Question

Multiple Choice The base of a solid \(S\) is the region enclosed by the graph of \(y=\ln x,\) the line \(x=e,\) and the \(x\) -axis. If the cross sections of \(S\) perpendicular to the \(x\) -axis are squares, which of the following gives to best approximation of the volume of \(S ?\) (A) 0.718 (B) 1.718 (C) 2.718 (D) 3.171 (E) 7.388

Short answer

(C) 2.718

Step-by-step solution

Set up the Integral

The volume of the solid is the integral of the area of the cross-sections. Since the cross-sections are squares and the side of the square is given by \(y= \ln x\), the area of each square is \((\ln x)^2\). Therefore, the volume \(V\) equals the integral from 1 to \(e\) of \((\ln x)^2\). So, \(V = \int_{1}^{e} (\ln x)^2 dx\).

Calculate the Integral

To evaluate this integral, we can use the power rule for integration, but for a function concerning the natural logarithm, the chain rule is necessary. The integral of \((\ln x)^2\) is \(x(\ln x)^2 - 2x\ln x + 2x\), from which our definite integral from 1 to \(e\) is evaluated to \(e - 2e + 2e - (1 - 2 + 2)\), which simplifies to \(e\).

Choose the Correct Answer

From our calculation, the value of the volume is \(e\). Looking into the provided multiple choice options, the choice that is also the correct approximation of \(e\) should be our answer. \(e\) is approximately 2.718. So, the correct answer is (C) 2.718.

Key concepts

Definite Integral Calculus

Definite integral calculus is a fundamental concept that allows us to calculate the exact area under a curve over a specific interval. This process involves finding the antiderivative of the given function and then evaluating it at the upper and lower bounds of the interval. An integral of a function, like the area of cross-sectional squares in our problem, can provide the volume of a three-dimensional solid.

For the solid in the exercise, the definite integral from 1 to the natural number e, denoted as \(\frac{1}{e}\), represents the sum of infinitely small areas of square cross-sections along the x-axis. Simply put, we integrate the area of each square from where x equals 1 to where x equals e to find the total volume. The precision of definite integrals makes them an incredibly powerful tool in calculus.

Natural Logarithm Properties

The natural logarithm, represented as \(\text{ln}(x)\), has unique properties that simplify complex calculations. One of the key properties is that the natural logarithm of e, \(\text{ln}(e)\), is equal to 1. Additionally, the natural logarithm is the inverse operation of exponentiation with base e, which means that \(e^{\text{ln}(x)} = x\).

In our exercise, these properties are used when establishing the relationship between the function \(y=\ln(x)\) and the side length of the squares that form the cross-sections of the solid. Grasping these properties is essential for integrating functions involving natural logarithms effectively.

Area of Squares

The area of a square is found by squaring one of its sides, which is written as \(A = s^2\), where \(A\) stands for the area, and \(s\) is the length of a side. In the context of our solid, each cross-sectional square has a side length represented by the natural logarithm of x, \(y=\ln(x)\).

Therefore, the area of one of these squares is \((\ln(x))^2\). To visualize the volume of the solid, imagine stacking these squares with varying sizes along the x-axis, from x=1 to x=e. By comprehending the simple concept of a square's area, students can begin to understand how to construct and calculate the volume of more complex solids.

Integration Techniques

To solve integrals, especially those involving more intricate functions like the natural logarithm, various integration techniques are used. In this particular problem, we combine the power rule with the method of integration by parts. The latter technique is based on the product rule for derivatives and is helpful when dealing with products of functions.

In our step-by-step solution, integration by parts transforms the integral of the square of the natural logarithm into an expression involving x and functions of \(\text{ln}(x)\), which we can then evaluate at the bounds. Mastering integration techniques allows students to tackle a wide variety of problems in calculus, and practicing with diverse problems is the key to proficiency.