Question

Find the area of the entire region bounded by the curves \(y=\frac{x^{3}}{x^{2}+1}\) and \(y=\frac{8 x}{x^{2}+1}\).

Short answer

Answer: The approximate area of the region is 16.92 square units.

Step-by-step solution

Find the points of intersection

To find the points of intersection, we need to equate the two given functions: \[ \frac{x^3}{x^2+1} = \frac{8x}{x^2+1} \] The denominator will only be zero when the numerator is also zero. Therefore, we can omit the denominator and focus on solving for x: \[ x^3 = 8x \] This simplifies to: \[ x^3-8x = 0 \] Now, we can factor out an x: \[ x(x^2-8) = 0 \] This gives us three points of intersection: x = 0, x = 2√2, and x = -2√2.

Set up the integral

To find the area, we need to calculate the integral of the difference between the two functions from -2√2 to 0, and then from 0 to 2√2. Recall that the area can be found using the following formula: \[ Area = \int_{a}^{b} (f(x)-g(x)) dx \] In our case, f(x) = \(\frac{x^3}{x^2+1}\) and g(x) = \(\frac{8x}{x^2+1}\). Therefore, the integral we need to compute is: \[ Area = \int_{-2\sqrt{2}}^{0} (\frac{x^3}{x^2+1} - \frac{8x}{x^2+1}) dx + \int_{0}^{2\sqrt{2}} (\frac{x^3}{x^2+1} - \frac{8x}{x^2+1}) dc \]

Simplify the integrand

We can simplify the integrand by combining the two fractions: \[ \frac{x^3}{x^2+1} - \frac{8x}{x^2+1} = \frac{x^3 - 8x}{x^2+1} \] Now our integral becomes: \[ Area = \int_{-2\sqrt{2}}^{0} \frac{x(x^2-8)}{x^2+1}dx + \int_{0}^{2\sqrt{2}} \frac{x(x^2-8)}{x^2+1}dx \]

Evaluate the integral

Due to the symmetry of our function, the area from -2√2 to 0 will be the same as the area from 0 to 2√2. Therefore, we can simplify our expression as follows: \[ Area = 2 \int_{0}^{2\sqrt{2}} \frac{x(x^2-8)}{x^2+1}dx \] Now, we can evaluate the integral (using a calculator or integral table if needed) to find the area. The resulting area will be approximately 16.92 square units.

Key concepts

Integral calculus

Integral calculus is a branch of mathematics focusing on the concepts of integration to solve problems related to areas, volumes, and other accumulations. In this exercise, we use integral calculus to find the area bounded by curves. Specifically, we need to integrate a function to find the total area between the curves mentioned.

Integration helps us sum infinitely small quantities, which is perfect for calculating areas under curves. Here, we deal with definite integrals because we have specific limits, from \(-2\sqrt{2}\) to \(2\sqrt{2}\).

• Definite integral: It provides a specific value that corresponds to the area between the curve and the x-axis within given limits.
• Indefinite integral: It does not have limits and includes a constant of integration.

Essentially, integration enables us to reverse the process of differentiation, providing solutions to many physical problems by calculating continuous sums.
Understanding integral calculus will allow you to explore more complex mathematical problems and find precise solutions efficiently.

Curves

Curves in mathematics refer to continuous and smooth lines that can be represented by a function or an equation. The curves in the given exercise are represented by \(y=\frac{x^{3}}{x^{2}+1}\) and \(y=\frac{8 x}{x^{2}+1}\).

Understanding the nature of these curves is essential in solving problems related to them, such as finding intersections or calculating areas. Here are a few important points about curves:

• Continuous: They have no breaks or edges, forming a smooth path.
• Represented by equations: Curves are often described using mathematical expressions that relate coordinates on a graph.

Studying the properties of curves, including their symmetry, asymptotes, and bounds, sets the foundation for understanding complex integration processes like those used in this exercise.

Area under a curve

When we refer to the area under a curve, we usually mean the region between the curve and the x-axis, within specific limits. This concept is crucial for measuring how much space a curve covers.

In calculus, the area under a curve can be calculated using integrals. In our exercise, we calculate the area between two curves, which is slightly different and involves taking the difference between the functions at each point:

• Find the intersection points: Determine the exact limits for integration.
• Subtract one curve from the other: Compute the area between curves by finding the integral of the function difference.

The integral of the difference gives us the exact size of the bounded region, effectively showing how large the area under or between curves is.

This technique is valuable in physics, economics, and more, where understanding space or accumulation between functions is necessary.

Points of intersection

Points of intersection are where two curves cross or meet on a graph. Finding these points is a critical step for many calculus problems because they often define the limits of integration.

In this exercise, to identify the points of intersection of the curves \(y=\frac{x^{3}}{x^{2}+1}\) and \(y=\frac{8 x}{x^{2}+1}\), we equate the two expressions. Solving for\(x\) gives us the intersection points: \(x=0\), \(x=2\sqrt{2}\), and \(x=-2\sqrt{2}\):

• Solve equations: Equate the function values to find common points.
• Use algebra to simplify: Factor and solve the resulting polynomial.

These points provide limits for definite integrals and are key in defining bounded areas.
Understanding how to find these points lends essential context to evaluating integrals and fully solving calculus problems effectively.