Finding the Area Enclosed by the Functions

Oct 13, 2025 by magicalceylontours.com 67 views

Let the Region ${$ R $}$ be the Area Enclosed by the Functions ${$ f(x) = 3x^{\frac{1}{3}} $}$ and ${$ g(x) = \frac{3}{2}x $}$

In this article, we will explore the concept of finding the area enclosed by two functions, ${$ f(x) = 3x^{\frac{1}{3}} $}$ and ${$ g(x) = \frac{3}{2}x $}$. We will start by defining the region ${$ R $}$ and then proceed to find the area enclosed by these two functions using integration.

Defining the Region ${$ R $}$

The region ${$ R $}$ is defined as the area enclosed by the functions ${$ f(x) = 3x^{\frac{1}{3}} $}$ and ${$ g(x) = \frac{3}{2}x $}$. To visualize this region, we can plot the two functions on a coordinate plane.

Plotting the Functions

To plot the functions, we can start by finding the x-intercepts of both functions. The x-intercept of ${$ f(x) = 3x^{\frac{1}{3}} $}$ is ${$ x = 0 $}$, and the x-intercept of ${$ g(x) = \frac{3}{2}x $}$ is ${$ x = 0 $}$. This means that both functions intersect at the point ${$ (0, 0) $}$.

Next, we can find the y-intercept of both functions. The y-intercept of ${$ f(x) = 3x^{\frac{1}{3}} $}$ is ${$ y = 0 $}$, and the y-intercept of ${$ g(x) = \frac{3}{2}x $}$ is ${$ y = 0 $}$. This means that both functions intersect at the point ${$ (0, 0) $}$.

Finding the Area Enclosed by the Functions

To find the area enclosed by the functions, we can use the concept of integration. We can start by finding the area under the curve of ${$ f(x) = 3x^{\frac{1}{3}} $}$ from ${$ x = 0 $}$ to ${$ x = 2 $}$.

Integral in Terms of ${$ x $}$

The area under the curve of ${$ f(x) = 3x^{\frac{1}{3}} $}$ from ${$ x = 0 $}$ to ${$ x = 2 $}$ can be found using the following integral:

${$ \int_{0}^{2} 3x^{\frac{1}{3}} dx $}$

To evaluate this integral, we can use the power rule of integration, which states that:

${$ \int x^n dx = \frac{x^{n+1}}{n+1} + C $}$

Applying this rule to the integral, we get:

${$ \int_{0}^{2} 3x^{\frac{1}{3}} dx = \left[ 3 \cdot \frac{x^{\frac{1}{3}+1}}{\frac{1}{3}+1} \right]_{0}^{2} $}$

Simplifying the expression, we get:

${$ \int_{0}^{2} 3x^{\frac{1}{3}} dx = \left[ 3 \cdot \frac{x^{\frac{4}{3}}}{\frac{4}{3}} \right]_{0}^{2} $}$

${$ \int_{0}^{2} 3x^{\frac{1}{3}} dx = \left[ 3 \cdot \frac{3}{4}x^{\frac{4}{3}} \right]_{0}^{2} $}$

${$ \int_{0}^{2} 3x^{\frac{1}{3}} dx = \left[ \frac{9}{4}x^{\frac{4}{3}} \right]_{0}^{2} $}$

Evaluating the expression at the limits of integration, we get:

${$ \int_{0}^{2} 3x^{\frac{1}{3}} dx = \frac{9}{4}(2)^{\frac{4}{3}} - \frac{9}{4}(0)^{\frac{4}{3}} $}$

${$ \int_{0}^{2} 3x^{\frac{1}{3}} dx = \frac{9}{4}(8) - 0 $}$

${$ \int_{0}^{2} 3x^{\frac{1}{3}} dx = 18 $}$

This means that the area under the curve of ${$ f(x) = 3x^{\frac{1}{3}} $}$ from ${$ x = 0 $}$ to ${$ x = 2 $}$ is ${$ 18 $}$.

Integral in Terms of ${$ y $}$

To find the area enclosed by the functions in terms of ${$ y $}$, we can use the concept of inverse functions. We can start by finding the inverse of ${$ f(x) = 3x^{\frac{1}{3}} $}$, which is ${$ f^{-1}(y) = \frac{3}{2}y^3 $}$.

Next, we can find the inverse of ${$ g(x) = \frac{3}{2}x $}$, which is ${$ g^{-1}(y) = \frac{2}{3}y $}$.

The area enclosed by the functions in terms of ${$ y $}$ can be found using the following integral:

${$ \int_{0}^{18} \left( \frac{3}{2}y^3 - \frac{2}{3}y \right) dy $}$

To evaluate this integral, we can use the power rule of integration, which states that:

${$ \int x^n dx = \frac{x^{n+1}}{n+1} + C $}$

Applying this rule to the integral, we get:

${$ \int_{0}^{18} \left( \frac{3}{2}y^3 - \frac{2}{3}y \right) dy = \left[ \frac{3}{2} \cdot \frac{y^4}{4} - \frac{2}{3} \cdot \frac{y^2}{2} \right]_{0}^{18} $}$

Simplifying the expression, we get:

${$ \int_{0}^{18} \left( \frac{3}{2}y^3 - \frac{2}{3}y \right) dy = \left[ \frac{3}{8}y^4 - \frac{1}{3}y^2 \right]_{0}^{18} $}$

Evaluating the expression at the limits of integration, we get:

${$ \int_{0}^{18} \left( \frac{3}{2}y^3 - \frac{2}{3}y \right) dy = \frac{3}{8}(18)^4 - \frac{1}{3}(18)^2 $}$

${$ \int_{0}^{18} \left( \frac{3}{2}y^3 - \frac{2}{3}y \right) dy = \frac{3}{8}(194,481) - \frac{1}{3}(324) $}$

${$ \int_{0}^{18} \left( \frac{3}{2}y^3 - \frac{2}{3}y \right) dy = 72,860.25 - 108 $}$

${$ \int_{0}^{18} \left( \frac{3}{2}y^3 - \frac{2}{3}y \right) dy = 72,752.25 $}$

This means that the area enclosed by the functions in terms of ${$ y $}$ is ${$ 72,752.25 $}$.

In our previous article, we explored the concept of finding the area enclosed by two functions, ${$ f(x) = 3x^{\frac{1}{3}} $}$ and ${$ g(x) = \frac{3}{2}x $}$. We defined the region ${$ R $}$ and then proceeded to find the area enclosed by these two functions using integration. In this article, we will answer some frequently asked questions about finding the area enclosed by two functions.

Q1: What is the region ${$ R $}$ defined as?

The region ${$ R $}$ is defined as the area enclosed by the functions ${$ f(x) = 3x^{\frac{1}{3}} $}$ and ${$ g(x) = \frac{3}{2}x $}$.

Q2: How do I find the area under the curve of ${$ f(x) = 3x^{\frac{1}{3}} $}$ from ${$ x = 0 $}$ to ${$ x = 2 $}$?

To find the area under the curve of ${$ f(x) = 3x^{\frac{1}{3}} $}$ from ${$ x = 0 $}$ to ${$ x = 2 $}$, you can use the following integral:

${$ \int_{0}^{2} 3x^{\frac{1}{3}} dx $}$

To evaluate this integral, you can use the power rule of integration, which states that:

${$ \int x^n dx = \frac{x^{n+1}}{n+1} + C $}$

Applying this rule to the integral, you get:

${$ \int_{0}^{2} 3x^{\frac{1}{3}} dx = \left[ 3 \cdot \frac{x^{\frac{1}{3}+1}}{\frac{1}{3}+1} \right]_{0}^{2} $}$

Simplifying the expression, you get:

${$ \int_{0}^{2} 3x^{\frac{1}{3}} dx = \left[ 3 \cdot \frac{x^{\frac{4}{3}}}{\frac{4}{3}} \right]_{0}^{2} $}$

${$ \int_{0}^{2} 3x^{\frac{1}{3}} dx = \left[ \frac{9}{4}x^{\frac{4}{3}} \right]_{0}^{2} $}$

Evaluating the expression at the limits of integration, you get:

${$ \int_{0}^{2} 3x^{\frac{1}{3}} dx = \frac{9}{4}(2)^{\frac{4}{3}} - \frac{9}{4}(0)^{\frac{4}{3}} $}$

${$ \int_{0}^{2} 3x^{\frac{1}{3}} dx = \frac{9}{4}(8) - 0 $}$

${$ \int_{0}^{2} 3x^{\frac{1}{3}} dx = 18 $}$

Q3: How do I find the area enclosed by the functions in terms of ${$ y $}$?

To find the area enclosed by the functions in terms of ${$ y $}$, you can use the concept of inverse functions. You can start by finding the inverse of ${$ f(x) = 3x^{\frac{1}{3}} $}$, which is ${$ f^{-1}(y) = \frac{3}{2}y^3 $}$.

Next, you can find the inverse of ${$ g(x) = \frac{3}{2}x $}$, which is ${$ g^{-1}(y) = \frac{2}{3}y $}$.

The area enclosed by the functions in terms of ${$ y $}$ can be found using the following integral:

${$ \int_{0}^{18} \left( \frac{3}{2}y^3 - \frac{2}{3}y \right) dy $}$

To evaluate this integral, you can use the power rule of integration, which states that:

${$ \int x^n dx = \frac{x^{n+1}}{n+1} + C $}$

Applying this rule to the integral, you get:

${$ \int_{0}^{18} \left( \frac{3}{2}y^3 - \frac{2}{3}y \right) dy = \left[ \frac{3}{2} \cdot \frac{y^4}{4} - \frac{2}{3} \cdot \frac{y^2}{2} \right]_{0}^{18} $}$

Simplifying the expression, you get:

${$ \int_{0}^{18} \left( \frac{3}{2}y^3 - \frac{2}{3}y \right) dy = \left[ \frac{3}{8}y^4 - \frac{1}{3}y^2 \right]_{0}^{18} $}$

Evaluating the expression at the limits of integration, you get:

${$ \int_{0}^{18} \left( \frac{3}{2}y^3 - \frac{2}{3}y \right) dy = \frac{3}{8}(18)^4 - \frac{1}{3}(18)^2 $}$

${$ \int_{0}^{18} \left( \frac{3}{2}y^3 - \frac{2}{3}y \right) dy = \frac{3}{8}(194,481) - \frac{1}{3}(324) $}$

${$ \int_{0}^{18} \left( \frac{3}{2}y^3 - \frac{2}{3}y \right) dy = 72,860.25 - 108 $}$

${$ \int_{0}^{18} \left( \frac{3}{2}y^3 - \frac{2}{3}y \right) dy = 72,752.25 $}$

Q4: What is the significance of finding the area enclosed by two functions?

Finding the area enclosed by two functions is significant in various fields such as physics, engineering, and economics. It can be used to model real-world problems and make predictions about the behavior of systems.

Q5: How can I apply the concept of finding the area enclosed by two functions to real-world problems?

You can apply the concept of finding the area enclosed by two functions to real-world problems by using it to model the behavior of systems. For example, you can use it to model the growth of a population, the flow of a fluid, or the behavior of a financial system.

In this article, we answered some frequently asked questions about finding the area enclosed by two functions. We provided step-by-step solutions to the problems and explained the significance of finding the area enclosed by two functions. We also provided examples of how to apply the concept of finding the area enclosed by two functions to real-world problems.