Problem 20 - Calculus Early Transcendentals

Question

Sketch the following regions (if a figure is not given) and then find the area. The regions bounded by \(y=x^{3}\) and \(y=9 x\)

Step-by-Step Solution

Answer

Answer: The area of the region is \(\frac{243}{2}\).

Step 1: Graph the functions

To start, let's graph the two functions \(y = x^3\) and \(y = 9x\). The first function is a typical cubic function, while the second function is a linear function with a positive slope. From their graphs, we can see where they intersect.

Step 2: Find the intersection points

The intersection points are the points in which the functions are equal, so we need to solve the equation \(x^3 = 9x\). For a non-zero x, we can divide both sides by x to get \(x^2 = 9\), which gives us two intersection points: \((-3, -27)\) and \((3,27)\). Also, we need to consider the case when x=0, so another intersection point is \((0,0)\).

Step 3: Set up the integral

To find the area between the two curves, we can subtract the lower function from the higher function and integrate it with respect to x, from the leftmost intersection point to the rightmost intersection point. In this case, the higher function is \(y=9x\) and the lower function is \(y=x^3\). Therefore, $$A = \int_{-3}^{3}(9x - x^3)dx$$

Step 4: Calculate the integral

Evaluating the integral, we get $$A = \left[\frac{9x^2}{2} - \frac{x^4}{4}\right]_{-3}^3$$ Plugging in x = 3, we find $$A_+ = \frac{9(3^2)}{2} - \frac{3^4}{4} = \frac{81}{2} - \frac{81}{4} = \frac{3}{4}\times \frac{81}{2} = \frac{243}{4}$$ Plugging in x = \(-3\), we find $$A_- = \frac{9(3^2)}{2} - \frac{(-3)^4}{4} = -\left(\frac{81}{2} - \frac{81}{4}\right) = -\frac{243}{4}$$ Now we can find the total area by summing \(A_-\) and \(A_+\) $$A = A_+ + A_- = \frac{243}{4} - \frac{243}{4} = \frac{486}{4} = \frac{243}{2}$$

Step 5: Write the conclusion

The area of the region bounded by the functions \(y=x^3\) and \(y=9x\) is $$A = \frac{243}{2}.$$

Key Concepts

Cubic Function

A cubic function is a polynomial of degree three, often written in the form \( y = ax^3 + bx^2 + cx + d \). The term 'cubic' indicates that the highest exponent of the variable \( x \) is 3. In any cubic function, the graph tends to have a unique shape characterized by a point of inflection, where the graph changes concavity.
In our problem, the cubic function is \( y = x^3 \). This is a simple cubic function with:

• a = 1
• b = 0
• c = 0
• d = 0

This specific form has no additional linear or constant terms, so it passes through the origin and has symmetry around the origin, meaning \( y(x) = -y(-x) \).
Cubic functions can have up to three real roots, and they always have one real root at least. They can increase and decrease at different intervals, allowing them to form intricate shapes.

Linear Function

A linear function is expressed in the form \( y = mx + c \), where \( m \) is the slope and \( c \) is the y-intercept. The graph of a linear function is always a straight line. Its slope determines how steep the line is, and the sign of \( m \) (positive or negative) dictates its direction.
Our linear function in this exercise is \( y = 9x \). Here, the slope \( m \) is 9, indicating a steep line that rises quickly from left to right. The y-intercept is \( c = 0 \), meaning it passes directly through the origin.
Key characteristics of this linear function include:

• It has a constant rate of change, which is the slope
• Every segment of the line has the same steepness
• The function is simple and straightforward with no curves

Intersection Points

Intersection points are where two graphs meet or cross each other. Determining these points involves setting the equations of the functions equal to one another and solving for \( x \). For our functions, we set \( x^3 = 9x \) to find intersections.
When solved, this becomes \( x^3 - 9x = 0 \). By factoring, \( x(x^2 - 9) = 0 \), giving the solutions \( x = 0, x = 3, \) and \( x = -3 \).

• \( x = 0 \): The origin where both graphs pass through
• \( x = -3 \): A point where both functions meet on the negative x-axis
• \( x = 3 \): A point where both functions meet on the positive x-axis

Each intersection point can be checked by plugging \( x \) back into the original equations to verify the corresponding \( y \) value. The \( y \)-values

• at \( x = 0 \) is \( y = 0 \)
• at \( x = -3 \) is \( y = -27 \)
• at \( x = 3 \) is \( y = 27 \)

Definite Integral

The definite integral helps find the area between curves. For this, we need to integrate the difference of the functions over the interval defined by their intersection points.
Generally, if \( f(x) > g(x) \), the area \( A \) between two curves from \( a \) to \( b \) is given by:\[A = \int_{a}^{b} (f(x) - g(x)) \, dx\]In this problem, the functions are \( f(x) = 9x \) and \( g(x) = x^3 \). They intersect at \( x = -3 \) and \( x = 3 \).
The integral expression becomes:\[A = \int_{-3}^{3} (9x - x^3) \, dx\]Evaluating this, we find:

• Antiderivative of \( 9x \) is \( \frac{9x^2}{2} \)
• Antiderivative of \( x^3 \) is \( \frac{x^4}{4} \)
• Substitute and calculate at the bounds, subtracting the results

Finally, we find that the total area \( A = \frac{243}{2} \), indicating the size of the bounded region by these curves.