( \ln(x) \), is a special logarithm with base \( e \), where \( e \) is an irrational number approximately equal to 2.71828. It has significant applications in mathematics due to its unique properties and its fundamental relationship with exponential functions. In this exercise, the natural logarithm is introduced in the context of a definite integral, \( \int_{1}^{x} \frac{1}{t} dt \), highlighting its role in calculus.

Mathematicians sometimes define the natural logarithm through the integral \( \ln x = \int_{1}^{x} \frac{1}{t} dt \) for \( x > 0 \). This definition provides a deeper insight into the behavior of \( \ln(x) \). Not only does it establish \( \ln(x) \) as a continuous and smooth function, but it also connects directly through calculus, illustrating growth as being accumulation of area under the curve \( \frac{1}{t} \).

• The natural logarithm reflects growth patterns, commonly used in continuous compounding and natural processes.
• It serves as the inverse function of the exponential function \( e^x \).
• Understanding \( \ln(x) \) through integrals showcases how log functions can be accumulating changes.

This integral-based definition of \( \ln(x) \) allows us to apply calculus operations, such as differentiation, to derive crucial characteristics of the logarithmic function.

Definite Integral

A definite integral is a concept in calculus that represents the accumulation of quantities, such as areas under curves. It is denoted as \( \int_{a}^{b} f(t) \, dt \), where \( f(t) \) is the function being integrated from \( a \) to \( b \). Definite integrals have a wide range of applications, including calculating areas, volumes, and in this context, the natural logarithm function.

In the given problem, \( f(x) = \int_{1}^{x} \frac{1}{t} dt \) establishes a specific use of a definite integral to define a function. Here, the definite integral captures the aggregate of small increments \( \frac{1}{t} \) over the interval from 1 to \( x \). This allows us to easily compute the values of \( f(x) \) at different points:

• \( f(1) = \int_{1}^{1} \frac{1}{t} dt = 0 \), as there is no interval to integrate over.
• \( f(5) = \ln 5 \), since \( \int_{1}^{5} \frac{1}{t} dt = \ln 5 \).
• \( f(10) = \ln 10 \), using a similar calculation technique.
• \( f(1/2) = -\ln 2 \), since the limits are reversed, effectively giving us a negative integral.

The provided problem and solutions further illuminate how the properties of definite integrals are applied to real-world scenarios—demonstrating transition from theoretical calculus to practical computation.

Derivatives

Derivatives represent the rate of change of a function with respect to its variables and are central to understanding dynamic systems. In calculus, the derivative of a function \( f(x) \), denoted as \( f'(x) \), provides insights into the function's behavior, such as increasing or decreasing trends and the nature of its concavity.

In the context of this exercise, the Fundamental Theorem of Calculus bridges differentiation and integration. It states that if a function \( f(x) \) is defined by an integral, then its derivative corresponds to the integrand. For this particular problem, the derivative of the function \( f(x) = \int_{1}^{x} \frac{1}{t} dt \) is found to be:

• \( f'(x) = \frac{1}{x} \), indicating that \( \ln(x) \) is increasing since \( \frac{1}{x} > 0 \) for \( x > 0 \).
• The second derivative \( f''(x) = -\frac{1}{x^2} \) shows that \( \ln(x) \) is concave down, as \( -\frac{1}{x^2} < 0 \).

• This increasing pattern of \( \ln(x) \) reflects its accumulation of areas and its reliability as a growth indicator.
• The concavity, on the other hand, indicates the diminishing rate of growth, which is a key insight for areas such as economics and physics.

Thus, derivatives do not only reveal instantaneous rates of change but also allow a deeper understanding of the graphical and functional forms of mathematical models.