Question

Let \(f(x)=x^{p}\) and \(g(x)=x^{1 / q},\) where \(p>1\) and \(q>1\) are positive integers. Let \(R_{1}\) be the region in the first quadrant between \(y=f(x)\) and \(y=x\) and let \(R_{2}\) be the region in the first quadrant between \(y=g(x)\) and \(y=x\) a. Find the area of \(R_{1}\) and \(R_{2}\) when \(p=q,\) and determine which region has the greater area. b. Find the area of \(R_{1}\) and \(R_{2}\) when \(p>q\), and determine which region has the greater area. c. Find the area of \(R_{1}\) and \(R_{2}\) when \(p

Short answer

Answer: When \(p=q\), the areas of both regions \(R_1\) and \(R_2\) are equal. When \(p>q\), the area of \(R_1\) is greater than the area of \(R_2\). When \(p<q\), the area of \(R_2\) is greater than the area of \(R_1\).

Step-by-step solution

Determine the intersections

First, we need to find the points of intersection between the curves \(y=f(x)\), \(y=g(x)\), and \(y=x\). To do this, we set \(f(x) = x\) and \(g(x) = x\). We will have: \(f(x) = x^p = x \Rightarrow x^{p-1} = 1\) \(g(x) = x^{1/q} = x \Rightarrow x^{1/q - 1} = 1\) The solution for both equations is \(x=1\). Therefore, the intersection points are \((0, 0)\) and \((1, 1)\).

Compute the area of \(R_{1}\) (area between \(y=x^p\) and \(y=x\))

To find the area of \(R_{1}\), we will integrate the function \(f(x) - x\) from \(0\) to \(1\): $$R_1 = \int_0^1 (x^p - x) dx$$

Compute the area of \(R_{2}\) (area between \(y=x^{1/q}\) and \(y=x\))

Similarly, to find the area of \(R_{2}\), we will integrate the function \(x - g(x)\) from \(0\) to \(1\): $$R_2 = \int_0^1 (x - x^{1/q}) dx$$

Computing the integrals

Since \(p=q\), we can compute the integrals: $$R_1 = \int_0^1 (x^p - x) dx = \frac{1}{p+1} - \frac{1}{2}$$ $$R_2 = \int_0^1 (x - x^{1/q}) dx = \frac{1}{2} - \frac{1}{q+1}$$

Comparing \(R_{1}\) and \(R_{2}\) for \(p=q\)

When \(p=q\), we have: $$R_1 = \frac{1}{p+1} - \frac{1}{2} = R_2$$ In this case, the areas of both regions are equal. #b. Finding the area of \(R_{1}\) and \(R_{2}\) when \(p>q\), and determine which region has the greater area#

Comparing \(R_{1}\) and \(R_{2}\) for \(p>q\)

We already found the formulas for \(R_1\) and \(R_2\). When \(p > q\), we have: $$R_1 = \frac{1}{p+1} - \frac{1}{2} > \frac{1}{2} - \frac{1}{q+1} = R_2$$ In this case, the area of \(R_{1}\) is greater than the area of \(R_{2}\). #c. Finding the area of \(R_{1}\) and \(R_{2}\) when \(p<q\), and determine which region has the greater area#

Comparing \(R_{1}\) and \(R_{2}\) for \(p

Using the same formulas for \(R_1\) and \(R_2\), when \(p < q\), we have: $$R_1 = \frac{1}{p+1} - \frac{1}{2} < \frac{1}{2} - \frac{1}{q+1} = R_2$$ In this case, the area of \(R_{2}\) is greater than the area of \(R_{1}\).

Key concepts

Area under curves

One of the fascinating applications of integration in calculus is finding the area under curves. This involves using definite integrals to determine the space between a curve and the x-axis or between two curves altogether. By calculating these areas, we can understand the relative magnitudes of different bounded regions.

In the given exercise, we are asked to find the area between two functions and the line \(y = x\). Specifically, we are interested in the regions between the graphs of the functions \(f(x) = x^p\) and \(g(x) = x^{1/q}\), with \(p > 1\) and \(q > 1\). For both functions, we are comparing them to the line \(y = x\), which serves as a common base in the established boundaries. Integrating in this context allows us to precisely measure the areas of these regions, referred to as \(R_1\) and \(R_2\).

The area of \(R_1\) is determined by integrating \(f(x) - x\) from 0 to 1, whereas \(R_2\) results from integrating \(x - g(x)\) across the same interval. These integrals help us capture the precise area sandwiched between the curves, providing valuable insights into these functions' relative behavior.

Definite integrals

Definite integrals play a crucial role when calculating the area under curves. By definition, a definite integral accumulates the total 'weight' of a function over a specified interval, providing an exact value for the area between the curve and the x-axis. This process involves evaluating the integral between two endpoints, often using the Fundamental Theorem of Calculus.

In our scenario, we aim to find out the integral of the functions \(x^p - x\) and \(x - x^{1/q}\) from 0 to 1. These definite integrals look like this:

• For \(R_1\): \[\int_0^1 (x^p - x)\, dx\]
• For \(R_2\): \[\int_0^1 (x - x^{1/q})\, dx\]

Evaluating these integrals provides us with concrete area values for \(R_1\) and \(R_2\), allowing for a meaningful comparison of these two regions.

Through the evaluation, we obtain \(R_1 = \frac{1}{p+1} - \frac{1}{2}\) and \(R_2 = \frac{1}{2} - \frac{1}{q+1}\). This positions us to further analyze the relationship between \(p\) and \(q\) and how it affects the relative areas of these regions.

Comparing regions

A key aspect of this problem is comparing the areas \(R_1\) and \(R_2\) under different conditions. Understanding how the parameters \(p\) and \(q\) influence these areas will help us gain a deeper comprehension of the interplay between power functions and linear functions.

Let's consider these scenarios:

• When \(p = q\): The calculated integrals equate the areas of \(R_1\) and \(R_2\), as their expressions simplify to the same value. Thus, the regions have the same size.
• When \(p > q\): The area of \(R_1\) is larger than that of \(R_2\), illustrated by \(\frac{1}{p+1} - \frac{1}{2} > \frac{1}{2} - \frac{1}{q+1}\). This condition indicates that the function \(g(x)\) is closer to \(y = x\) than \(f(x)\).
• When \(p < q\): Conversely, the area of \(R_2\) becomes greater than that of \(R_1\), suggesting that the function \(f(x)\) approximates \(y = x\) more closely within the established interval.

These comparisons highlight the sensitivity of area calculations to the exponents involved, showing how integral calculus can be employed to deduce precise quantitative differences between regions.