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Magazine Chapter 6: Problem 8
Sketch the region enclosed by the curves and find its area. $$ y=x^{3}-4 x, y=0, x=0, x=2 $$
Short Answer
Expert verified
The area of the region is 4 square units.
Step by step solution
01
Identify the Curves and Intersection Points
We are given the curves \(y = x^3 - 4x\) and \(y = 0\). Additionally, the region is bounded by \(x = 0\) and \(x = 2\). The points of intersection of the curve \(y = x^3 - 4x\) with \(y = 0\) are found by setting \(x^3 - 4x = 0\) and solving for \(x\). Factoring gives \(x(x^2 - 4) = 0\), leading to solutions \(x = 0, x = 2\), and \(x = -2\). Since \(x = -2\) is not within our given bounds, we only consider \(x = 0\) and \(x = 2\).
02
Sketch the Region
Draw the curve \(y = x^3 - 4x\) on a graph, which is a cubic function crossing the x-axis at \(x = -2, 0, 2\). The line \(y = 0\) represents the x-axis. The vertical lines \(x = 0\) and \(x = 2\) define the interval of interest. The region of interest is the area between the curve and the x-axis from \(x = 0\) to \(x = 2\).
03
Set up the Integral for the Area
The area \(A\) under the curve from \(x = 0\) to \(x = 2\) is calculated using the definite integral formula: \[A = \int_{0}^{2} (x^3 - 4x) \, dx.\] This integral represents the area between the curve and the x-axis over the interval \([0, 2]\).
04
Compute the Integral
Evaluate the integral \[A = \int_{0}^{2} (x^3 - 4x) \, dx = \left[ \frac{x^4}{4} - 2x^2 \right]_{0}^{2}.\]Calculate it by substituting the limits:\[= \left( \frac{2^4}{4} - 2(2)^2 \right) - \left( \frac{0^4}{4} - 2(0)^2 \right) = (4 - 8) - (0)= -4.\]
05
Interpret the Negative Area
The negative sign indicates that the curve \(y = x^3 - 4x\) is below the x-axis between \(x = 0\) and \(x = 2\). The absolute value of the integral result gives the area \[A = 4.\]The area is always positive when regarding physical regions.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integral
A definite integral is a powerful tool in calculus that helps us find the area under a curve between two points on a graph. It's represented in the form \[ \int_{a}^{b} f(x) \, dx \]. The integral sign \(\int\) tells us to integrate, which is a form of summing up an infinite number of infinities small quantities to find a total.
For the exercise given, the definite integral \[\int_{0}^{2} (x^3 - 4x) \, dx\] is set up to calculate the area between the function \(y = x^3 - 4x\) and the x-axis from \(x = 0\) to \(x = 2\).
Calculating this integral helps find the region's area, taking into account that the area can sometimes result as negative if the curve is underneath the x-axis. But practically, area itself is always a positive value.
- \(a\) and \(b\) are the limits of integration, indicating the start and end points on the x-axis.
- \(f(x)\) is the function being integrated which forms a part of our curve or line.
For the exercise given, the definite integral \[\int_{0}^{2} (x^3 - 4x) \, dx\] is set up to calculate the area between the function \(y = x^3 - 4x\) and the x-axis from \(x = 0\) to \(x = 2\).
Calculating this integral helps find the region's area, taking into account that the area can sometimes result as negative if the curve is underneath the x-axis. But practically, area itself is always a positive value.
Area Under a Curve
The area under a curve provides a way to quantify the space enclosed by a specific function and the x-axis over an interval. This helps in understanding the behavior of a function within that boundary.
When computing the area, if the curve lies above the x-axis, the integral typically gives a positive value. Conversely, if it's below, the integral yields a negative result, signifying the direction but not affecting the actual size of the area.For our problem, the integral evaluates to \(-4\), indicating the curve is beneath the x-axis from \(x = 0\) to \(x = 2\).
By taking the absolute value, we find the area of the region without concern for its position relative to the x-axis, giving us a final area of \(4\).
- It is visually the space between the line (or curve) and the x-axis from one x-value to another.
- Mathematically, it is represented by a definite integral.
When computing the area, if the curve lies above the x-axis, the integral typically gives a positive value. Conversely, if it's below, the integral yields a negative result, signifying the direction but not affecting the actual size of the area.For our problem, the integral evaluates to \(-4\), indicating the curve is beneath the x-axis from \(x = 0\) to \(x = 2\).
By taking the absolute value, we find the area of the region without concern for its position relative to the x-axis, giving us a final area of \(4\).
Cubic Function
A cubic function is a polynomial of degree three and is represented generically by \(f(x) = ax^3 + bx^2 + cx + d\). These functions display unique characteristics:
This results in an intersection with the x-axis at \(x = -2, 0,\) and \(2\), of which \(x = 0\) and \(x = 2\) are within our region of interest. These points are crucial as they help determine where the function crosses the x-axis, thus segmenting our area for calculation. Understanding this enables us to accurately set up integrals and visually comprehend the graph's behavior and transitions across the axis.
- They can have up to three real roots (values of \(x\) where the function equals zero).
- Their graph usually has an "S" shaped curve with turning points depending on the function's specific terms.
This results in an intersection with the x-axis at \(x = -2, 0,\) and \(2\), of which \(x = 0\) and \(x = 2\) are within our region of interest. These points are crucial as they help determine where the function crosses the x-axis, thus segmenting our area for calculation. Understanding this enables us to accurately set up integrals and visually comprehend the graph's behavior and transitions across the axis.
