Problem 9 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the area of the region under the graph of the function \(f\) on the interval \([a, b]\). $f(x)=\frac{1}{x} ;[1,2]$$

Step-by-Step Solution

Answer

The area of the region under the graph of the function \(f(x) = \frac{1}{x}\) on the interval [1, 2] is \(\ln(2)\).

Step 1: (Step 1: Write down the integral)

Integrate f(x) over the interval [1, 2]. \(\int_{1}^{2} \frac{1}{x} dx\)

Step 2: (Step 2: Find the antiderivative of f(x))

To compute the integral, first find the antiderivative of f(x) = 1/x. The antiderivative of 1/x is ln|x|. So, \(\int_{1}^{2} \frac{1}{x} dx = \left[ \ln|x| \right]_{1}^{2}\)

Step 3: (Step 3: Evaluate the antiderivative at the limits of the interval, and then subtract)

Now, evaluate the antiderivative at the upper limit of the interval (2), and then subtract the value of the antiderivative evaluated at the lower limit (1). \(\left[ \ln|x| \right]_{1}^{2} = \ln(2) - \ln(1)\)

Step 4: (Step 4: Simplify the result)

Simplify the result using the fact that ln(1) = 0. \(\ln(2) - \ln(1) = \ln(2) - 0\)

Step 5: (Step 5: Determine the area under the curve)

The area under the curve is the value of the integral on the interval [1, 2]. Area under the curve = \(\ln(2)\) The area of the region under the graph of the function f(x) = 1/x on the interval [1, 2] is \(\ln(2)\).

Key Concepts

Definite Integral

To find the area under a curve in calculus, we often use a concept called the definite integral. The definite integral of a function between two points gives us the exact area under the curve between those two points along the x-axis. It doesn't just calculate the area, but also accounts for parts below the x-axis as negative area. This makes it a powerful tool for a variety of applications.

When dealing with the definite integral, you need to pay attention to:

• The function you are integrating, known as the integrand.
• The limits of integration, which are the values that define the interval over which you are calculating the area.
• The method of integration—this can involve finding an antiderivative (which we'll discuss next).

In the original exercise, we calculated the definite integral of the function \(f(x) = \frac{1}{x}\) over the interval \([1, 2]\). The solution provides a step-by-step method from setting up the integral to evaluating it.

It's fascinating how this integral results in a value of \(\ln(2)\), showing that integration isn't just about curves and shapes but also interconnected with functions like logarithms.

Antiderivative

An antiderivative is a function that reverses differentiation, effectively moving backwards from a derivative to an original function. When we seek the antiderivative, we're looking for the original function whose derivative gives us the function we started with.

Let's illustrate this with the function \(f(x) = \frac{1}{x}\). The antiderivative of \(\frac{1}{x}\) is the natural logarithm function, specifically \(\ln|x|\).

• The symbol \(\ln|x|\) represents the natural logarithm function.
• It allows us to calculate the integral of functions like \(\frac{1}{x}\).

In the step-by-step solution, finding this antiderivative was a crucial step in solving the definite integral. Once we have the antiderivative, evaluating it over the specific interval \"gives us the area under the curve from point \(a\) to point \(b\).

Natural Logarithm

The natural logarithm, denoted as \(\ln(x)\), is a special logarithm because it has a base of \(e\) where \(e\) is approximately 2.718. It's one of the most common and useful logarithms in mathematics, particularly because of its relationship to exponential functions and calculus.

Why is it called "natural"? The natural logarithm results in simple derivatives and integrals, which means it's very appealing for solving calculus problems.

• The derivative of \(\ln(x)\) is \(1/x\).
• The integral of \(1/x\) is \(\ln|x|\), which is precisely what was used in the original exercise.

In practical terms, the natural logarithm is used to unravel the exponential growth and decay in various scenarios. In the exercise, we used the natural logarithm to find the area under the curve by integrating \(\frac{1}{x}\) from 1 to 2, concluding with the result \(\ln(2)\). This showcases how the natural logarithm acts as a bridge between algebra and calculus, simplifying complex mathematical relationships.