Problem 11 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find (exactly) the area bounded by \(x=1 / e, y=\ln x\), and \(y=1\).

Step-by-Step Solution

Answer

The exact area bounded by \(x=1 / e, y=\ln x\), and \(y=1\) is \(1 + 1/e\).

Step 1: Identify the Bounded Region

Sketch the graphs of \(y=\ln x\), \(x=1/e\) and \(y=1\). The area that you are seeking to find is the one bounded by these three: a rectangle with one side along the vertical line \(x=1/e\), and a curve along the function \(y=\ln x\).

Step 2: Set Up the Integral

The differential element for area under a curve with respect to x-axis is given as \(dx\). So, the integral to find the area will be a definite integral from \(x=1/e\) to \(x=1\) (from intersection point of \(y=\ln x\) and \(y=1\) with \(x=1/e\)) of \(1 - \ln x\) dx. Subtracting \(\ln x\) because the area needed is beneath the line \(y=1\).

Step 3: Evaluate the Integral

Evaluate the definite integral: \[\int_{1/e}^{1} (1 - \ln x) dx\]. To do this, first find the antiderivative of \(1 - \ln x\) which is \(x - x \ln x -x\). Then, apply Fundamental Theorem of Calculus by plugging the bounds of integration into the antiderivative function.

Step 4: Calculate the Area

Subtract the value of the antiderivative at \(x=1/e\) from the value of the antiderivative at \(x=1\) to find the exact area. The resulting value represents the area bounded by the given lines and curves.

Key Concepts

Definite Integral

In calculus, the definite integral is a fundamental concept that helps us find the total amount of "something" accumulated over an interval. This "something" could be area, volume, or any other quantity of interest. For this case, we're tasked with finding the area under a curve by computing a definite integral.
The definite integral of a continuous function \(f(x)\) from \(a\) to \(b\) is denoted as \(\int_{a}^{b} f(x) \, dx\). This expression represents the accumulation of the function's values over the interval from \(a\) to \(b\).

• The bounds \(a\) and \(b\) are the points where you begin and end the accumulation.
• The function \(f(x)\) is the expression you're calculating the area under.
• \(dx\) signifies integration with respect to \(x\).

To solve the exercise, we integrated the function \(1 - \ln x\) from \(x = 1/e\) to \(x = 1\), meaning we're calculating the area between these bounds under the specified curve.

Area Under a Curve

The concept of the area under a curve is a visualization of the integral. When you see a curve on a graph, think of the area under it between two points as the tilt of a surface.
Finding this area helps understand relationships in physics, economics, and more.
In the context of our problem, the area under the curve is bound between two curves: `y = ln x` and `y = 1`, alongside the vertical line `x = 1/e`. The area looks somewhat like a trapezoid. By integrating `1 - ln x`, we're measuring the space that lies directly under `y = 1`, down to the curve `y = ln x`.

• The section between the curve and the \(x\)-axis helps visualize the volume of a bowl with a curve as its rim.
• The use of bounds \(x=1/e\) and \(x=1\) ensures we only calculate the area in the specified range.

Understanding this area helps frame bigger mathematical connections like probability distributions.

Antiderivative

An antiderivative is closely linked to integration. When you're asked to find an integral, you often seek the antiderivative of the function involved.
It's essentially a function whose derivative equals the original function. In our problem, to solve the integral \(\int (1 - \ln x) \, dx\), we find its antiderivative.
The antiderivative of \(1 - \ln x\) is \(x - x \ln x - x\). Here's the breakdown:

• The derivative of \( - x\ln x\) with respect to \(x\) gives \(- \ln x\), capturing part of the original function.
• The \(x\) term when derived gives 1, aligning with the remaining part of the function.

Finding this antiderivative is essential for applying the Fundamental Theorem of Calculus, which utilizes these expressions for evaluation.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) is a powerful tool that connects differentiation and integration. It makes the process of evaluating definite integrals straightforward.
Essentially, the FTC tells us that we can find the accumulation (integral) of a function by plugging the interval's bounds into its antiderivative.
The theorem can be outlined in two main parts:

• The first part states that if a function is continuous, its indefinite integral can be related to its antiderivative.
• The second part states that to evaluate a definite integral, use \[ F(b) - F(a) \] where \(F(x)\) is the antiderivative of \(f(x)\), and \(a\) and \(b\) are the limits of integration.

For our exercise, after finding that the antiderivative of \(1 - \ln x\) is \(x - x \ln x - x\), we applied the FTC. We inserted our bounds \(x=1/e\) and \(x=1\) into the antiderivative, then subtracted to find the exact area of the region.