Problem 18 - Calculus of a Single Variable

Question

Finding the Area of a Region In Exercises \(17-30,\) sketch the region bounded by the graphs of the equations and find the area of the region. $$ y=-x^{3}+2, \quad y=x-3, \quad x=-1, \quad x=1 $$

Step-by-Step Solution

Answer

The area of the region bounded by the given curves is 1 square unit.

Step 1: Sketching the Graphs

Firstly, we sketch the four curves on a graph. \(y=-x^{3}+2\) is a cubic function and \(y=x-3\) is a linear function. Both \(x=-1\) and \(x=1\) are vertical lines.

Step 2: Identifying the Boundaries

We can see that the region of interest is bounded on the left by the vertical line \(x=-1\), on the right by the vertical line \(x=1\), on the bottom by the line \(y=x-3\), and on the top by the curve \(y=-x^{3}+2\). This is a continuous region, hence the total area can be obtained by subtracting the lower curve from the upper curve for each \(x\) in the range of \([-1, 1]\)

Step 3: Setting up the Integral

The area \(A\) of a region \(S\) in the x-y plane bounded by \(y=f(x), y=g(x)\), and \(x=a, x=b\), where \(f(x)\) and \(g(x)\) are continuous and \(f(x) ≥ g(x)\) for all \(x\) in \([a, b]\), is given by \(A = \int_{a}^{b} f(x) - g(x) dx\). Applying this here, we have \(A = \int_{-1}^{1} (-x^{3}+2 - (x-3)) dx\) = \( \int_{-1}^{1} (-x^{3}+x+1) dx\).

Step 4: Evaluating the Integral

We evaluate the integral using the power rule. The antiderivative of the integrand is \(-\frac{1}{4}x^{4} + \frac{1}{2}x^{2} + x\). Evaluate at \(1\) and \(-1\) and subtract to find the area: \(-\frac{1}{4} + \frac{1}{2} + 1 - (\frac{1}{4} - \frac{1}{2} + 1) = 1\).

Key Concepts

Integral Calculus

Integral calculus is a fundamental part of mathematics that helps us calculate areas, volumes, and other quantities. It is essentially a way to sum up infinitely small pieces to find the whole. One of the core operations in integral calculus is the integral, which is signified by the symbol \(\int\). This operation accumulates the quantity across an interval, often producing an area under a curve, relative to an axis.

• Understanding the Integral: An integral can be imagined as the area under a curve. It tells us how the function behaves over a specific interval. For functions, there are two types of integrals - indefinite and definite.
• Indefinite Integrals: This type finds the general form of an antiderivative of the function.
• Definite Integrals: These give the total value accumulated by the function across an interval from \(x = a\) to \(x = b\), often represented geometrically as an area.

Let's look at how integral calculus was applied to find the area between the curves in the exercise. We used a definite integral to determine the bounded area between the cubic and linear functions, which was the objective of the problem.

Cubic Functions

Cubic functions are polynomial functions of degree three, generally expressed in the form \(y = ax^3 + bx^2 + cx + d\). These functions can take on an S-shape and have one or two critical points where they change direction.

• Characteristics: A key property of cubic functions is their ability to display a turning point, or point of inflection, depending on the sign and relative sizes of their coefficients.
• Cubic Function in the Problem: In our problem, we worked with the cubic function \(y = -x^3 + 2\). This is a simple cubic function that shifts the basic cubic shape and flips it upside down, with its turning point adjusted by the constant \(+2\).
• Importance: Understanding the shape and behavior of cubic functions helps us visualize areas when determining the area between curves, especially when they serve as upper bounds.

Linear Functions

Linear functions are the simplest form of polynomial functions. They can be expressed in the general form \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept.

• Simplicity: Linear functions form straight lines on a graph, characterized by their constant rate of change. This makes calculating regions involving them straightforward.
• Role in the Exercise: In our original exercise, we dealt with the linear function \(y = x - 3\). This line slopes upwards with a unit slope and crosses the y-axis at \(-3\).
• Relevance: When finding areas between curves, linear functions often act as boundaries either on the top or bottom, making them a critical part of such calculus problems.

Definite Integral

The definite integral is a concept used to compute the precise difference in quantities between two points, \(a\) and \(b\), along the x-axis of a function's curve. Represented mathematically as \(\int_{a}^{b} f(x) \,dx\), it gives us a numeric value with distinct geometric interpretation, such as area or net change.

• Evaluating Definite Integrals: We calculate definite integrals by finding the antiderivative of the function, then computing the difference between its values at the upper and lower bounds.
• Application in the Exercise: In solving the problem, the definite integral \(\int_{-1}^{1} (-x^3 + x + 1) \,dx\) was used, with the antiderivative being evaluated at both \(-1\) and \(1\).
• Understanding Results: The outcome of the definite integral in this context provided the net area between our cubic and linear functions, resulting in a total area of \(1\) square unit.

By understanding definite integrals, we can measure shapes accurately in the x-y plane and solve real-world problems related to accumulation.