Problem 33 - Applied Calculus

Question

Use a definite integral to find the area under each curve between the given \(x\) -values. For Exercises \(19-24\) also make a sketch of the curve showing the region. $$ f(x)=\frac{1}{x} \text { from } x=1 \text { to } x=5 $$

Step-by-Step Solution

Answer

The area is \( \ln(5) \).

Step 1: Understand the Problem

We need to determine the area under the curve defined by the function \( f(x) = \frac{1}{x} \) between \( x = 1 \) and \( x = 5 \). This requires computing a definite integral.

Step 2: Set Up the Integral

The area under the curve for the function \( f(x) = \frac{1}{x} \) from \( x = 1 \) to \( x = 5 \) is found using the integral:\[ \int_{1}^{5} \frac{1}{x} \, dx \]

Step 3: Calculate the Integral

The integral of \( \frac{1}{x} \) is \( \ln|x| \). Therefore, we have:\[ \int \frac{1}{x} \, dx = \ln|x| + C \]Evaluating the definite integral:\[ \left[ \ln|x| \right]_{1}^{5} = \ln(5) - \ln(1) \]

Step 4: Simplify the Result

Since \( \ln(1) = 0 \), the expression simplifies to:\[ \ln(5) - 0 = \ln(5) \]

Step 5: Sketch the Curve

The curve \( y = \frac{1}{x} \) is a hyperbola that approaches the x-axis as \( x \to \infty \) and the y-axis as \( x \to 0 \). For the interval from \( x = 1 \) to \( x = 5 \), draw the curve decreasing smoothly and draw vertical lines at \( x=1 \) and \( x=5 \) to mark the area under the curve.

Key Concepts

Area Under a Curve

The area under a curve in calculus is a fundamental concept that helps us understand how a quantity accumulates over a given interval. Imagine you are looking at the graph of a function, and you want to find out how much space is enclosed between the curve and the x-axis over a particular range of x-values.
This is exactly what we're doing in this exercise with the function \( f(x) = \frac{1}{x} \) from \( x = 1 \) to \( x = 5 \).

To find the area under a curve:

• Identify the function and the interval over which you need to find the area.
• Use integration, which is the process of finding the integral, to determine this area.
• Evaluate the definite integral, which will give the exact "accumulated" value, or area, under the curve.

This method is incredibly powerful because it gives us precise measurements of irregularly shaped areas that cannot be easily calculated using basic geometry.
In this case, the result of the definite integral represents the area in the specified range under the curve, and for our function, that turns out to be \( \ln(5) \).

Integral of 1/x

The integral of \( \frac{1}{x} \) is a standard result in calculus that leads us to a well-known mathematical function: the natural logarithm. When you take the integral of \( \frac{1}{x} \), you're essentially finding the function whose derivative brings you back to \( \frac{1}{x} \).

In formal terms, the integral is given by:

• \[ \int \frac{1}{x} \, dx = \ln|x| + C \] where \( C \) is the constant of integration for indefinite integrals.

This represents an important bridge between algebra and calculus because it introduces logarithmic behavior through integration.
For our definite integral from \( x = 1 \) to \( x = 5 \), we can simplify further:

• Calculate the difference: \( \left[ \ln|x| \right]_{1}^{5} = \ln(5) - \ln(1) \).
• Since \( \ln(1) = 0 \), the area simplifies directly to \( \ln(5) \).

This showcases how the integral provides a succinct and precise way to compute areas under such curves.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is an essential mathematical function that pops up frequently in calculus and beyond. It is the inverse of the exponential function \( e^x \), where \( e \) is Euler's number, approximately equal to 2.718.

Natural logarithms are crucial because:

• They appear naturally in various real-world scenarios, such as compound interest and population growth.
• Their properties allow for simplifying complex expressions, especially those involving growth or decay.
• The derivative and integral of functions involving the natural logarithm are straightforward, making it an invaluable tool in calculus.

In our case, the result of the integral \( \ln(5) \) is a natural logarithm. This result effectively describes the "growth" observed under the curve \( f(x) = \frac{1}{x} \) from \( x = 1 \) to \( x = 5 \).
This logarithmic interpretation is not just abstract; it reflects how such quantities and areas in graphs "grow" proportionally over specified intervals.

Calculus Sketching

Sketching curves in calculus goes hand-in-hand with understanding the behavior of functions over different intervals. It helps visualize how a function behaves and assists in interpreting more abstract concepts like integrals.

For the function \( y = \frac{1}{x} \) from \( x = 1 \) to \( x = 5 \):

• The curve is a hyperbola that shows reciprocal behavior, meaning as \( x \) increases, \( y \) decreases.
• The curve trends closer to the axes but never actually touches them, due to its asymptotic nature.
• By marking vertical lines at \( x=1 \) and \( x=5 \), you can clearly see the area under the curve that we're interested in, essentially forming a trapezoidal region between these points and the x-axis.

Sketching such curves isn't just an artistic endeavor. It serves as an aid in understanding the abstract notions of calculus --- it transforms numerical solutions into a visual narrative that can be often more intuitive.
In this way, calculus sketching acts as a bridge from numbers to insights, assisting us in grasping how integrals encompass areas within graphs.