Problem 58 - Calculus Early Transcendentals

Question

Find the area of the following regions, expressing your results in terms of the positive integer \(n \geq 2\) The region bounded by \(f(x)=x^{1 / n}\) and \(g(x)=x^{n},\) for \(x \geq 0\)

Step-by-Step Solution

Answer

Answer: The area of the region bounded by these functions is given by \(\frac{n^2 - n}{n+1}\).

Step 1: Find points of intersection

To find the points of intersection, we need to solve the equation \(f(x) = g(x)\): \(x^{\frac{1}{n}} = x^{n}\) By taking the logarithm on both sides: \(\frac{1}{n}\log{x} = n\log{x}\) Since \(x>0\), we can divide by \(\log{x}\): \(\frac{1}{n} = n\) Hence, \(n = 1\) but we're given that \(n \geq 2\). Therefore, there's no intersection for \(x \geq 0\) and \(n \geq 2\).

Step 2: Integrate the difference between the functions

The area between the functions will be formed by integrating the difference of the functions \(f(x)\) and \(g(x)\) over the range \(x=[0,1]\) since given \(x \geq 0\) and observing that \(x^n\) will be greater than \(x^{\frac{1}{n}}\) for \(x \ge 1\) : \(\text{Area} = \int_0^1 \left(f(x) - g(x)\right) dx\) Plug in the functions: \(\text{Area} = \int_0^1 \left(x^{\frac{1}{n}} - x^n\right) dx\)

Step 3: Evaluate the integral

To evaluate this integral, we need to find the antiderivative of each part: \(\int{ x^{\frac{1}{n}} - x^n} dx = \frac{n x^{1+\frac{1}{n}}}{1+\frac{1}{n}} - \frac{x^{1+n}}{1+n} + C\) Now, we can plug in the limits: \(\text{Area} = \left[\frac{n x^{(1+\frac{1}{n})}}{1+\frac{1}{n}} - \frac{x^{1+n}}{1+n}\right]_0^1\) \(\text{Area} = \left[\frac{n \cdot 1^{(1+\frac{1}{n})}}{1+\frac{1}{n}} - \frac{1^{1+n}}{1+n}\right] - \left[\frac{n \cdot 0^{(1+\frac{1}{n})}}{1+\frac{1}{n}} - \frac{0^{1+n}}{1+n}\right]\)

Step 4: Simplify the expression

Simplify the expression to get the final answer: \(\text{Area} = \left[\frac{n}{1+\frac{1}{n}} - \frac{1}{1+n}\right]\) Now, solve for the common denominator: \(\text{Area} = \frac{n^2}{n+1} - \frac{n}{n+1}\) Combining the fractions, we get the final answer: \(\text{Area} = \frac{n^2 - n}{n+1}\) So, the area of the region bounded by \(f(x) = x^{\frac{1}{n}}\) and \(g(x) = x^n\) is \(\frac{n^2 - n}{n+1}\) for \(n \ge 2\).

Key Concepts

Integration

Understanding the concept of integration is crucial when determining the area between curves. Integration is a fundamental tool in calculus that allows us to calculate the 'total accumulation' of a quantity. In the context of geometry, it can be thought of as adding up an infinite number of infinitesimally small area elements.

When solving for the area between two functions, we essentially subtract one function from the other and integrate over the interval of interest. The integral sign \(\int\) symbolizes the accumulation of area, and the 'dx' at the end of the integral stands for a small change in x, which, when multiplied by the function, gives a small rectangle's area. As the rectangles get infinitesimally small, the sum approaches the actual area.

For instance, in our exercise, the total area between the curves \(f(x) = x^{1/n}\) and \(g(x) = x^n\) from \(x=0\) to \(x=1\) is found by setting up the integral \(\int_0^1 (x^{1/n} - x^n) dx\). By doing this, we are summing all the tiny differences between \(x^{1/n}\) and \(x^n\) for each value of x in the range \(0\) to \(1\), which gives us the enclosed area.

Antiderivative

The antiderivative, also known as the indefinite integral, is essentially the reverse process of differentiation. If we are given a derivative, the antiderivative is the original function before it was differentiated. Antiderivatives play a pivotal role in evaluating definite integrals, like the one in our example problem.

In order to find the area between two curves, we first need to find the antiderivative of the function that defines the top curve, and subtract the antiderivative of the function that defines the bottom curve. Remember that the antiderivative is not unique; it includes a constant of integration \(C\), which can take any value. However, when evaluating a definite integral - that is, an integral with limits - this constant cancels out.

In our exercise, we found the antiderivatives \( \int x^{1/n} dx \) and \( \int x^n dx \) to then evaluate the definite integral from \(0\) to \(1\). The solution involved finding the antiderivative of each term and then applying the fundamental theorem of calculus by substituting the upper and lower bounds of the interval.

Properties of Exponents

Working with exponents is an essential skill when dealing with integration, as many functions we integrate will include exponential terms. The properties of exponents tell us how to manipulate expressions that contain powers.

Some key properties include:

• The product of powers property: \(x^a \cdot x^b = x^{a+b}\), which says that when we multiply like bases, we add the exponents.
• The power of a power property: \( (x^a)^b = x^{a \cdot b}\), which indicates that when taking a power to another power, we multiply the exponents.
• The power of a product property: \( (xy)^a = x^a y^a\), showing that when raising a product to a power, we can apply the power to each factor individually.

These properties allow us to simplify expressions before integrating. In our example, we relied on these properties to understand why for \(x > 0\) and \(n \geq 2\), the term \(x^n\) will always be greater than \(x^{1/n}\). Hence, no intersection points exist within this domain. Furthermore, they help us when we integrate functions with exponential terms and need to simplify the result.